The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither?Net force and its conceptWhy is acceleration constant in this example?Why can velocity and acceleration be negative?(+)/(-) Acceleration & (+)(-) VelocityEffect on speed when decreasing the magnitude of accelerationTerminology for time derivative of speed (not velocity)
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The instant an accelerating object has zero speed, is it speeding up, slowing down, or neither?
Net force and its conceptWhy is acceleration constant in this example?Why can velocity and acceleration be negative?(+)/(-) Acceleration & (+)(-) VelocityEffect on speed when decreasing the magnitude of accelerationTerminology for time derivative of speed (not velocity)
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This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down.
However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?
kinematics acceleration terminology velocity speed
edited Jun 14 at 10:19
knzhou
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user532874user532874
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show 9 more comments
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This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down.
However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?
kinematics acceleration terminology velocity speed
edited Jun 14 at 10:19
knzhou
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asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
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7
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Suppose you are walking, and are at the top of a hill. Are you going up or down?
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– Peter Shor
Jun 13 at 9:09
4
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You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
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– garyp
Jun 13 at 11:08
33
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I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
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– Derek Elkins
Jun 14 at 9:22
15
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This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
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– knzhou
Jun 14 at 10:22
5
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
$endgroup$
– Peter A. Schneider
Jun 14 at 10:30
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show 9 more comments
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This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down.
However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?
kinematics acceleration terminology velocity speed
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
$endgroup$
This problem is from Khan Academy. Specifically for the blue point circled in red, the answer is that at this blue point, the object is neither speeding up nor slowing down. When I think about the rule about the signs of velocity and acceleration and what this means for change in speed, this makes sense: if velocity and acceleration and have the same sign, the object is speeding up, and if velocity and acceleration have opposite signs, the object is slowing down. At the blue point, the instantaneous velocity is zero and because zero is neither positive nor negative, the object is neither speeding up nor slowing down.
However, intuitively, this doesn't make sense. At the blue point circled in red, the velocity is zero so the speed must be zero. Also, the slope of the function at this blue point is negative so acceleration is negative, meaning that velocity is decreasing every second. If velocity is decreasing every second, then right after 6 seconds, the velocity will turn negative but the speed will have increased. So at the blue point, the object, even though it has zero speed, is in the process of speeding up. Why is this thinking wrong?
kinematics acceleration terminology velocity speed
kinematics acceleration terminology velocity speed
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
edited Jun 14 at 10:19
knzhou
56.9k14 gold badges163 silver badges276 bronze badges
56.9k14 gold badges163 silver badges276 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
asked Jun 13 at 9:04
user532874user532874
2491 silver badge8 bronze badges
2491 silver badge8 bronze badges
7
$begingroup$
Suppose you are walking, and are at the top of a hill. Are you going up or down?
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– Peter Shor
Jun 13 at 9:09
4
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You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
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– garyp
Jun 13 at 11:08
33
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I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
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– Derek Elkins
Jun 14 at 9:22
15
$begingroup$
This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
$endgroup$
– knzhou
Jun 14 at 10:22
5
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
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– Peter A. Schneider
Jun 14 at 10:30
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show 9 more comments
7
$begingroup$
Suppose you are walking, and are at the top of a hill. Are you going up or down?
$endgroup$
– Peter Shor
Jun 13 at 9:09
4
$begingroup$
You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
$endgroup$
– garyp
Jun 13 at 11:08
33
$begingroup$
I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
$endgroup$
– Derek Elkins
Jun 14 at 9:22
15
$begingroup$
This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
$endgroup$
– knzhou
Jun 14 at 10:22
5
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
$endgroup$
– Peter A. Schneider
Jun 14 at 10:30
7
7
$begingroup$
Suppose you are walking, and are at the top of a hill. Are you going up or down?
$endgroup$
– Peter Shor
Jun 13 at 9:09
$begingroup$
Suppose you are walking, and are at the top of a hill. Are you going up or down?
$endgroup$
– Peter Shor
Jun 13 at 9:09
4
4
$begingroup$
You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
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– garyp
Jun 13 at 11:08
$begingroup$
You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
$endgroup$
– garyp
Jun 13 at 11:08
33
33
$begingroup$
I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
$endgroup$
– Derek Elkins
Jun 14 at 9:22
$begingroup$
I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
$endgroup$
– Derek Elkins
Jun 14 at 9:22
15
15
$begingroup$
This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
$endgroup$
– knzhou
Jun 14 at 10:22
$begingroup$
This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
$endgroup$
– knzhou
Jun 14 at 10:22
5
5
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
$endgroup$
– Peter A. Schneider
Jun 14 at 10:30
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
$endgroup$
– Peter A. Schneider
Jun 14 at 10:30
|
show 9 more comments
9 Answers
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At the first point prior to the one you circled, the object is slowing down. It's velocity instantaneously is dropping and approaching the zero axis. However, at 0 m/s the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than 0 m/s) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting the ground it is moving 0 m/s and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing dowm"). The mathematical explanation for this is that the derivative of the magnitude of v (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of v, not the magnitude of v) is non zero and pointing upwards, acting to change the ball's direction of motion.
answered Jun 13 at 9:30
AndrewAndrew
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Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
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– Andrew
Jun 13 at 15:13
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“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
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– Bob Knighton
Jun 13 at 15:40
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The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
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– Bob Knighton
Jun 13 at 16:13
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I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
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– Dan Staley
Jun 13 at 17:55
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@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
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– Nuclear Wang
Jun 13 at 20:43
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The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:
The above is the graph of $$ y = left lvert 4 - left ( fracx - 22 right ) ^2 right rvert text,$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,
a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
answered Jun 13 at 23:14
jarhill0jarhill0
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"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
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– Nuclear Wang
Jun 14 at 13:00
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Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
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– Rad80
Jun 14 at 15:07
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@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
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– jarhill0
Jun 14 at 17:45
3
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@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
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– Nuclear Wang
Jun 14 at 18:41
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My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
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– Michael Grant
Jun 16 at 14:39
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I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up").
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| equiv sqrtv^2 = (v^2)^1/2
$$
as a definition. We take the derivative now:
$$
beginalign
fracdt &= fracd(v^2)^1/2dt\
&=frac12(v^2)^-1/2cdot 2vcdot fracdvdt \
&= fracvsqrtv^2cdot fracdvdt \
&=fracvsqrtv^2cdot a
endalign
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/sqrtv^2$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
answered Jun 13 at 14:40
gabegabe
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Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
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– dmckee♦
Jun 13 at 14:58
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@dmckee No worries, just making sure I wasn't mistaken.
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– gabe
Jun 13 at 15:00
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t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
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– Nuclear Wang
Jun 13 at 18:18
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@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
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– gabe
Jun 13 at 18:37
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@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
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– Michael Grant
Jun 16 at 14:44
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Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point.
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero.
Hope this helps.
answered Jun 13 at 9:27
Bob DBob D
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So you are agreeing with what Andrew said basically?
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– user532874
Jun 13 at 9:48
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You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
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– user532874
Jun 13 at 9:50
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@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
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– Bob D
Jun 13 at 10:13
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Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
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– user532874
Jun 13 at 10:45
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The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
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– dmckee♦
Jun 13 at 20:13
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Speed is the magnitude of the velocity and hence always a positive quantity.
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".
If one looked at the speedometer (a devise which measures speed) just before the
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
answered Jun 13 at 10:49
FarcherFarcher
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The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
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Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
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– cmaster
Jun 14 at 16:20
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This is the chart of the movement of an airplane which transitions into "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear. acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.
The only thing that makes time 6 interesting is this is when velocity crosses the zero line. That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration. At time 6/velocity 0, it is definitely accelerating.
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?) So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes. Sounds like increasing to me.
No, you say? Then the question of "speed increasing" boils down to semantics. It seems like a trick/dodge question.
answered Jun 13 at 23:50
HarperHarper
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In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$
that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
so before the moment it stops it speeds down.
after the moment it stops it speeds up
and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy=0$$
Edit:
Thanks to dmckee to suggest that actually the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
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The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
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– dmckee♦
Jun 13 at 16:38
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@dmckee the curve smooth at $t = 6$?
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– jim
Jun 13 at 20:07
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$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
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– dmckee♦
Jun 13 at 20:08
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Kinks are possible in the speed graph
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– Neil G
Jun 14 at 20:08
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There is nothing wrong with the kink; the problem is people's interpretation of it.
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– Michael Grant
Jun 17 at 0:11
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Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
answered Jun 13 at 20:19
jimjim
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At the first point prior to the one you circled, the object is slowing down. It's velocity instantaneously is dropping and approaching the zero axis. However, at 0 m/s the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than 0 m/s) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting the ground it is moving 0 m/s and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing dowm"). The mathematical explanation for this is that the derivative of the magnitude of v (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of v, not the magnitude of v) is non zero and pointing upwards, acting to change the ball's direction of motion.
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
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Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
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– Andrew
Jun 13 at 15:13
1
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“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
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– Bob Knighton
Jun 13 at 15:40
1
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The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
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– Bob Knighton
Jun 13 at 16:13
21
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I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
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– Dan Staley
Jun 13 at 17:55
3
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
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– Nuclear Wang
Jun 13 at 20:43
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show 14 more comments
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At the first point prior to the one you circled, the object is slowing down. It's velocity instantaneously is dropping and approaching the zero axis. However, at 0 m/s the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than 0 m/s) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting the ground it is moving 0 m/s and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing dowm"). The mathematical explanation for this is that the derivative of the magnitude of v (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of v, not the magnitude of v) is non zero and pointing upwards, acting to change the ball's direction of motion.
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
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Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
$endgroup$
– Andrew
Jun 13 at 15:13
1
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“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
$endgroup$
– Bob Knighton
Jun 13 at 15:40
1
$begingroup$
The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
$endgroup$
– Bob Knighton
Jun 13 at 16:13
21
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I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
$endgroup$
– Dan Staley
Jun 13 at 17:55
3
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
$endgroup$
– Nuclear Wang
Jun 13 at 20:43
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show 14 more comments
$begingroup$
At the first point prior to the one you circled, the object is slowing down. It's velocity instantaneously is dropping and approaching the zero axis. However, at 0 m/s the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than 0 m/s) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting the ground it is moving 0 m/s and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing dowm"). The mathematical explanation for this is that the derivative of the magnitude of v (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of v, not the magnitude of v) is non zero and pointing upwards, acting to change the ball's direction of motion.
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
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At the first point prior to the one you circled, the object is slowing down. It's velocity instantaneously is dropping and approaching the zero axis. However, at 0 m/s the instantaneous velocity has ceased dropping (because it has now reached zero, it can't slow down more than 0 m/s) but hasn't yet begun speeding up in the negative direction.
You may wonder, "but how can it not be speeding up or slowing down if its acceleration is not zero?" As we know, acceleration is the slope of the graph. At the circled point acceleration is non-zero because the object is changing directions from the positive to the negative direction, not because it is speeding up/slowing down.
Positive and negative signs here don't refer to slowing down and speeding up; they refer to two directions - the positive direction and negative direction. If this was a position vs. time graph then negative would refer to a negative position relative to the zero position and vice versa for positive. Positive and negative as signs are used here to give you a 1 dimensional line along which you can move in two directions with the origin being an arbitrary point we call zero.
A good physical one dimensional analogy to this question (although its velocity curve would be linear and not curved) is a ball that is dropped vertically. After impacting the ground it is moving 0 m/s and has lost all its downward speed but hasn't gained any upwards speed yet in that instant (it's in the "in between stage of speeding up and slowing dowm"). The mathematical explanation for this is that the derivative of the magnitude of v (which determines if the object is speeding up/slowing down) is undefined. At that instant acceleration (the derivative of v, not the magnitude of v) is non zero and pointing upwards, acting to change the ball's direction of motion.
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
edited Jun 16 at 16:04
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
answered Jun 13 at 9:30
AndrewAndrew
3259 bronze badges
3259 bronze badges
1
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Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
$endgroup$
– Andrew
Jun 13 at 15:13
1
$begingroup$
“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
$endgroup$
– Bob Knighton
Jun 13 at 15:40
1
$begingroup$
The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
$endgroup$
– Bob Knighton
Jun 13 at 16:13
21
$begingroup$
I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
$endgroup$
– Dan Staley
Jun 13 at 17:55
3
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
$endgroup$
– Nuclear Wang
Jun 13 at 20:43
|
show 14 more comments
1
$begingroup$
Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
$endgroup$
– Andrew
Jun 13 at 15:13
1
$begingroup$
“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
$endgroup$
– Bob Knighton
Jun 13 at 15:40
1
$begingroup$
The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
$endgroup$
– Bob Knighton
Jun 13 at 16:13
21
$begingroup$
I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
$endgroup$
– Dan Staley
Jun 13 at 17:55
3
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
$endgroup$
– Nuclear Wang
Jun 13 at 20:43
1
1
$begingroup$
Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
$endgroup$
– Andrew
Jun 13 at 15:13
$begingroup$
Thanks! Speed by definition cannot include negative values. Speed is simply the magnitude of the velocity. We can explain decreasing NEGATIVE velocity as speeding up in the negative direction - two bits of information there, its growing magnitude (speed/magnitude of velocity) & direction of movement (sign of velocity). We can use increasing NEGATIVE velocity as slowing down in the negative direction (& vice versa for the positive direction). If we allow speed to include negative numbers indicating direction then we are really just talking about velocity & speed loses its explanatory usefulness
$endgroup$
– Andrew
Jun 13 at 15:13
1
1
$begingroup$
“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
$endgroup$
– Bob Knighton
Jun 13 at 15:40
$begingroup$
“1 dimension is the only case where you can accelerate without moving (at a speed of zero).” This simply isn’t true. One can easily write down a velocity vector which becomes zero at some time but has no zero acceleration at that time.
$endgroup$
– Bob Knighton
Jun 13 at 15:40
1
1
$begingroup$
The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
$endgroup$
– Bob Knighton
Jun 13 at 16:13
$begingroup$
The particle motion given by $textbfx(t)=cos(t^2),hati+sint^2,hatj$ has zero velocity but nonzero acceleration at $t=0$. Almost any example of a particle trajectory which has vanishing velocity at some point will have nonzero acceleration at that point.
$endgroup$
– Bob Knighton
Jun 13 at 16:13
21
21
$begingroup$
I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
$endgroup$
– Dan Staley
Jun 13 at 17:55
$begingroup$
I think that the key thing here is that the object's speed is not differentiable. As the absolute value of velocity (in this 1-dimensional case), the speed has a kink at the circled point. So if "speeding up" means "positive derivative of speed", "slowing down" means "negative derivative of speed", and "not changing speed" means "zero derivative of speed", then the key observation is that it's possible for none of these three things to be true.
$endgroup$
– Dan Staley
Jun 13 at 17:55
3
3
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
$endgroup$
– Nuclear Wang
Jun 13 at 20:43
$begingroup$
@DanStaley I think you've identified the key here - the derivative of the object's speed is undefined at t=6. In my view, that means trying to interpret the derivative at t=6 is a meaningless exercise. It's not positive, it's not negative, it's not zero - it's simply undefined. You can argue that the object is not speeding up, but you can equally well argue that the object is not not speeding up. Saying the object is both speeding up and slowing down is exactly as valid as saying it's neither speeding up nor slowing down.
$endgroup$
– Nuclear Wang
Jun 13 at 20:43
|
show 14 more comments
$begingroup$
The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:
The above is the graph of $$ y = left lvert 4 - left ( fracx - 22 right ) ^2 right rvert text,$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,
a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
$endgroup$
3
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
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– Nuclear Wang
Jun 14 at 13:00
7
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Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
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– Rad80
Jun 14 at 15:07
2
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
3
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
|
show 12 more comments
$begingroup$
The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:
The above is the graph of $$ y = left lvert 4 - left ( fracx - 22 right ) ^2 right rvert text,$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,
a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
$endgroup$
3
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
$endgroup$
– Nuclear Wang
Jun 14 at 13:00
7
$begingroup$
Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
$endgroup$
– Rad80
Jun 14 at 15:07
2
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
3
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
|
show 12 more comments
$begingroup$
The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:
The above is the graph of $$ y = left lvert 4 - left ( fracx - 22 right ) ^2 right rvert text,$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,
a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
$endgroup$
The tricky part of this question is that you are given a graph of velocity but asked a question about speed.
Several others have said essentially the same thing, but what really makes this clear for me is a graph of speed:
The above is the graph of $$ y = left lvert 4 - left ( fracx - 22 right ) ^2 right rvert text,$$ which is just the absolute value of the velocity graph in your screenshot.
This represents the fact that speed is the absolute value of velocity.
We understand "slowing down" to mean that the slope of the speed is negative, and "speeding up" to mean that the slope of the speed is positive. What is the slope of point $(6, 0)$ on the graph (which corresponds to your circled dot)?
This point is a cusp. The notion of "slope" only exists for differentiable points, and as Wikipedia says,
a function with a bend, cusp, or vertical tangent may be continuous, but fails to be differentiable at the location of the anomaly.
Thus the slope of speed does not exist at this point, and so the object is neither speeding up nor slowing down in this instant.
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
edited Jun 14 at 17:41
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
answered Jun 13 at 23:14
jarhill0jarhill0
4131 silver badge5 bronze badges
4131 silver badge5 bronze badges
3
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
$endgroup$
– Nuclear Wang
Jun 14 at 13:00
7
$begingroup$
Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
$endgroup$
– Rad80
Jun 14 at 15:07
2
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
3
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
|
show 12 more comments
3
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
$endgroup$
– Nuclear Wang
Jun 14 at 13:00
7
$begingroup$
Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
$endgroup$
– Rad80
Jun 14 at 15:07
2
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
3
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
3
3
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
$endgroup$
– Nuclear Wang
Jun 14 at 13:00
$begingroup$
"Slope > 0" means speeding up, "Slope < 0" means slowing down. Since slope is undefined at the point in question, we wind up with the comparisons of "Undefined > 0" and "Undefined < 0". Your answer suggests both of these statements are false, but really they cannot be evaluated. It's like asking "is this sandwich positive, negative, or neither?". Saying an undefined value is nonpositive is like saying a sandwich is nonpositive - it's a meaningless statement because you're using qualifiers that simply don't apply. I agree "neither" is probably the correct answer, but it's a terrible question.
$endgroup$
– Nuclear Wang
Jun 14 at 13:00
7
7
$begingroup$
Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
$endgroup$
– Rad80
Jun 14 at 15:07
$begingroup$
Rather he is saying the statements are not true. Which is correct. This is the best answer IMHO.
$endgroup$
– Rad80
Jun 14 at 15:07
2
2
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
$begingroup$
@NuclearWang you raise a good point. Perhaps to be more precise we define "speeding up" as "slope exists and is positive." Then, we have no issue with undefined comparison. This is similar to definitions in calculus, for example the limit definition of continuity, which requires a limit to exist and be equal to a certain value.
$endgroup$
– jarhill0
Jun 14 at 17:45
3
3
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
@Rad80 "Not true" and "false" are identical for statements with well-defined truth values. By applying the "less than" sign in a situation where it is not applicable, we have crafted a logically self-inconsistent statement that is neither true nor false. The resolution, as in the Liar Paradox, is that we cannot meaningfully assign truth values of true (not false) or false (not true) to the original statement
$endgroup$
– Nuclear Wang
Jun 14 at 18:41
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
$begingroup$
My issue with these answers based on the derivative is that the standard derivative assumes no directionality on the independent variable. But time does have a direction. For this reason, it is best to consider a one-sided derivative (see, for example, this Wikipedia article). And in this context, there is an unambiguous answer.
$endgroup$
– Michael Grant
Jun 16 at 14:39
|
show 12 more comments
$begingroup$
I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up").
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| equiv sqrtv^2 = (v^2)^1/2
$$
as a definition. We take the derivative now:
$$
beginalign
fracdt &= fracd(v^2)^1/2dt\
&=frac12(v^2)^-1/2cdot 2vcdot fracdvdt \
&= fracvsqrtv^2cdot fracdvdt \
&=fracvsqrtv^2cdot a
endalign
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/sqrtv^2$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
$endgroup$
$begingroup$
Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
$endgroup$
– dmckee♦
Jun 13 at 14:58
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
$endgroup$
– gabe
Jun 13 at 15:00
1
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
$endgroup$
– Michael Grant
Jun 16 at 14:44
add a comment
|
$begingroup$
I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up").
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| equiv sqrtv^2 = (v^2)^1/2
$$
as a definition. We take the derivative now:
$$
beginalign
fracdt &= fracd(v^2)^1/2dt\
&=frac12(v^2)^-1/2cdot 2vcdot fracdvdt \
&= fracvsqrtv^2cdot fracdvdt \
&=fracvsqrtv^2cdot a
endalign
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/sqrtv^2$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
$endgroup$
$begingroup$
Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
$endgroup$
– dmckee♦
Jun 13 at 14:58
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
$endgroup$
– gabe
Jun 13 at 15:00
1
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
$endgroup$
– Michael Grant
Jun 16 at 14:44
add a comment
|
$begingroup$
I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up").
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| equiv sqrtv^2 = (v^2)^1/2
$$
as a definition. We take the derivative now:
$$
beginalign
fracdt &= fracd(v^2)^1/2dt\
&=frac12(v^2)^-1/2cdot 2vcdot fracdvdt \
&= fracvsqrtv^2cdot fracdvdt \
&=fracvsqrtv^2cdot a
endalign
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/sqrtv^2$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
$endgroup$
I will provide a bit more formal of an answer. "Speeding up" or "slowing down" typically refers to whether the speed of an object is increasing or decreasing. Imagine you are in a boat speeding down a canal (so that you may only move in one dimension - the canal is very narrow). At $t=2$, you flip a switch and your engine starts running in reverse. Here there is an instant where your acceleration is 0 before becoming negative, and this corresponds to the maximum on your velocity-time plot. Now, your engine is running in reverse and your boat is "slowing down" in the traditional sense. This corresponds to $2<t<6$ in your plot. There comes an instant where you have killed off all of your velocity, and you start running in reverse. After you kill off your velocity, your speed begins to increase (you are "speeding up").
In math, we can explain this as follows. The speed of an object is defined as the magnitude of its velocity. In 1 dimension, this is saying
$$
s = |v|
$$
i.e. speed is the absolute value of velocity. If we are interested in whether or not you are speeding up or slowing down, we want to find $ds/dt$. This can be done using the chain rule of ordinary calculus. First, we note that:
$$
|v| equiv sqrtv^2 = (v^2)^1/2
$$
as a definition. We take the derivative now:
$$
beginalign
fracdt &= fracd(v^2)^1/2dt\
&=frac12(v^2)^-1/2cdot 2vcdot fracdvdt \
&= fracvsqrtv^2cdot fracdvdt \
&=fracvsqrtv^2cdot a
endalign
$$
This final expression tells us a few things. First, we recover the rule that you are familiar with: namely, that if $v$ and $a$ have the same sign, then $ds/dt$ will be positive. If they have different signs, it will be negative. However, we also note that we have a discontinuity at $v=0$, which is the situation considered here. At zero, $v/sqrtv^2$ jumps from $-1$ to $1$ and the derivative $ds/dt$ does not exist - the speed is formally undefined. This is known as the sign function $sgn(v)$, which returns the sign of the argument. Since the derivative of the speed does not exist at $v=0$ in one dimension, we are justified in saying that we are neither speeding up or slowing down. However, the velocity is decreasing this whole time, as evidence by the constant negative acceleration.
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
edited Jun 13 at 18:43
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
answered Jun 13 at 14:40
gabegabe
8405 silver badges22 bronze badges
8405 silver badges22 bronze badges
$begingroup$
Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
$endgroup$
– dmckee♦
Jun 13 at 14:58
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
$endgroup$
– gabe
Jun 13 at 15:00
1
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
$endgroup$
– Michael Grant
Jun 16 at 14:44
add a comment
|
$begingroup$
Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
$endgroup$
– dmckee♦
Jun 13 at 14:58
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
$endgroup$
– gabe
Jun 13 at 15:00
1
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
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– Michael Grant
Jun 16 at 14:44
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Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
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– dmckee♦
Jun 13 at 14:58
$begingroup$
Oh crap. You're talking about 2 seconds? Sorry. I totally flaked on that. I withdraw it all.
$endgroup$
– dmckee♦
Jun 13 at 14:58
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
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– gabe
Jun 13 at 15:00
$begingroup$
@dmckee No worries, just making sure I wasn't mistaken.
$endgroup$
– gabe
Jun 13 at 15:00
1
1
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
t=6 is the last moment at which the object is slowing down (according to the left-sided derivative), and the first moment at which it speeds up (according to the right-sided derivative). It wasn't an option as an answer, but I think "both" is as defensible an answer as "neither". The object isn't speeding up, nor slowing down, nor staying at constant speed, which are the only three possibilities - the real answer is that the change in speed at t=6 is undefined and cannot be meaningfully interpreted.
$endgroup$
– Nuclear Wang
Jun 13 at 18:18
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang Your point about the speed being undefined is a good one, and I will make it explicit. I agree that "both" should be a valid interpretation as well as "neither". Ultimately, it cannot be interpreted (though the answer given by Khan academy is at least an understandable one in this light).
$endgroup$
– gabe
Jun 13 at 18:37
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
$endgroup$
– Michael Grant
Jun 16 at 14:44
$begingroup$
@NuclearWang you're on the right track. But surely you see there is a reason to favor the right-derivative over the left when the independent variable is physical time, a quantity that evolves in a singular direction.
$endgroup$
– Michael Grant
Jun 16 at 14:44
add a comment
|
$begingroup$
Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point.
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero.
Hope this helps.
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
$endgroup$
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
1
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
7
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
|
show 2 more comments
$begingroup$
Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point.
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero.
Hope this helps.
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
$endgroup$
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
1
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
7
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
|
show 2 more comments
$begingroup$
Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point.
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero.
Hope this helps.
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
$endgroup$
Something doesn’t seem correct. For a curve of velocity versus time the acceleration at any point on the curve is the derivative of the function, that is, the instantaneous slope of the curve at the point.
At the circled point the slope is negative and not zero, indicating negative acceleration. So while the velocity at the circled point is zero it is still changing, in this case changing direction.
The slope of the curve corresponding to t=2 seconds, on the other hand, appears to be zero. That’s where the acceleration is zero.
Hope this helps.
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
edited Jun 13 at 9:38
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
answered Jun 13 at 9:27
Bob DBob D
15.5k3 gold badges13 silver badges40 bronze badges
15.5k3 gold badges13 silver badges40 bronze badges
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
1
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
7
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
|
show 2 more comments
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
1
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
7
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
So you are agreeing with what Andrew said basically?
$endgroup$
– user532874
Jun 13 at 9:48
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
You are emphasizing how the answer of "neither speeding up nor slowing down" only applies to t=2 seconds for the graph above
$endgroup$
– user532874
Jun 13 at 9:50
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
$begingroup$
@user532874 Regarding your first comment, I agree with what Andrew said. We both posted answers at nearly the same time and I did not see his answer when I posted mine. My answer differs mainly in that I emphasize acceleration is the derivative of the velocity vs time function (slope of the curve) and that clearly the slope is not zero when the velocity is zero, therefore the acceleration is not zero.
$endgroup$
– Bob D
Jun 13 at 10:13
1
1
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
$begingroup$
Actually, Andrew realized Khan Academy is right because their justification is that the object is changing directions at the blue point circled in red so it is not speeding up and it is not slowing down.
$endgroup$
– user532874
Jun 13 at 10:45
7
7
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
$begingroup$
The mixing of technical language and casual language in kinematics is one of the most fruitful sources of trick questions in all of physics. Unfortunately giving trick questions to students who haven't had time to even start getting the subject organized in their heads is a bad, bad idea. (And, students often generate such questions for themselves.) But instructors can at least (a) be clear about the distinction between casual and technical language, (b) clearly identify trick questions as such, and (c) not grade introductory students on them
$endgroup$
– dmckee♦
Jun 13 at 20:13
|
show 2 more comments
$begingroup$
Speed is the magnitude of the velocity and hence always a positive quantity.
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".
If one looked at the speedometer (a devise which measures speed) just before the
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
$endgroup$
add a comment
|
$begingroup$
Speed is the magnitude of the velocity and hence always a positive quantity.
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".
If one looked at the speedometer (a devise which measures speed) just before the
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
$endgroup$
add a comment
|
$begingroup$
Speed is the magnitude of the velocity and hence always a positive quantity.
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".
If one looked at the speedometer (a devise which measures speed) just before the
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
$endgroup$
Speed is the magnitude of the velocity and hence always a positive quantity.
If the terms “speeding up” and “slowing down” refer to the speed then at the time indicated on the graph the speed is zero and having reached a minimum (zero) the speed would have been increasing in the future.
The problem is that prior to that time the speed was deceasing to eventually reach zero and that means that a graph of speed against time is discontinuous at the time in question ie the gradient of the graph is not defined at that time.
So perhaps that is why the response "neither" was given as being correct?
The rate of change of velocity (acceleration), the gradient of the velocity against time graph which is well defined, is negative and so at that time the rate of change of velocity (acceleration) is negative.
What was happening was that the direction of motion of the body changed at that time from moving in the positive direction to moving in the negative direction.
So one could say that the component of velocity, in the direction which was chosen to be positive, changed from being positive to being negative.
The label on the graph "velocity" is an abbreviation for "component of velocity in a chosen direction".
If one looked at the speedometer (a devise which measures speed) just before the
that the speed was decreasing and just after that time the speed was increasing but at the instant of time in question which one of the two options do you choose?
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
answered Jun 13 at 10:49
FarcherFarcher
57.7k3 gold badges46 silver badges125 bronze badges
57.7k3 gold badges46 silver badges125 bronze badges
add a comment
|
add a comment
|
$begingroup$
The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
$endgroup$
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
add a comment
|
$begingroup$
The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
$endgroup$
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
add a comment
|
$begingroup$
The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
$endgroup$
The object is neither speeding up nor slowing down, or both speeding up and slowing down. The graph you show plots velocity vs time, which can be converted to speed by taking the absolute value. The rate of change in speed is the derivative of the speed function. On the graph of speed, we can see a discontinuous corner at t=6 where the graph touches the x-axis and leaves again. The derivative at this point is undefined, making its interpretation rather nebulous - it's not negative, it's not non-negative, it's not positive, it's not non-positive, it's not zero, it's simply undefined.
We can argue that this undefined derivative isn't positive (i.e. speeding up), nor is it negative (i.e. slowing down), so the object is neither speeding up nor slowing down. But we can equally well argue that the derivative is not negative and not zero (i.e. speeding up) and that the derivative is not positive and not zero (i.e. slowing down), so the object is both speeding up and slowing down.
But really, the rate change of the object's speed is undefined at t=6. We cannot say anything meaningful about the rate of change in the object's speed at t=6, as we literally cannot define it. Even the reasoning I use the previous paragraph is rather specious - there isn't really any basis to claim that an undefined quantity is not positive or non-zero. Undefined > 0 and Undefined < 0 aren't false or true statements, they simply cannot be evaluated. It's like asking if a sandwich is positive or negative - the term just doesn't apply.
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
edited Jun 14 at 13:02
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
answered Jun 13 at 20:58
Nuclear WangNuclear Wang
1,2535 silver badges11 bronze badges
1,2535 silver badges11 bronze badges
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
add a comment
|
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
$begingroup$
Yep. And in most cases, we just shouldn't talk about "is speeding up" or "is slowing down", and just say "is accelerating in this direction". Only friction works on speed (as in brakes), all other forces work on velocity (more precisely: momentum).
$endgroup$
– cmaster
Jun 14 at 16:20
add a comment
|
$begingroup$
This is the chart of the movement of an airplane which transitions into "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear. acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.
The only thing that makes time 6 interesting is this is when velocity crosses the zero line. That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration. At time 6/velocity 0, it is definitely accelerating.
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?) So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes. Sounds like increasing to me.
No, you say? Then the question of "speed increasing" boils down to semantics. It seems like a trick/dodge question.
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
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add a comment
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$begingroup$
This is the chart of the movement of an airplane which transitions into "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear. acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.
The only thing that makes time 6 interesting is this is when velocity crosses the zero line. That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration. At time 6/velocity 0, it is definitely accelerating.
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?) So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes. Sounds like increasing to me.
No, you say? Then the question of "speed increasing" boils down to semantics. It seems like a trick/dodge question.
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is the chart of the movement of an airplane which transitions into "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear. acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.
The only thing that makes time 6 interesting is this is when velocity crosses the zero line. That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration. At time 6/velocity 0, it is definitely accelerating.
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?) So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes. Sounds like increasing to me.
No, you say? Then the question of "speed increasing" boils down to semantics. It seems like a trick/dodge question.
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
$endgroup$
This is the chart of the movement of an airplane which transitions into "beta" or thrust reverse at time 2, and steadily increases thrust reverse from then onward. The aircraft slows to a stop, but does not deselect thrust reverse but continues advancing it. So the instant it stops, it heels back and starts rolling backwards down the runway.
At speed 0, the circled point, the reverse thrust levers have not been zeroed and continue to be advanced steadily.
That's pretty clear. acceleration is a linear function (straight line) here, crossed zero at time 2, and it’s definitely nonzero at time 6.
The only thing that makes time 6 interesting is this is when velocity crosses the zero line. That, plus $6, will get you a small coffee at Starbucks, but it has no bearing on acceleration. At time 6/velocity 0, it is definitely accelerating.
The plane is stopped, the thrust reversers are howling, and the tower is wondering what the pilot intends. (Missed exit?) So is velocity increasing at the circled instant? It is zero and in the next instant his velocity will be more, so yes. Sounds like increasing to me.
No, you say? Then the question of "speed increasing" boils down to semantics. It seems like a trick/dodge question.
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
edited Jun 14 at 0:01
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
answered Jun 13 at 23:50
HarperHarper
1713 bronze badges
1713 bronze badges
add a comment
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add a comment
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$begingroup$
In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$
that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
so before the moment it stops it speeds down.
after the moment it stops it speeds up
and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy=0$$
Edit:
Thanks to dmckee to suggest that actually the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
$endgroup$
2
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
@dmckee the curve smooth at $t = 6$?
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– jim
Jun 13 at 20:07
2
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
1
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
1
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
|
show 1 more comment
$begingroup$
In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$
that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
so before the moment it stops it speeds down.
after the moment it stops it speeds up
and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy=0$$
Edit:
Thanks to dmckee to suggest that actually the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
$endgroup$
2
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
@dmckee the curve smooth at $t = 6$?
$endgroup$
– jim
Jun 13 at 20:07
2
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
1
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
1
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
|
show 1 more comment
$begingroup$
In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$
that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
so before the moment it stops it speeds down.
after the moment it stops it speeds up
and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy=0$$
Edit:
Thanks to dmckee to suggest that actually the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
$endgroup$
In my opinion khan academy is majorly correct . You may think that the acceleration at $$v=0$$
that is the derivative at the point where the velocity is zero is negative then it must be slowing down.but the term speeding up or speeding down means that if the modulus of velocity is increasing or decreasing.
Think physically
this graph shows the body speeding down in a direction then stopping and then reversing it's direction to speed up in an opposite direction.
so before the moment it stops it speeds down.
after the moment it stops it speeds up
and at the moment it stops it neither speeds up or speeds down.
That is $$d/dy=0$$
Edit:
Thanks to dmckee to suggest that actually the graph of |v| would have kink at v=0
And thus the graph is non differentiable at 0
So I'd want to add
I didn't thought in mathematical way but physical way and in any real physical system kinks in graph or undefined limits are not possible
various cases
.so yeah I should not use the word absolutely .I think the question was made to think about the change in velocity in more physical way not mathematical one . But we should suggest khan to modify the graph such that there would not be any kinks in |v|
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
edited Jun 14 at 5:47
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
answered Jun 13 at 16:08
gaurang agarwalgaurang agarwal
356 bronze badges
356 bronze badges
2
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
@dmckee the curve smooth at $t = 6$?
$endgroup$
– jim
Jun 13 at 20:07
2
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
1
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
1
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
|
show 1 more comment
2
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
@dmckee the curve smooth at $t = 6$?
$endgroup$
– jim
Jun 13 at 20:07
2
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
1
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
1
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
2
2
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
The derivative of speed is not zero but undefined at that $t=6$ (the left and right limits are different). That does make Khan's answer defensible, but it makes the question marginal in the first place.
$endgroup$
– dmckee♦
Jun 13 at 16:38
$begingroup$
@dmckee the curve smooth at $t = 6$?
$endgroup$
– jim
Jun 13 at 20:07
$begingroup$
@dmckee the curve smooth at $t = 6$?
$endgroup$
– jim
Jun 13 at 20:07
2
2
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
$begingroup$
$v(t)$ is smooth at $t = 6$, but $|v(t)|$ is not.
$endgroup$
– dmckee♦
Jun 13 at 20:08
1
1
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
$begingroup$
Kinks are possible in the speed graph
$endgroup$
– Neil G
Jun 14 at 20:08
1
1
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
$begingroup$
There is nothing wrong with the kink; the problem is people's interpretation of it.
$endgroup$
– Michael Grant
Jun 17 at 0:11
|
show 1 more comment
$begingroup$
Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
$endgroup$
add a comment
|
$begingroup$
Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
$endgroup$
add a comment
|
$begingroup$
Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
$endgroup$
Am assuming one dimensional motion so that when the speed is $+ve$ it is moving away from some fixed point and when it is $-ve$ it is moving in the opposite direction, back towards the fixed point.
Between $t = 0 s$ and $t = 2 s$ the slope of speed versus time is $+ve$ so that the particle is increasing its speed. At $t= 2 s$ the speed is $+ 4 m/s$ but it's acceleration is zero. Its speed is neither increasing or decreasing at this time.
For $t > 2 s$ the slope is negative so that the speed of the body is decreasing. It comes to a momentarily stop at $t = 6 s$ but its velocity appears to become more $-ve$ as time increases. Remember, velocity is "speed + direction" so that it's actual speed (magnitude of velocity) is getting larger.
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
edited Jun 13 at 20:28
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
answered Jun 13 at 20:19
jimjim
2,5787 silver badges23 bronze badges
2,5787 silver badges23 bronze badges
add a comment
|
add a comment
|
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$begingroup$
Suppose you are walking, and are at the top of a hill. Are you going up or down?
$endgroup$
– Peter Shor
Jun 13 at 9:09
4
$begingroup$
You have the answer in your statement of the problem. You define speeding up: velocity and acceleration have the same sign (and similar for slowing down). What is the sign of zero? Undefined. When $v=0$, the object does not meet the conditions for either definition, speeding up or slowing down. It's neither.
$endgroup$
– garyp
Jun 13 at 11:08
33
$begingroup$
I feel like this question (from Khan Academy, not the OP's) is confusing in all the wrong ways assuming the course is an introduction to physics...
$endgroup$
– Derek Elkins
Jun 14 at 9:22
15
$begingroup$
This kind of question annoys me -- it all comes down to defining what you mean by "speeding up/down", which is not even very important. It's just some technicality that depends on who's making the definition, not real physics. High school physics seems to be almost entirely composed of remembering a host of technicalities like this.
$endgroup$
– knzhou
Jun 14 at 10:22
5
$begingroup$
@DerekElkins Yes! An artificial distinction with no basis in Physics which only knows acceleration. After all, the specific velocity depends on the (arbitrary) observer's reference inertial system; acceleration doesn't. To understand that "slowing down" is a form of acceleration requires a mental change of every-day paradigms which are unfortunately reinforced here. Sad.
$endgroup$
– Peter A. Schneider
Jun 14 at 10:30