Bsrgvty

The problem statement:

Two trains move towards each other at a speed of $34 km/h$ in the same rectilinear road. A certain bird can fly at a speed of $58 km/h$ and starts flying from the front of one of the trains to the other, when they're $102 km$ apart. When the bird reaches the front of the other train, it starts flying back to the first train, and so on.

• How many of these trips can the bird make before the two trains meet?


• What is the total distance the bird travels?

Commentary:

The second question of the problem seems relatively simple, since one only has to notice that the trains will take 1.5 hours to meet, therefore, the bird travels $58cdot1.5=87 km$. However, the first question baffles me. How can one calculate how many trips the bird makes? If I'm correct, in order to obtain the time the bird will take to make its first trip, we have to add the bird's speed and the speed at which the distance of the trains is being reduced ($68 km/h$).

This means the bird will take $frac102126approx0.809$ hours to finish the first trip, and the trains will be $frac98621approx 46.95 km$ apart. If I continue this way (now finding how long will the bird take to travel those 46.95 km), it seems that I'll never stop or that at least it will take a huge amount of trips that cannot be computed by hand. Is there a way to find a 'quick' answer to this problem? Am I making it more complicated than it actually is?

Thanks in advance!

The bird will make infinitely many trips, that get smaller and smaller in distance.

In fact, because of this, this question is often asked as a kind of 'trick' question. That is, like you did in the second part of your post, people trying to answer the second question will often try and calculate how much time the first trip takes, how far the bird flew during that first trip, and how far the trains are still apart at that point. Then, they'll try and compute the same for the second trip, third, etc .... but of course you never get done with this ... and the numbers are intentionally chosen to be 'ugly' as well (as they are in this case). So, many people will throw up their hands when asked the total distance made by the bird, because they try and calculate the sum of all these distances, and the calculation just gets too nasty for them.

Now, of course you could use an infinite series to do this ... or you do what you did! First calculate how much time it takes for the trains to reach each other, and that tells you how much time the bird is flying back and forth, and that'll immediately tell you the answer to the total distance question.

So, good for you for not being tripped up by this ... but maybe that's exactly because you didn't realize that the bird would take infinitely many trips? :)

Part 1 surely has no answer. The number of trips would be infinite if we replace the bird by a point mass capable of infinite acceleration, but for a real bird we need to know something about the bird and what it can do.

There is a nice story that Part 2 was posed to John von Neumann who gave the correct numerical answer after brief thought. "Ah" said the questioner, "I should have known I couldn't trick you. Most people take a lot of time trying to sum the infinite series." Von Neumann said "What? Is there another way?"

The quick answer to this well-known puzzle is infinite number of trips, which is shown explicitly below.

Let $v_b$ and $v_t$ be the speeds of the bird and train, respectively. For the initial distance $d_0$ between the train, it takes the bird time $t_1$, given by $ (v_b+v_t)t_1 = d_0$, to reach the other train. Then, the new distance between the trains becomes,

$$d_1= d_0-2v_tt_1=d_0fracv_b-v_tv_b+v_t$$

Likewise,

$$d_2 =d_1fracv_b-v_tv_b+v_t=d_0left(fracv_b-v_tv_b+v_tright)^2$$

$$ ... $$

$$d_n =d_0left(fracv_b-v_tv_b+v_tright)^n$$

As can be seen from the above expression, the distance $d_n$ can be arbitrarily small, but it will never be zero, which indicates that it requires infinite number of trips.

Similarly, the distances covered by the bird on each trip can be expressed as

$$D_1 =fracv_bv_b+v_td_0$$
$$D_2 =fracv_bv_b+v_td_1$$

$$...$$

$$D_n =fracv_bv_b+v_t d_n-1$$

The total distance is a converging geometric sum, which yields,

$$D = fracv_bv_b+v_t(d_0+d_1+d_2+>...)=v_bfracd_02v_t=frac5868times 102 = 87km$$

The result for $D$ has a simple interpretation: the distance is simply the time taken by the two train to meet multiplied by the speed of the bird.

In the stationary reference frame of the train the bird does not start on, they are 102 km apart and the far train moves at 68 km/h. The bird flies at 92 km/h.

It reaches the 'stationary' train in $frac10292 h$, after which the far train has moved $68 * frac10292 km$.

If the trains started X apart, then it reaches the 'stationary' train in $fracX92 h$ and the trains are now $X-frac68X92 km = (1-frac6892) X km$ apart.

This then applies recursively, so after $K$ flights, the trains are $102 * (1-frac6892)^K km$ apart. For no number of flights $K$ does this reach 0, so the bird must fly an infinite number of times for the trains to collide, assuming an infinitely small bird.

For a sanity check, we'll now take this solution and see if we can get the proper flight distance from it.

On flight K, the bird has flown the distance between the cars after flight $K$, times $frac5858+34$ to account for frame-shift. So the bird flies the sum, from $1$ to $infty$, of $frac5892 * 102 * (1-68/92)^K$ km.

Using the identity $1+x+x^2+x^3+... = frac11-x$ this is $102 km * frac11 - (1-frac6892) * frac5892$, aka $102km * frac9268 * frac5892$, or $87 km$.

Just the first question: Every time the bird reaches the front of the other train, their distance has shrunk by a factor 24/92 (distance between bird and oncoming train shrinks at a rate of 92 km/h, distance between trains shrinks at a rate of 68 km/h).

After 7 trips, the bird reaches the front of the other train when the trains are about 8.4 metres apart. They crash within 0.44 seconds. There's no way the bird can stop, turn around, and reach a speed of 58 km/h again within 0.44 seconds, so the seventh trip is the last one.