Two trains move towards each other, a bird moves between them. How many trips can the bird make?

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,

$$d_1= d_0-2v_tt_1=d_0\frac{v_b-v_t}{v_b+v_t}$$

Likewise,

$$d_2 =d_1\frac{v_b-v_t}{v_b+v_t}=d_0\left(\frac{v_b-v_t}{v_b+v_t}\right)^{2}$$

$$ ... $$

$$d_{n} =d_0\left(\frac{v_b-v_t}{v_b+v_t}\right)^{n}$$

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

$$D_1 =\frac{v_b}{v_b+v_t}d_0$$ $$D_2 =\frac{v_b}{v_b+v_t}d_1$$

$$...$$

$$D_n =\frac{v_b}{v_b+v_t} d_{n-1}$$

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

$$D = \frac{v_b}{v_b+v_t}(d_0+d_1+d_2+\>...)=v_b\frac{d_0}{2v_t}=\frac{58}{68}\times 102 = 87km$$

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.