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So that the problem:

There are 2 trains at speed of 30km/h in the same trail and opposite directions and moving for a collision at 60km of distance of each other. One bird in front of one of the trains fly at 60km/h toward of other train. When he reaches he flies back to the other train until the trains collide.

I got a question: as the bird travels in all the way to the collision?

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  • $\begingroup$ Except the obvious $d=vt$ answer, I would be interested to see the series solution as well. $\endgroup$
    – Ali
    Commented Mar 17, 2014 at 14:00

2 Answers 2

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This is a classic question. The trains travel one hour till the collision. So the bird flies a distance of 60 km. Which is in contradiction with the other answer.

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  • $\begingroup$ A Google search for "Von Neumann" and "train" will lead to further information on this classic problem... $\endgroup$
    – DJohnM
    Commented Mar 17, 2014 at 17:18
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    $\begingroup$ There is a story about this relates to Feynman (I think). When asked the question he paused briefly and then gave the correct answer. "Ah", said the questioner, "I see you got the quick method." "No", he replied, "what quick method? It was just a simple convergent series." $\endgroup$
    – Dr Chuck
    Commented Mar 17, 2014 at 17:20
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Can you rephrase the question? I don't understand what you're asking.

[Would comment, but insufficient reputation]

So maybe a good way to start is to describe the distance remaining between the trains as a function of time, which seems to me like it should be

$$ d(t) = 60 - 60t, 0<t<1$$

since in the rest frame of one train the other is coming at it at 60km/h.

Moreover, in this frame the bird is moving forward at 30km/h, so it will meet the other train exactly when

$$ 30t_1 = 60 - 60t_1 $$ $$ 90t_1 = 60$$ $$ t_1 = 2/3 $$

At this point the trains are 20km apart. It's convenient to switch to the rest frame of the other train now. Again the bird moves forward at 30km/h while the original train now approaches at 60km/h. If $t_2$ is the time between when the bird hits the second train and when it reaches the first train again, we require that

$$ 30t_2 = 20 - 60t_2 $$ $$ 90t_2 = 20 $$ $$ t_2 = 2/9 $$.

You may be starting to notice a pattern. If there's any justice in the world, $t_3$ better come out to be $2/3^3 = 2/27$ and $t_n$ should be $2/3^n$. So we just need to sum them up:

$$ t_{total} = \sum_1^\infty{t_n} = \sum_1^\infty{\frac{2}{3^n}} $$ $$ = 2 \sum_1^\infty{(\frac{1}{3})^n} $$

This sum is geometric. We find

$$ t_{total} = 2 (\frac{1}{1-1/3}-1) = 2 \frac{1}{2} = 1 $$

So it takes the bird 1 hour total, traveling at 30km/h = 1/120 km/s, so its total distance is

$$ d_{total} = 3*(1/120) = 1/40 km $$

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  • $\begingroup$ how far the bird fly at collision :) $\endgroup$
    – Rogers
    Commented Mar 17, 2014 at 13:45
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    $\begingroup$ The time you calculate is in hours , not seconds. $\endgroup$
    – KvdLingen
    Commented Mar 17, 2014 at 15:57
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    $\begingroup$ Your summation of the series is wrong. The value of 3/2 you've given is $\sum_0^\infty{(\frac{1}{3})^n}$ not $\sum_1^\infty{(\frac{1}{3})^n}$. $\endgroup$
    – John Rennie
    Commented Mar 17, 2014 at 16:43
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    $\begingroup$ Correcting for the first term gives 1 h indeed for the time flown. $\endgroup$
    – KvdLingen
    Commented Mar 17, 2014 at 16:59
  • $\begingroup$ For extra points: which way is the bird facing when it is... compressed? $\endgroup$
    – DJohnM
    Commented Mar 17, 2014 at 17:20

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