Bird flying b/w colliding trains

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The discussion revolves around the classic problem of calculating the distance a bird travels between two trains on a collision course. The simplest solution involves determining the time until the trains collide and multiplying that by the bird's constant speed, which is 60 km/h. The problem assumes zero turn-around time for the bird and constant speeds for both the trains and the bird. A humorous anecdote is shared about a famous mathematician who approached the problem differently, leading to a misunderstanding about the method of summation. Ultimately, the problem highlights the simplicity of the solution despite its seemingly complex nature.
mbrmbrg
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I'm sure there's already a thread about how else to solve the famous problem re: the distance covered by a bird flying between two trains on a collision course (other than adding up the sums, of course), but I can't find it.
Can anyone give me a quick link?
Thanks!
 
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The elapsed time is how long it takes for the trains to collide from the starting point. Distance covered is the bird's average speed times the elasped time.
 
It might be famous in the US, but I've sure never heard of it. What the hell are you talking about? :confused:
 
One way to do this problem is to calculate distance the bird, after leaving the first train, must fly to reach the second, then the distance back to the first (taking into account the motion of the trains during the flights), etc. thus getting an infinite series to be summed.

The easy way is to calculate the time until the trains collide, the multiply the bird's speed by that time.

What makes it "famous" is that there is a story that a person once asked (Von Neumann, Weyl, Wiener, ... pick your favorite big brain) that question. The mathematician thought for a second and gave the correct answer. The first person chuckled and said "many people try to do it by summing an infinite series." At which (Von Neumann, Weyl, Wiener, ... ) looked puzzled and said "but I did it by summing an infinite series!"
 
:smile:
Okay, I see what the original problem was. Does the question assume a constant speed for the bird with zero turn-around time, or are positive and negative accelerations included?
 
Jeff Reid said:
The elapsed time is how long it takes for the trains to collide from the starting point. Distance covered is the bird's average speed times the elasped time.

:hits self over head repeatedly:

Right... Talk about a brain twister. I knew there was an easy solution and I thought for SO LONG, and yet--

Well, thanks for untwisting my brain!
 
Danger said:
:smile:
Okay, I see what the original problem was. Does the question assume a constant speed for the bird with zero turn-around time, or are positive and negative accelerations included?
It does indeed assume zero turn-around time and constant speed throughout (and no wind :biggrin: ) which is why it can be done by just calculating he time until collision and multiplying this by the speed of teh bird.
 
Sorry, Danger. Here's the problem as was given to me:

Two trains, each having a speed of 25 km/h, are headed at each other on the same straight track. A bird that can fly 60 km/h flies off the front of one train when they are 58 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth. (We have no idea why a bird would behave in this way.) What is the total distance the bird travels before the trains collide?
 
mbrmbrg said:
A bird that can fly 60 km/h flies off the front of one train when they are 58 km apart and heads directly for the other train. On reaching the other train it flies directly back to the first train, and so forth.
Not the most efficient way to play badminton, but I suppose it would be good for a laugh. :rolleyes:
 

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