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Rail Road Bridge

What is it what we want to solve, calculate, etc. I just tells the facts, never asks a puzzle. Am I missing something

Pankaj Gupta
Friday, May 17, 2002

He was so concerned with laying out the assumptions that he forgot the question!
The question is, "how fast is the train going?"  Since the man's speed isn't given, you have to solve in terms of it.
I'll post the answer seperately to not spoil.

Brian McKeever
Friday, May 17, 2002


Assume the length of the tunnel is L, the man's speed is Vm and the train's speed is Vt.  Assume also that the train is a distance X behind the start of the tunnel.

Since the man just escapes in both cases, the time taken by him to cover either distance should be same as the time taken by the train to do the same.

If he runs to the beginning, he escapes, so we have:
L/4Vm = X/Vt

If he runs to the other end, he escapes also, so we have:
3L/4Vm = (X+L)/Vt = X/Vt + L/Vt = L/4Vm (from above) + L/Vt

=> 3L/4Vm - L/4Vm = L/Vt
=> L/2Vm = L/Vt
=> Vt = 2Vm

so the train is traveling at the twice the man's speed.

Vin
Friday, May 17, 2002

I fought for a bit with the algebra of it, before coming up with a more aha!-like solution...

The man can run back to the start of the tunnel, and cover some distance L/4 in some time T. Or he can run forward, covering 3L/4 in 3T. Subtracting the two, we see that the change of time by 2T corresponds to a change of distance by L/2 for the man.

But in the same delta T, the train will cover the entirety of the tunnel, or L. Since the train covers twice more distance in the same delta T, its speed is twice as much.

levik
Friday, May 17, 2002

From the information that he barely escapes by running backwards, we know that the train reaches the tunnel entrance in the time it takes the man to run 1/4th of a tunnel lenth.

So, if he runs forwards, he makes it to the middle of the tunnel when the train reaches the beginning of the tunnel.  Both the train and the man then get to the end at the same time, so the train is going twice as fast as the man.

Billy Martin
Saturday, May 18, 2002

hi,
I thought some cool solution, no mathematics involved,

  the man has covered 1/4 of the tunnel, the train arrives and
1) he runs back and just reaches, so in the time he covers 1/4 of the tunnel, the train just reaches the tunnel.

2) he runs forward and by the time he covers 1/2 of the tunnel, the train just reaches the tunnel.  (see point 1) .

now, in the time he covers rest half of the tunnel, the train reaches from the starting to the end of the tunnel,
so, speed of train is double to the speed of man

Vipul Goyal
Tuesday, May 21, 2002

(I had posted this as a new topic. Didn't notice the existing one!)

Let's assume that there are 2 persons in the tunnel, instead of one and that they run at the same speed, one-fourth from the tunnel entrance.

The first person turns around and runs towards the entrance, covering one-fourth of the tunnel length. At the same time the second person runs towards the exit of the tunnel with the same speed. By the time the first person reaches the entrance, the train is there; the second person is at the mid-point of the tunnel.

By the time the second person reaches the exit of the tunnel, covering one-half the tunnel length, the train would have covered the length of the tunnel and just misses him.

So, the train's speed is twice that of the man's.

smnaha
Tuesday, June 04, 2002

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