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Topic: Railway Station Duellists (Read 340 times) |
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ThudnBlunder
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Railway Station Duellists
« on: Oct 16th, 2004, 8:13pm » |
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Two duellists, who decide to settle their differences immediately, are standing back-to-back on a railway station platform. When the engine of a passing train reaches them they start walking in opposite directions along the platform. The end of the train passes them when they have walked D and d metres, respectively. (D > d) If they are both walking at the same speed, how long is the train in terms of D and d?
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THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
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TenaliRaman
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I am no special. I am only passionately curious.
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Re: Railway Station Duellists
« Reply #1 on: Oct 17th, 2004, 10:54am » |
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::
i get,
L = [(D+d)2/d] - D
a small goof up,
L = 2Dd/(D-d)
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| « Last Edit: Oct 17th, 2004, 11:50am by TenaliRaman » |
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Mayank
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Re: Railway Station Duellists
« Reply #2 on: Oct 20th, 2004, 5:32am » |
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Explanation:
Assume the following:
1. The length of the train is X.
2. Train is travelling at V mph.
3. The speed of men is W mph.
4. So the distance travelled by the men walking in the direction of movement of train is D and the other man's distance travelled is d.
5. The time take for the train to cross the nearest man be T1.
6. The time take for the train to cross the nearest man be T2.
So for the train T1 = (x - d)/V
For the nearest man T1 = d/W
hence: (x - d)/V = d/W
or (x - d)/d = V/W <= (1)
Similarly for other man: T2 = (x + D)/V
and T2 = D/W
hence: (x + D)/V = D/W
or (x + D)/D = V/W <= (2)
From (1) and (2)
(x - d)/d = (x + D)/D
Solving it will reduce it to:
x = 2dD/(D - d)
I know it was simple but there are some people out there who want to know the explanation 
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ThudnBlunder
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Re: Railway Station Duellists
« Reply #3 on: Oct 20th, 2004, 5:52am » |
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Well done, Mayank. 
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