well the solution is pretty straight forward.
let the distance between stations be 'd' and speed of trains starting at A
and B (lets call them X and Y) be 'u' and 'v' resp.
A-----------------------------------------------------B
(X) u-> <-v(Y) at t=0
(X)|(Y) meet each
other at t= d/(u+v)
So distance left to cover for X = dv/(u+v)
and distance left to cover for Y= du/(u+v)
time X will take to cover this distance= dv/(u*(u+v)) = a
time Y will take to cover this distance= du/(v*(u+v)) = b
=> a : b = v^2 : u^2
=> u : v = b^1/2 : a^1/2
hope its clear
Anurag Sharma
On Thu, Jun 10, 2010 at 11:05 PM, Raj N <[email protected]> wrote:
> Can someone help me deriving this ?
> If 2 trains start at the same time from points A and B towards each
> other and after crossing they take a and b sec in reaching B and A
> respectively, then A's speed:B's speed=b^1/2:a^1/2
>
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