Yeah got it !! I know, its simple. Dunno what I didn't get it when I posted
:-)

On Fri, Jun 11, 2010 at 9:26 AM, Anurag Sharma <anuragvic...@gmail.com>wrote:

> 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 <rajn...@gmail.com> 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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