Two boys are standing at the ends A and B of a ground, where `AB=a`. The boy at B starts running in a direction perpendicular to AB with velocity `v_(1)`. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :

Options

A. `a/(v+v_(1))`
B. `a/(v-v_(1))`
C. `a/(sqrt(v^(2)+v_(1)^(2)))`
D. `a/(sqrt(v^(2)-v_(1)^(2)))`

Answer

D

Explanation

`a^(2)+v_(1)^(2)t^(2)=v^(2)t^(2)`
`t_(2)(v^(2)-v_(1)^(2))=a^(2)`
`t=a/(sqrt(v^(2)-v_(1)^(2)))`