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

Answer

B

Explanation

Ac=vt , BC =` v_(1)t`
`:' AB= sqrt(AC^(2)-BC^(2))`
`:. a=sqrt(v^(2)t^(2)-v_(1)t^(2))impliest=(a)/(sqrt(v^(2)-v_(1)^(2)))`