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//(v^2 - v_1^2))`
C. `a//sqrt(v - v_1)`
D. `a//(v + v_1)`

Answer

B

Explanation

Let two boys meet at point `C` after time 't' from the starting. Then `AC = vt, BC = v_1 t`
`(AC)^2 = (AB)^2 + (BC)^2 rArr v^2 t^2 = a^2 + v_1^2 t^2`
By solving we get
`sqrt((a^2)/(v^2 - v_1^2))`.