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(upsilon^(2)+upsilon_(1)^(2))`
B. `sqrt(a^(2)//(upsilon^(2)-upsilon_(1)^(2)))`
C. `a//(upsilon-upsilon_(1))`
D. `a//(upsilon+upsilon_(1))`

Answer

B

Explanation

Let two boys meet at point C after time 't' from the starting. Then `AC=ut, BC=upsilon_(1)t` (see figure)

`(AC)^(2)=(AB)^(2)+(BC)^(2)rArr upsilon^(2)t^(2)=a^(2)+upsilon_(1)^(2)t^(2)`
By solving, we get `t= sqrt((a^(2))/(upsilon^(2)-upsilon_(1)^(2)))`