Question

# 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 v1. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is

Solution

## The correct option is B Let two boys meet at point C after time 't' from the starting. Then AC=vt,BC=v1t   (AC)2=(AB)2+(BC)2⇒v2t2=a2+v21t2 By solving we get t=√a2v2−v21

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