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 in a time t, where t is
- a v2 + v12
- av + v1
- av - v1
- a2v2 - v12
Answer
(4) a2v2 - v12
Solution
In time t, boy B moves v1 t perpendicular to AB, while boy A runs straight to the meeting point, covering vt.
AB, B's path and A's path form a right-angled triangle with A's path as hypotenuse:
(vt)2 = a2 + (v1 t)2
t2(v2 - v12) = a2
t = a2v2 - v12
Answer: (4)