Two boys are standing at points A and B on ground where distance AB = a. The boy at point B starts running perpendicular to line AB with velocity ' V₁ '. The boy at point A starts running simultaneously with velocity 'V' and catches the other boy in time 't'. The value of 't' is

Step-by-step solution

Let the meeting point be C. In the right-angled triangle ABC, the initial distance AB is a . Since the boy at B runs perpendicular to AB, the angle at B is 90^ . The distance travelled by the boy from B is BC = V₁ t . The distance travelled by the boy from A is AC = V t . Using Pythagoras theorem in ABC : (AC)^2 = (AB)^2 + (BC)^2 (V t)^2 = a^2 + (V₁ t)^2 V^2 t^2 - V₁^2 t^2 = a^2 t^2 (V^2 - V₁^2) = a^2 t = [ a^2 V^2 - V₁^2 ]^ 1 2 Answer: [ a^2 (V^2 - V₁^2) ]^ 1 2