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 $v_1$. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

• A. $ \frac{a}{ \sqrt{ v^2 + v_1^2}}$
• B. $ \frac{a}{v + v_1}$
• C. $ \frac{a}{ v - v_1}$
• D. \(\sqrt{ \frac{a^2}{v^2 - v_1^2 }}\)

Answer 1

The Correct Option is D

If the boy A catches boy B in time t,
then vt₂ = (v₁t)² + a²
⇒ \(t^2=\frac{a^2}{v^2-v_{1}^2}\)
⇒ \(t=\frac{a}{\sqrt{v^2-v^{2}_1}}\)

Therefore , the correct option is (D) : \(\sqrt{ \frac{a^2}{v^2 - v_1^2 }}\).