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

Answer: (D)

$\sqrt{ \frac{a^2}{v^2 - v_1^2 }}$

Answer 2

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) : $\sqrt{ \frac{a^2}{v^2 - v_1^2 }}$.