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 :

• A. $\frac{a}{\sqrt{v^2+v^2_1}}$
• B. $\sqrt{\frac{a^2}{v^2-v_1^2}}$
• C. $\frac{a}{v-v_1}$
• D. $\frac{a}{v+v_1}$

Answer 1

Answer: (B)

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

Answer 2

If boy A catches boy B in time t,

$\Rightarrow$ (vt)2=(v1t)2+a2

$\Rightarrow$ t2 = ${\frac{a^2}{v^2-v_1^2}}$

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

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