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

The Correct Option is B

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): \(\sqrt{\frac{a^2}{v^2-v_1^2}}\)