Two boys are standing at the ends $A$ and $B$ of a ground, where $A B=a$. The boy at $B$ starts running in a direction perpendicular to $A B$ with velocity $v_1$. The boy at $A$ starts running simultaneously with velocity $v$ and catches the other boy in a time $t$, where $t$ is

• A

  $\frac{a}{\sqrt{v^2+v_1^2}}$
• Correct answer
  B

  $\sqrt{\frac{a^2}{v^2-v_1^2}}$
• C

  $\frac{a}{\left(v-v_1\right)}$
• D

  $\frac{a}{\left(v+v_1\right)}$

Solution

Distance covered by boy $A$ in time $t$ $A C=v t...(i)$ Distance covered by boy $B$ in time $t$ $B C=v_1 t...(ii)$ Using Pythagorus theorem $\begin{aligned}& & A C^2 & =A B^2+B C^2 \\& \text { or } & (v t)^2 & =a^2+\left(v_1 t\right)^2 \\& \text { or } & v^2 t^2-v_1^2 t^2 & =a^2 \\& \text { or } & t^2\left(v^2-v_1^2\right) & =a^2\end{aligned}$ $\therefore$ $t=\sqrt{\frac{a^2}{\left(v^2-v_1^2\right)}}$

Hence, the correct answer is option: 'B'.