Question

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. $\frac{a}{\sqrt{v^{2}+v_{1}^{2}}}$
• B. $\sqrt{\frac{a^{2}}{v^{2}-v_{1}^{2}}}$
• C. $\frac{a}{\left(v-v_{1}\right)}$
• D. $\frac{a}{\left(v+v_{1}\right)}$

Answer 1

Answer: (B)

$\sqrt{\frac{a^{2}}{v^{2}-v_{1}^{2}}}$

Answer 2

Distance covered by boy $A$ in time $t$ $AC = vt$ Distance covered by boy $B$ in time $t$ $B C=v_{1} t$ Using Pythagorus theorem $A C^{2} =A B^{2}+B C^{2}$ or $(v t)^{2} =a^{2}+\left(v_{1} t\right)^{2}$ or $v^{2} t^{2}-v_{1}^{2} t^{2} =a^{2}$ or $t^{2}\left(v^{2}-v_{1}^{2}\right) =a^{2}$ $\therefore t =\sqrt{\frac{a^{2}}{\left(v^{2}-v_{1}^{2}\right)}}$