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 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)

Distance covered by boy $A$ in time $t$
$AC=vt$

Distance covered by boy $B$ in time $t$
$BC=v_{1}t$
Using Pythagorus theorem
$A{C}^2=A{B}^2+B{C}^2$
or $(vt{)}^2=a^2+{\left(v_{1}t\right)}^2$
or $v^2{t}^2-v_1^2{t}^2=a^2$
or $t^2\left(v^2-v_1^2\right)=a^2$
$\therefore t=\sqrt{\frac{a^2}{\left(v^2-v_1^2\right)}}$