Q. 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 $t,$ when $t$ is
A) $\frac{a}{\sqrt{\left( v^{2} + v_{1}^{2} \right)}}$
B) $\sqrt{a^{2}\ /\ \left( v^{2} - v_{1}^{2} \right)}$
C) $a\ /\ (v - v_{1})$
D) $a\ /\ \left( v\ + v_{1} \right)$
Explanation
(b) Let two boys meet at point $C$ after time $t$ from the starting. Then,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ AC = vt,\ BC = v_{1}t$
${\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (AC)}^{2} = (AB)^{2} + (BC)^{2}$
$\Rightarrow \ \ \ \ \ \ v^{2}t^{2} = a^{2} + v_{1}^{2}t^{2}$
By showing, we get $t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$
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Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.