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 $t$, where $t$ is
A) $a/\sqrt{v^{2} + v_{1}^{2}}$
B) $\sqrt{a^{2}/(v^{2} - v_{1}^{2})}$
C) $a/(v - v_{1})$
D) $a/(v + v_{1})$
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 solving we get $t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.