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

Answer 1

Answer: (B)

$\sqrt{a^{2}/(v^{2} - v_{1}^{2})}$

Answer 2

Let two boys meet at point C after time 't' from the starting. Then $AC = vt$, $BC = v_{1}t$

$(AC)^{2} = (AB)^{2} + (BC)^{2}$ ⇒ $v^{2}t^{2} = a^{2} + v_{1}^{2}t^{2}$

By solving we get

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