Two boys are standing at the end 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 simultanieously with velocity V and catches the other boy in a timet, where t is

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$\frac{a}{\sqrt{V^{2} + V_{1}^{2}}}$

$\frac{a}{V + V_{1}}$

$\frac{a}{V - V_{1}}$

$\sqrt{\frac{a^{2}}{(V^{2} - V_{1}^{2})}}$

answer is D.

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Detailed Solution

Velocity of A relative to B is given by

$V_{A B} = \bar{V_{A}} - \bar{V_{B}} = \bar{V} - \bar{V_{1}} \to (1)$ by taking x-components of eq(1), we get
$O = v \sin θ - v_{1} \Rightarrow \sin θ = \frac{v_{1}}{v} \to (2)$
By taking y-components of eq(1), we get $V_{y} = V \cos θ \to (3)$
Time taken by boy at A to catch the boy B is
$t = \frac{r e l a t i v e s}{r e l a t i v e s} \frac{d i s p l a c e m e n t s}{v e l o c i t y} \frac{a l o n g}{a l o n g} \frac{y - a x i s}{y - a x i s} = \frac{a}{v \cos θ} = \frac{a}{v \sqrt{1 - \sin^{2} θ}}$
$= \frac{a}{v \sqrt{1 - {(\frac{v_{1}}{v})}^{2}}} = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$