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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:

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

$\frac{a}{(v + v_{1})}$

$\frac{a}{(v - v_{1})}$

$\frac{a}{\sqrt{v^{2} + {v_{1}}^{2}}}$

answer is B.

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Detailed Solution

Distance covered by boy A in time t

AC = vt -------(i)

Distance covered by boy B in time t

BC = $v_{1} t - - - - (ii)$

Using Pythagoras theorem

${AC}^{2} = {AB}^{2} + {BC}^{2}$

or ${(vt)}^{2} = a^{2} + {(v_{1} t)}^{2}$

or $v^{2} t^{2} - {v^{2}}_{1} t^{2} = a^{2}$

or $t^{2} (v^{2} - {v_{1}}^{2}) = a^{2} \Rightarrow t = \sqrt{\frac{a^{2}}{(v^{2} - {v_{1}}^{2})}}$

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