Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

AC2=AB2+BC2(vt)2=a2+v1t2t2v2−v12=a2t2=a2v2−v12t=a2v2−v12

Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

AC2=AB2+BC2(vt)2=a2+v1t2t2v2−v12=a2t2=a2v2−v12t=a2v2−v12

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Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts running in a direction perpendicular to AB with velocity v₁. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

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

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

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

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

answer is D.

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$\begin{array}{l} {AC}^{2} = {AB}^{2} + {BC}^{2} \\ (vt)^{2} = a^{2} + {(v_{1} t)}^{2} \\ t^{2} (v^{2} - v_{1}^{2}) = a^{2} \\ t^{2} = \frac{a^{2}}{v^{2} - v_{1}^{2}} \\ t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}} \end{array}$

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Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts running in a direction perpendicular to AB with velocity v1. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is

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Two boys are standing at the ends A and B of a ground where AB = a. They boy at B starts running in a direction perpendicular to AB with velocity v₁. The boy at A starts running simultaneously with velocity v and catches the other in a time t, where t is