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Two boys are standing at ends A and B of a ground where AB=a. The boy at B starts running in a direction perpendicular to AB with a velocity v$_1$. The boy at A starts running simultaneously with velocity v and catches the other boy in a boy in a time t. Calculate t in terms of a, v and v$_1$.

$\begin{array}{1 1} \sqrt {\large\frac{a^2}{v^2+v_1^2}} \\ \sqrt {\large\frac{a^2}{v^2-v_1^2}} \\a^2\sqrt {\large\frac{v^2}{v^2-v_1^2}} \\ a^2\sqrt {\large\frac{v^2}{v^2+v_1^2}}\end{array}$

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Answer: $t=\sqrt {\large\frac{a^2}{v^2-v_1^2}}$
The problem is illustrated below.
Let two boys meet at c in time 't'
Then $AC=vt$, $BC=v_1t$ $\rightarrow AC^2=BC^2+AB^2$
$\Rightarrow v^2t^2=v_1^2t^2+a^2$
$\Rightarrow t=\sqrt {\large\frac{a^2}{v^2-v_1^2}}$
answered Jun 21, 2013 by meena.p
edited Aug 11, 2014 by balaji.thirumalai

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