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Standard XII

Physics

Velocity

Two boys are ...

Question

Two boys are standing at the ends A and B of a ground where $A B = a$ . The boy at B starts running in a direction perpendicular to $A B$ 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

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

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

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

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

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Solution

The correct option is B $\sqrt{\frac{a^{2}}{(v^{2} - v_{1}^{2})}}$
Let two boys meet at point $C$ after time $^{'} t^{'}$ from the starting. Then
$A C = v t, B C = v_{1} t$

$(A C)^{2} = (A B)^{2} + (B C)^{2} \Rightarrow v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
By solving we get $t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$

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