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Standard XII

Physics

Relative Velocity

Two boys are ...

Question

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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Solution

The correct option is C.

Let two boys meet at point C after time 't' from the starting.Then $A C = v t, B C = v_{1} t$

Therefore in a right angle triangle ABC,
$(A C)^{2} = (A B)^{2} + (B C)^{2} \Rightarrow v^{2} t^{2} = a^{2} + v_{1}^{2} t^{2}$
$v^{2} t^{2} - v_{1}^{2} t^{2} = a^{2}$
$t^{2} (v^{2} - v_{1}^{2}) = a^{2}$
$\Rightarrow t = \sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}$ .

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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 v1. 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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