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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


A
av2+v21
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B
a2v2v21
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C
a(vv1)
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D
a(v+v1)
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Solution

The correct option is C $$\displaystyle \sqrt{\frac{a^2}{v^2-v_1^2}}$$
If boy A catches boy B in time t, 
$$\implies (vt)^2=(v_1t)^2+a^2$$
$$\implies t^2=\dfrac{a^2}{v^2-v_1^2}$$
$$\implies t=\dfrac{a}{\sqrt{v^2-v_1^2}}$$

504096_466846_ans_f93ad6aaac1c4d87a1199f8f4562ac9b.png

Physics

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