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
B
a2v2v21
C
a(vv1)
D
a(v+v1)

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, $$\implies (vt)^2=(v_1t)^2+a^2$$$$\implies t^2=\dfrac{a^2}{v^2-v_1^2}$$$$\implies t=\dfrac{a}{\sqrt{v^2-v_1^2}}$$Physics

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