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 simultaneously with velocity y as shown in figure and catches the other boy in a time t, where t is

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av+v1

av2+v12

a2v2−v121/2

av−v1

answer is C.

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Detailed Solution

The velocity of A relative to B is given byvAB=v2−v12Now, VAB×t=aor t=avAB=av2−v12

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