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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a
av+v1
b
av2+v12
c
a2v2−v121/2
d
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