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Question

Three particles A,B and C are situated at the vertices of an equilateral triangle ABC of side d at t=0. Each of the particles moves with constant speed v. A always has its velocity along AB i.e. along the line joining the two particles, B along BC and C along CA. At what time will the particles meet each other?

A
t=d3v
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B
t=2d3v
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C
t=2dv
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D
t=dv
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Solution

The correct option is B t=2d3v

Motion of the particle is shown in the figure. Let they meet at point O
Considering motion of particle B w.r.t C, the relative velocity of B w.r.t C along the line joining them is VBC=VBVC=v(v cos60)
VBC=3v2
time taken in reducing the separation BC from d to zero is
t=dVBC=d3v2=2d3v

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