Question

Three particles $$A,B,C$$ are situated at the vertices of an equilateral triangle of side $$l$$. Each of the particle starts moving with a constant velocity $$v$$ such that $$A$$ is always directed towards $$B$$, $$B$$ towards $$C$$ and $$C$$ towards $$A$$. Find the time when they meet.

A
l3v
B
2lv
C
2l3v
D
none

Solution

The correct option is C $$\displaystyle \frac{2l}{3v}$$We look into it as a problem of relative velocity and find $$v_{BA}$$ in the direction of $$B$$.Thus,$$t=\displaystyle \frac{l}{v_{A}-v_{B}\cos 120}$$$$=\displaystyle \frac{l}{v+v/2}$$$$=\displaystyle \frac{2l}{3v}$$Physics

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