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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
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B
2lv
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C
2l3v
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D
none
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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,
$$t=\displaystyle \frac{l}{v_{A}-v_{B}\cos 120}$$
$$=\displaystyle \frac{l}{v+v/2}$$
$$=\displaystyle \frac{2l}{3v}$$
129895_134053_ans.png

Physics

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