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

t = \frac{d}{3 v}

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t = \frac{2 d}{3 v}

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t = \frac{2 d}{v}

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t = \frac{d}{v}

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Solution

The correct option is B $t = \frac{2 d}{3 v}$

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 $V_{B C} = V_{B} - V_{C} = v - (- v c o s 60^{\circ})$
$⟹ V_{B C} = \frac{3 v}{2}$
$∴$ time taken in reducing the separation $B C$ from $d$ to zero is
$t = \frac{d}{V_{B C}} = \frac{d}{\frac{3 v}{2}} = \frac{2 d}{3 v}$

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