Three particles are initially at the verticle of an equilateral triangle. The particle $A$ and $B$ start to move with constant velocities $v_{A}$ and $v_{B}$ in any direction. Which of following relations satisfy the speed of the $3 rd$ particle $C$ such that all particles lie on an equilateral triangle?

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$v_{C} \leq v_{A} + v_{B}$

$v_{C} \geq v_{A} + v_{B}$

$v_{C} < v_{A} + v_{B}$

$v_{C} \leq \frac{1}{2} (v_{A} + v_{B})$

answer is A.

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Detailed Solution

According to addition law of vector,

$v_{A} + v_{C} = v_{B}$
$v_{C} = v_{B} - v_{A}$
Squaring on both sides, we get
$v_{C}^{2} = {(v_{B} - v_{A})}^{2}$
or $v_{C}^{2} = v_{B}^{2} + v_{A}^{2} - 2 v_{A} v_{B} \cos 60 °$
$v_{C} \leq v_{A} + v_{B}$

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