A particle moves with constant speed v along | vedclass

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Question

$\left|\vec{v}_{a v}ight|=\left|\frac{	ext {displacement}}{	ext {timetaken}}ight|$

$\left|\vec{v}_{a v}ight|=\frac{X}{5 X / V}=\frac{v}{5}$

Vedclass · Accepted Answer

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Vedclass · Answer

$\left|\vec{v}_{a v}ight|=\left|\frac{	ext {displacement}}{	ext {timetaken}}ight|$

$\left|\vec{v}_{a v}ight|=\frac{X}{5 X / V}=\frac{v}{5}$

$\left|\vec{v}_{a v}ight|=\frac{x}{3 x / v}=\frac{2 v}{3}$

$\left|\vec{v}_{a v}ight|=\frac{2 x \cos 30^{\circ}}{2 x / v}=\frac{\sqrt{3} v}{2}$

A particle moves with constant speed $v$ along a regular hexagon $ABCDEF$ in the same order. Then the magnitude of the average velocity for its motion from $A$ to

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