A particle moves with constant speed $v$ along a regular hexagon $A B C D E F$ in the same order. Then the magnitude of the average velocity for it’s motion from $A$ to

F

\frac{v}{5}

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D

\frac{v}{3}

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C

\frac{v \sqrt{3}}{2}

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B

v

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Solution

The correct option is D $B$ is $v$
(A) $A$ to $F$
Average velocity $= \frac{Total displacement}{Total time}$
$= \frac{a}{5 a / v} = \frac{v}{5}$
(B) $A$ to $D = \frac{2 a}{3 a / v} = \frac{2}{3} v$
(C) $A$ to $C = \frac{a \sqrt{3}}{2 a / v} = \frac{v \sqrt{3}}{2}$
(D) $A$ to $B = \frac{a}{a / v} = v$

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