MCQ
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
- A$F$ is $v/5$
- B$B$ is $v$
- C$\frac{{\sqrt 3 v}}{2}$
- ✓All of the above
✓
Answer
Correct option: D.
All of the above
d
$\left|\vec{v}_{a v}\right|=\left|\frac{\text {displacement}}{\text {timetaken}}\right|$
$\left|\vec{v}_{a v}\right|=\left|\frac{\text {displacement}}{\text {timetaken}}\right|$
$\left|\vec{v}_{a v}\right|=\frac{X}{5 X / V}=\frac{v}{5}$
$\left|\vec{v}_{a v}\right|=\frac{x}{3 x / v}=\frac{2 v}{3}$
$\left|\vec{v}_{a v}\right|=\frac{2 x \cos 30^{\circ}}{2 x / v}=\frac{\sqrt{3} v}{2}$
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