Question 21 of 308

As shown in the figure, a particle is moving with constant speed $\pi m{s}^{-1}$. Considering its motion from $A$ to $B$, the magnitude of the average velocity is:

A$\sqrt{3}m{s}^{-1}$B$\pi m{s}^{-1}$C$1.5\sqrt{3}m{s}^{-1}$D$2\sqrt{3}m{s}^{-1}$

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Answer: (C) $1.5\sqrt{3}m{s}^{-1}$

The given data is
$v=\pi m{s}^{-1}$
By using trigonometric ratios in the triangle shown above the displacement can be found out by
$\sin 60^{\circ}=\dfrac{AB}{2R} \Rightarrow AB=R\sqrt{3}$
The formula for distance is
$d=vt \Rightarrow t=\dfrac{d}{v} \Rightarrow t=\dfrac{120^{\circ}}{360^{\circ}}\times \dfrac{2\pi R}{\pi }=\dfrac{2R}{3}$
So, the average velocity is
${v}_{avg}=\dfrac{\sqrt{3}R}{\dfrac{2R}{3}}=\dfrac{3\sqrt{3}}{2}$
$=1.5\sqrt{3}m{s}^{-1}$

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