Question

A particle moves with a uniform speed $v$ in a circle of radius $R$.


The $x-$component of the average velocity of the particle in a half revolution will be (at final position as shown in figure)

• A. $\frac{-v}{\pi }$
• B. $\frac{-v}{2\pi }$
• C. $\frac{-v}{4\pi }$
• D. $\frac{-2v}{\pi }$

Answer 1

The correct option is A $\frac{-v}{\pi }$

When particle reaches to the final position after half revolution, the net displacement in $x-$direction will be

$\Delta x=x_2-x_1$

$\Rightarrow \Delta x=(-R\cos {60}^{\circ })-\left(R\cos {60}^{\circ }\right)$

$\Rightarrow \Delta x=-2R\cos {60}^{\circ }=-R$

Time interval for the travel is given by

$\Delta t=\frac{T}{2}=\frac{\left(\frac{2\pi R}{v}\right)}{2}=\frac{\pi R}{v}$

We know that

$(v_{avg}{)}_x=\frac{\Delta x}{\Delta t}$

$\therefore (v_{avg}{)}_x=\frac{(-R)}{\pi R/v}=-\frac{v}{\pi }$

Hence, option $\left(a\right)$ is correct.

Why this Question? Tip: Average velocity is a vector quantity and in problems asking for $v_{avg}$, be very careful about the direction of displacement during a given time interval. Bottom line: The direction of displacement can be represented by $+\&-$ sign for a single direction i.e., $x$ or $y-$ direction.