# Average Velocity of a body moving in a circle with constant speed $v$ [closed]

A Body is moving with constant speed $v$ along a circle of radius $R$. Find the average velocity of the body from time $t = 0$ to $t= \frac{R}{3V}$.

My attempt at the question:

Let distance traveled along the circle be $d$. $$d= \left(\frac{R}{3v}\right)\cdot v \Rightarrow d= \frac{R}{3}$$ Let angle subtended by arc be $\theta$. Then $$\theta = \frac13 rad.$$ By sine rule I can find out the displacement $S$:
$$S = 2R\sin(1/6)$$

Hence average velocity : $$\text{average velocity} = \frac{S}{t} \Rightarrow \text{average velocity}= 6v\sin(1/6)$$ But the correct answer is $\dfrac{3v}{\pi}$.
I need some help here as to how to solve it.

• Curiously enough, your answer is within a significant figure or two of the correct answer, if you plug in sin(1/6) Commented Aug 5, 2014 at 14:51
• @Tesla, do not worry, your work is impeccable, the result they gave is probably incorrect, it happens Commented Aug 6, 2014 at 12:45

Just off the top of my head, $T=\frac{2\pi R}{v}$ so the time $t=3R/v$ is a fraction $3/2\pi<1/2$ of the total period. Just less than one half, which means that the angle sub tended is not $\theta=1/3$ actually $\theta = 3rad\sim\pi$ so that most of the one direction of motion along the semi circle will have canceled out and the other component will be responsible for most of the motion.
I think you should set up a vector quantity for the position and derive the velocity by taking derivatives. find the position after a time $t'=3R/v$, find the change in position between the starting point and ending point, find the distance by squaring the change in position then divide by the time.
$$\vec r=Rcos(\omega t)\hat x + Rsin(\omega t)\hat y$$ $$v=\omega R$$