A particle moves in a circle of radius $R$ with a constant speed $v$. What is the magnitude of the average acceleration of the particle over a time interval of $\frac{\pi R}{2v}$?

Solution

1. The time interval $\Delta t=\frac{\pi R}{2v}$ corresponds to a rotation of $90^{\circ }$, meaning the velocity vector changes from ${\vec{v}}_i=v\hat{i}$ to ${\vec{v}}_f=v\hat{j}$.
2. The magnitude of the change in velocity is $$\begin{array}{rl}|\Delta \vec{v}|& =|{\vec{v}}_f-{\vec{v}}_i|\\ & =\sqrt{v^2+v^2}\\ & =v\sqrt{2}\end{array}$$.
3. The average acceleration is defined as $$\begin{array}{rl}{\vec{a}}_{avg}& =\frac{|\Delta \vec{v}|}{\Delta t}\\ & =\frac{v\sqrt{2}}{\pi R/2v}\end{array}$$.
4. Simplifying the expression yields ${\vec{a}}_{avg}=\frac{2\sqrt{2}{v}^2}{\pi R}$.
5. Hence the answer is (B).