## Solution to 2001 Problem 23

 This is an example of nonuniform circular motion.The component of the particle's acceleration in the tangential direction is given by\begin{align*}a_{tan} = \frac{d \left| \mathbf{v} \right|}{dt}\end{align*}where $\mathbf{v}$ is the particle's velocity. The component of the particle's acceleration in the radial direction is\begin{align*}a_{R} = \frac{\left|\mathbf{v}\right|^2}{r}\end{align*}We are given that $\left|\mathbf{v}\right| = 10 \mbox{ m/s}$ and $r= 10 \mbox{ m}$. Therefore,\begin{align*}a_R = 10 \;\mathrm{m}\mathrm{/}\mathrm{s}^2\end{align*}We are also given that $\frac{d \left| \mathbf{v} \right|}{dt} = 10 \;\mathrm{m}\mathrm{/}\mathrm{s}^2$, therefore,\begin{align*}a_{tan} = 10 \;\mathrm{m}\mathrm{/}\mathrm{s}^2\end{align*}So, the tangential and the radial component of the particle's acceleration have the same magnitude. Therefore, the resultant acceleration vector makes a $45^{\circ}$ angle with each of them. The particle's velocity vector also points in the tangential direction, therefore, the resultant acceleration vector makes a $45^{\circ}$ angle with it also. Therefore, answer (C) is correct.