Answer
The acceleration vector is
${\bf{a}}\left( t \right) = - 2\left( {\cos \frac{t}{2},\sin \frac{t}{2}} \right)$
At $t = \frac{\pi }{4}$, we get ${\bf{a}}\left( {\frac{\pi }{4}} \right) \simeq \left( { - 1.85, - 0.77} \right)$.
Please see the figure attached.
Work Step by Step
We have $R=8$ and $v=4$. From Example 5, we get
${\bf{r}}\left( t \right) = R\left( {\cos \omega t,\sin \omega t} \right)$
${\bf{r}}\left( {\frac{\pi }{4}} \right) \simeq \left( {7.39,3.06} \right)$
The angular velocity $\omega$ is given by $\left| \omega \right| = \frac{v}{R} = \frac{1}{2}$.
Also from Example 5, the acceleration vector is given by
${\bf{a}}\left( t \right) = - R{\omega ^2}\left( {\cos \omega t,\sin \omega t} \right)$
${\bf{a}}\left( t \right) = - 8\cdot\frac{1}{4}\left( {\cos \frac{t}{2},\sin \frac{t}{2}} \right) = - 2\left( {\cos \frac{t}{2},\sin \frac{t}{2}} \right)$
At $t = \frac{\pi }{4}$, we get ${\bf{a}}\left( {\frac{\pi }{4}} \right) = - 2\left( {\cos \frac{\pi }{8},\sin \frac{\pi }{8}} \right) \simeq \left( { - 1.85, - 0.77} \right)$.
