Compute the velocity and acceleration of a particle

A mechanism moves a particle along a straight line with displacement (from an initial position 0) at time $t$ given by the function

$f(t) = \frac{1}{2} t^2 + 2t \sin t.$

At time $t = \pi$ the propelling mechanism stops working, and the particle then moves with constant velocity (the velocity imparted by the mechanism at time $t = \pi$ ). Compute the following.

The velocity of the particle at time $t = \pi$ .
The acceleration of the particle at time $t = \frac{1}{2} \pi$ .
The acceleration of the particle at time $t = \frac{3}{2} \pi$ .
The displacement of the particle from its initial position 0 at time $t = \frac{5}{2} \pi$ .
Find a time $t > \pi$ such that the particle returns to its initial position, or prove that no such time exists.

Since the velocity of the particle is given by the derivative of the position we can compute,
$\begin{align*} &&v(t) &= f'(t) = t + 2 \sin t + 2t \cos t\\ \implies && v(\pi) &= \pi + 2 \sin (\pi) + 2\pi cos (\pi) = \pi - 2 \pi = - \pi. \end{align*}$
Next, the acceleration is the derivative of the velocity so using part (a) we have
$\begin{align*} && a(t) &= v'(t) = 1 + 2 \cos t + 2 \cos t - 2t \sin t \\ &&&= 1 + 4 \cos t - 2t \sin t \\ \implies && a \left( \frac{\pi}{2} \right) &= 1 + 4 \cos \left( \frac{\pi}{2} \right) - \pi \sin \left( \frac{\pi}{2} \right) \\ \implies &&&= 1 - \pi. \end{align*}$
We know from the problem statement that $v(t) = c$ for $t > \pi$ where $c$ is a constant. Hence, $v'(t) = 0$ for $t > \pi$ . Thus, the acceleration at time $t = \frac{3}{2} \pi$ is 0.
To find the position at time $t = \frac{5}{2} \pi$ we consider the motion of the particle over two time intervals: the time from 0 to $\pi$ and the time from $\pi$ to $\frac{5}{2} \pi$ . During the time from 0 to $\pi$ the position is given by the function
$f(t) = \frac{1}{2} t^2 + 2t \sin t.$

From time $t = \pi$ we know the particle then moves with constant velocity of $f'(\pi)$ so its position changes by $\frac{3}{2} \pi f'(\pi)$ during the time interval from $\pi$ to $\frac{5}{2} \pi$ . Therefore, the position at time $t = \frac{5}{2} \pi$ is given by

$\begin{align*} f \left( \frac{5 \pi}{2} \right) &= f(\pi) + \left( \frac{3 \pi}{2} \right) f'(\pi) \\ &= \frac{\pi^2}{2} - \frac{3 \pi^2}{2} \\ &= -\pi^2. \end{align*}$
Finally, the position of the particle at time $t > \pi$ is given by
$f(\pi) - \pi(t-\pi).$

Setting this equal to 0 and solving for time,

$\begin{align*} f(\pi) - \pi(t- \pi) = 0 && \implies && \frac{\pi^2}{2} - \pi t + \pi^2 &= 0 \\ && \implies && t &= \frac{3}{2} \pi. \end{align*}$