The velocity of a particle, $P$, moving along the $x$-axis is given by the differentiable function $v_P$, where $v_P(t)$ is measured in meters per hour and $t$ is measured in hours. Selected values of $v_P(t)$ are shown in the table below. Particle $P$ is at the origin at time $t = 0$.

\begin{center}
\begin{tabular}{ | c | | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
$t$ \\
(hours) \\
\end{tabular} & 0 & 0.3 & 1.7 & 2.8 & 4 \\
\hline
\begin{tabular}{ c }
$v_P(t)$ \\
(meters per hour) \\
\end{tabular} & 0 & 55 & -29 & 55 & 48 \\
\hline
\end{tabular}
\end{center}

(a) Justify why there must be at least one time $t$, for $0.3 \leq t \leq 2.8$, at which $v_P'(t)$, the acceleration of particle $P$, equals 0 meters per hour per hour.

(b) Use a trapezoidal sum with the three subintervals $[0, 0.3]$, $[0.3, 1.7]$, and $[1.7, 2.8]$ to approximate the value of $\int_0^{2.8} v_P(t)\, dt$.

(c) A second particle, $Q$, also moves along the $x$-axis so that its velocity for $0 \leq t \leq 4$ is given by $v_Q(t) = 45\sqrt{t}\cos\left(0.063t^2\right)$ meters per hour. Find the time interval during which the velocity of particle $Q$ is at least 60 meters per hour. Find the distance traveled by particle $Q$ during the interval when the velocity of particle $Q$ is at least 60 meters per hour.

(d) At time $t = 0$, particle $Q$ is at position $x = -90$. Using the result from part (b) and the function $v_Q$ from part (c), approximate the distance between particles $P$ and $Q$ at time $t = 2.8$.