A particle $P$ of mass 2 kg can only move along the straight line segment $OA$, where $OA$ is on a rough horizontal surface. The particle is initially at rest at $O$ and the distance $OA$ is 0.9 m.

When the time is $t$ seconds the displacement of $P$ from $O$ is $x$ m and the velocity of $P$ is $v$ ms$^{-1}$.
$P$ is subject to a force of magnitude $4e^{-2t}$ N in the direction of $A$ for any $t \geqslant 0$. The resistance to the motion of $P$ is modelled as being proportional to $v$.

At the instant when $t = \ln 2$, $v = 0.5$ and the resultant force on $P$ is 0 N.

\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model, $\frac{dv}{dt} + v = 2e^{-2t}$. [3]

\item Find an expression for $v$ in terms of $t$ for $t \geqslant 0$. [5]

\item By considering the behaviour of $v$ as $t$ becomes large explain why, according to the model, $P$'s speed must reach a maximum value for some $t > 0$. [2]

\item Determine the maximum speed considered in part (c). [2]
\end{enumerate}

\hfill \mbox{\textit{SPS SPS FM Pure 2022 Q11 [12]}}