A particle $P$ of mass $m$ kg is attached to the end $A$ of a light elastic string $AB$, of natural length $a$ metres and modulus of elasticity $9ma$ newtons. Initially the particle and the string lie at rest on a smooth horizontal plane with $AB = a$ metres. At time $t = 0$ the end $B$ of the string is set in motion and moves at a constant speed $U$ m s$^{-1}$ in the direction $AB$. The air resistance acting on $P$ has magnitude $6mv$ newtons, where $v$ m s$^{-1}$ is the speed of $P$. At time $t$ seconds, the extension of the string is $x$ metres and the displacement of $P$ from its initial position is $y$ metres.

Show that, while the string is taut,

\begin{enumerate}[label=(\alph*)]
\item $x + y = Ut$
[2]

\item $\frac{d^2x}{dt^2} + 6\frac{dx}{dt} + 9x = 6U$
[5]
\end{enumerate}

You are given that the general solution of the differential equation in (b) is
$$x = (A + Bt)e^{-3t} + \frac{2U}{3}$$

where $A$ and $B$ are arbitrary constants.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the value of $A$ and the value of $B$.
[5]

\item Find the speed of $P$ at time $t$ seconds.
[2]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M4 2013 Q6 [14]}}