A car $C$ is moving horizontally in a straight line with velocity $v \text{ms}^{-1}$ at time $t$ seconds, where $v > 0$ and $t \geq 0$. The acceleration, $a \text{ms}^{-2}$, of $C$ is modelled by the equation
$$a = v\left(\frac{8t}{7 + 4t^2} - \frac{1}{2}\right).$$

\begin{enumerate}[label=(\alph*)]
\item \textbf{In this question you must show detailed reasoning.}

Find the times when the acceleration of $C$ is zero. [3]

At $t = 0$ the velocity of $C$ is $17.5 \text{ms}^{-1}$ and at $t = T$ the velocity of $C$ is $5 \text{ms}^{-1}$.

\item By setting up and solving a differential equation, show that $T$ satisfies the equation
$$T = 2 \ln\left(\frac{7 + 4T^2}{2}\right).$$ [6]

\item Use an iterative formula, based on the equation in part (b), to find the value of $T$, giving your answer correct to 4 significant figures. Use an initial value of 11.25 and show the result of each step of the iteration process. [2]

\item The diagram below shows the velocity-time graph for the motion of $C$.

\includegraphics{figure_7d}

Find the time taken for $C$ to decelerate from travelling at its maximum speed until it is travelling at $5 \text{ms}^{-1}$. [1]
\end{enumerate}

\hfill \mbox{\textit{OCR H240/03 2023 Q7 [12]}}