A particle is moving vertically downwards in a liquid. Initially its velocity is zero, and after $t$ seconds it is $v$ m s$^{-1}$. Its terminal (long-term) velocity is 5 m s$^{-1}$.

A model of the particle's motion is proposed. In this model, $v = 5(1 - e^{-2t})$.

\begin{enumerate}[label=(\roman*)]
\item Show that this equation is consistent with the initial and terminal velocities. Calculate the velocity after 0.5 seconds as given by this model. [3]

\item Verify that $v$ satisfies the differential equation $\frac{dv}{dt} = 10 - 2v$. [3]
\end{enumerate}

In a second model, $v$ satisfies the differential equation
$$\frac{dv}{dt} = 10 - 0.4v^2.$$

As before, when $t = 0$, $v = 0$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{2}
\item Show that this differential equation may be written as
$$\frac{10}{(5-v)(5+v)} \frac{dv}{dt} = 4.$$

Using partial fractions, solve this differential equation to show that
$$t = \frac{1}{4} \ln\left(\frac{5+v}{5-v}\right).$$ [8]

This can be re-arranged to give $v = \frac{5(1-e^{-4t})}{1+e^{-4t}}$. [You are not required to show this result.]

\item Verify that this model also gives a terminal velocity of 5 m s$^{-1}$.

Calculate the velocity after 0.5 seconds as given by this model. [3]
\end{enumerate}

The velocity of the particle after 0.5 seconds is measured as 3 m s$^{-1}$.

\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{4}
\item Which of the two models fits the data better? [1]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI C4  Q4 [18]}}