Problem source

The maximum power that can be developed by the engine 
of train $A$, of mass $m$, when travelling at speed $v$ is $Pv^{3/2}\,$, where $P$ is a constant. The maximum power that can be developed by the engine of train $B$, of mass $2m$, when travelling at speed $v$ is  $2Pv^{3/2}.$ For both $A$ and $B$ resistance to motion is equal to $kv$, where $k$ is a constant. For $t\le0$, the engines are crawling along at very low equal speeds. 
At $t = 0\,$, both drivers switch on full power and at time $t$ the speeds of $A$ and $B$ are $v_{\vphantom{\dot A}\!A}$ and $v_{\vphantom{\dot B}\hspace{-1pt}B},$ respectively.
\begin{questionparts}
\item
Show that 
\[
v_{\vphantom{\dot A}\!A} = \frac{P^2 \left(1-\e^{-kt/2m}\right)^2}{k^2}
\]
and write down the corresponding result for $v_{\vphantom{\dot B}B}$.
\item Find  $v_{\vphantom{\dot B}A}$ and  $v_{\vphantom{\dot B}B}$ when $9  v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\;$.
[Not on original paper]
Show that $1 < v_{\vphantom{\dot B}\hspace{-1pt}B} /v_{\vphantom{\dot A}\!A} < 4$ for $t > 0\,$.
\item Both engines are switched off when  $9  v_{\vphantom{\dot B}A} =4v_{\vphantom{\dot B}B}\,$.
Show that thereafter $k^2 v_{\vphantom{\dot B}B}^2 = 4 P^2  v_{\vphantom{\dot B}A}\,$.
 
\end{questionparts}

Solution source

\begin{questionparts}
\item $\,$
\begin{align*}
&& P &= Fv \\
\text{N2}(\rightarrow): && Pv^{1/2} - kv &= ma \\
\Rightarrow && \dot{v} &= \frac{P}{m} \sqrt{v}-\frac{k}{m}v \\
\Rightarrow && \int \d t & = \int \frac{m}{\sqrt{v}\left (P-k\sqrt{v} \right)} \d v \\
&&t &= -\frac{2m}k\ln\left (P - k\sqrt{v_A}\right) + C \\
t = 0, v_A \approx 0: && 0 & =-\frac{2m}{k} \ln P+ C \\
\Rightarrow && C &= \frac{2m}{k} \ln P
\\
\Rightarrow && e^{-kt/2m} &= \frac{P- k \sqrt{v_A}}{P} \\
\Rightarrow && v_A &= \frac{P^2(1-e^{-kt/2m})^2}{k^2}
\end{align*}

The equation of motion for $B$ is $\dot{v_B} = \frac{P}{m}\sqrt{v} - \frac{k}{2m} v$, ie $k \to \frac{k}{2}$, so

\[ v_B = \frac{4P^2(1-e^{-kt/4m})^2}{k^2} \]

\item Suppose $9v_A = 4v_B$, then and let $e^{-kt/4m} = X$
\begin{align*}
&& 9 \frac{P^2(1-e^{-kt/2m})^2}{k^2} &= 4  \frac{4P^2(1-e^{-kt/4m})^2}{k^2} \\
\Rightarrow && \frac{3}{4} &= \frac{1-X}{1-X^2} \\
\Rightarrow && 0 &= 3X^2-4X+1 \\
&&&= (3X-1)(X-1) \\
\Rightarrow && X &= 1, \frac13 \\
X = 1: && t &= 0 \\
X = \frac13: && e^{-kt/4m} &= \frac13\\
\Rightarrow && t &= \frac{4m}{k}\ln 3 \\
&& v_A &= \frac{P^2(1-\frac19)^2}{k^2} \\
&&&= \frac{64P^2}{81k^2} \\
&& v_B &= \frac{P^2(1-\frac13)^2}{k^2} \\
&&&= \frac{4P^2}{9k^2}
\end{align*}

Notice also that
\begin{align*}
&& \frac{v_B}{v_A} &= 4\left ( \frac{1-X}{1-X^2} \right)^2 \\
&&&= 4 \frac{1}{(1+X)^2}
\end{align*}

Since $X \in (0,1)$ $\frac{v_B}{v_A} \in (1, 4)$

\item Once the engines are switched off, the equation of motion for $A$ is (where $t$ is measured from that point)
\begin{align*}
&& \dot{v} &= -\frac{k}{m}v \\
\Rightarrow && v &= Ae^{-kt/m} \\
\Rightarrow && v_A &= \frac{64P^2}{81k^2}e^{-kt/m}
\end{align*}

Similarly, $v_B = \frac{4P^2}{9k^2}e^{-kt/2m}$

so
\begin{align*}
&& \frac{v_A}{v_B^2} &= \frac{64P^2}{81k^2} \cdot \frac{81k^4}{16P^4}  = \frac{4k^2}{P^2}
\end{align*}
as required.

\end{questionparts}