Problem source

A train starts from a station. The tractive force exerted by the engine is at first constant and equal to $F$. However, after the speed attains the value $u$, the engine works at constant rate $P,$ where $P=Fu.$
The mass of the engine and the train together is $M.$ Forces opposing motion may be neglected. Show that the engine will attain a speed $v$, with $v\geqslant u,$ after a time 
\[
t=\frac{M}{2P}\left(u^{2}+v^{2}\right).
\]
Show also that it will have travelled a distance 
\[
\frac{M}{6P}(2v^{3}+u^{3})
\]
in this time.

Solution source

While the force is constant, the train is accelerating at $\frac{F}{M}$, and since $u = \frac{F}{M} t_1 \Rightarrow t_1 = \frac{Mu^2}{Fu} = \frac{Mu^2}{P}$.

Once the train is being driven at a constant rate, we can observe that change in energy will be power times time, ie $Pt_2 = \frac{1}{2}M(v^2 - u^2) \Rightarrow t_2 = \frac{M}{2P} ( v^2 - u^2)$.

Therefore the total time will be $t_1 + t_2 = \frac{M}{2P} ( u^2 + v^2)$.

During the first period, the distance will be:

$s_1 = \frac12 \frac{F}{M} t_1^2 = \frac12 \frac{F}{M} \frac{M^2u^2}{F^2} = \frac{Mu^3}{2P}$

In the second period, $P = Fu$ and so $\text{Force} = \frac{P}{v} \Rightarrow M v \frac{\d v}{\d x} = \frac{P}{v} \Rightarrow M \l \frac{v^3}{3} - \frac{u^3}{3}\r = Ps_2$

and therefore total distance will be: $\frac{M}{6P}(2v^{3}+u^{3})$