A car of mass \(m\) makes a journey of distance \(2d\)
in a straight line.
It experiences air resistance and rolling resistance
so that the total resistance to motion when it is moving with
speed \(v\) is \(Av^2 +R\), where \(A\) and \(R\) are constants.
The car starts from rest and
moves with constant acceleration \(a\) for a distance \(d\).
Show that the work done by the engine
for this half of the journey
is
\[
\int_0^d (ma+R+Av^2) \, \d x
\]
and that it can be written
in the form
\[
\int_0^w \frac {(ma+R+Av^2)v}a\; \d v
\,,
\]
where \(w =\sqrt {2ad\,}\,\).
For the second half of the journey, the acceleration of
the car is \(-a\).
In the case \(R>ma\),
show that the work done by the
engine for the whole journey is
\[
2Aad^2 + 2Rd
\,.
\]
In the case \(ma-2Aad< R< ma\), show that at a certain speed
the driving
force required to maintain the constant acceleration
falls to zero.
Thereafter, the engine does no work
(and the driver applies the brakes to maintain
the constant acceleration).
Show that the work done by the engine for the whole journey is
\[
2Aad^2 + 2 Rd
+ \frac{(ma-R)^2}{4Aa}
\,
.\]
Solution: The force delivered by the engine must be \(ma + R + Av^2\), (so the net force is \(ma\)). Therefore the work done is \(\displaystyle \int_0^d F \d x = \int_0^d (ma + R + Av^2) \d x\)
Notice that \(a = v \frac{\d v}{\d x} \Rightarrow \frac{a}{v} = \frac{\d v}{\d x}\) and so
\begin{align*}
&& WD &= \int_0^d (ma + R + Av^2) \d x \\
&&&= \int_{x=0}^{x=d} (ma + R + Av^2) \frac{v}{a} \frac{\d v}{\d x} \d x \\
&&&= \int_{x=0}^{x=d} \frac{ (ma + R + Av^2)v}{a} \d v \\
\end{align*}
Also notice that if we move with constant acceleration from rest for a distance \(d\) the final speed is \(v^2 = 2ad \Rightarrow v = \sqrt{2ad}\)
For the second part of the journey, the engine will be putting out a force of \(-ma+R+Av^2>0\), and the car will have a final speed of \(0\)
\begin{align*}
WD &= \int_0^{w} \frac{(ma+R+Av^2)v}{a} \d v + \int_w^0 \frac{(-ma+R+Av^2)v}{-a} \d v \\
&= \int_0^w \frac{2(Rv+Av^3)}{a} \d v \\
&= \frac{Rw^2+\frac12Aw^4}{a} \\
&= \frac{R2ad+\frac12A4a^2d^2}{a} \\
&= 2Rd + 2Aad^2
\end{align*}
If \(ma - 2Aad < R < ma\) then the driving force is still \(-ma+R+Av^2\) which is positive when \(v = \sqrt{2as}\) but negative when \(v = 0\), and therefore at some point in-between the driving force must be \(0\).
The engine will stop working when \(-ma+R+Av^2 =0 \Rightarrow v = \sqrt{\frac{ma-R}{A}}\) so
\begin{align*}
WD &= \int_0^w \frac{(ma+R+Av^2)v}{a} \d v + \int_w^{ \sqrt{\frac{ma-R}{A}}} \frac{(-ma+R+Av^2)v}{-a} \d v \\
&= \int_0^w \frac{2(R+Av^2)v}{a} \d v - \int_0^{\sqrt{\frac{ma-R}{A}}} \frac{(-ma+R+Av^2)v}{a}\d v \\
&= 2Aad^2+2Rd + \frac1a\left (\frac12(R-ma)\frac{ma-R}{A} + \frac{A}{4}\left ( \frac{ma-R}{A}\right)^2 \right) \\
&= 2Aad^2+2Rd - \frac{(ma-R)^2}{Aa}\left (-\frac12+ \frac14 \right) \\
&= 2Aad^2+2Rd - \frac{(ma-R)^2}{4Aa}
\end{align*}