Problem source

A spaceship of mass $M$ is travelling at constant speed $V$ in a straight line when it enters a force field which applies a resistive force acting directly backwards and of magnitude $M\omega(v^{2}+V^{2})/v$, where $v$ is the instantaneous speed of the spaceship, and $\omega$
is a positive constant. No other forces act on the spaceship. Find the distance travelled from the edge of the force field until the speed is reduced to $\frac{1}{2}V$. 
As soon as the spaceship has travelled this distance within the force field, the field is altered to a constant resistive force, acting directly backwards, whose magnitude is within 10% of that of the force acting on the spaceship immediately before the change. If $z$ is the extra distance travelled by the spaceship before coming instantaneously to rest, determine limits between which $z$ must lie.

Solution source

Using Newton's second law, we have:

\begin{align*}
&& -M\omega(v^2+V^2)/v &= M v \frac{\d v}{\d x} \\
\Rightarrow && \frac{v^2}{v^2+V^2} \frac{\d v}{\d x} &= -\omega \\
\Rightarrow && \omega X &= \int_{V/2}^V \frac{v^2}{v^2+V^2} \d v \\
&&&= \int_{V/2}^V \l 1 - \frac{V^2}{v^2+V^2} \r \d v \\
&&&= \left [v - V\tan^{-1} \frac{v}{V} \right]_{V/2}^V \\
&&&= V \l \frac12 - \tan^{-1} 1 + \tan^{-1} \frac12 \r \\
\Rightarrow X &= \frac{V}{\omega} \l \tan^{-1} \frac12 + \frac12 - \frac{\pi}{4} \r
\end{align*}.

The resistive force just before the field changes is $M \omega (\frac{V^2}{4} + V^2)/\frac{V}{2} = \frac52MV\omega$. Therefor the constant resistive force is between $\frac{11}4MV\omega$ and $\frac{9}{4}MV \omega$ and acceleration is $\frac{11}{4}V\omega, \frac{9}{4}V\omega$.

Since $v^2 = u^2 + 2as \Rightarrow s = \frac{v^2-u^2}{2a} = \frac{\frac{V^2}{4}}{2kV\omega} = \frac{V}{8k\omega}$

therefore $z \in \left [ \frac{V}{22\omega},\frac{V}{18 \omega} \right]$