A particle moves along the \(x\)-axis in such a way that its acceleration is \(kx \dot{x}\,\) where \(k\) is a positive constant.
When \(t = 0\), \(x = d\) (where \(d>0\)) and \(\dot{x} =U\,\).
Find \(x\) as a function of \(t\) in the case \(U = kd^2\) and show that \(x\) tends to infinity as \(t\) tends to \(\displaystyle \frac{\pi }{2 dk}\,\).
If \(U < 0\), find \(x\) as a function of \(t\) and show that it tends to a limit, which you should state in terms of \(d\) and \(U\,\), as \(t\) tends to infinity.
Solution:
\(\,\) \begin{align*}
&& \ddot{x} &= kx \dot{x} \\
\Rightarrow && \frac{\d v}{\d x} \dot{x} &= k x \dot{x} \\
\Rightarrow && \int \d v &= \int k x \d x \\
\Rightarrow && v &= \frac12kx^2 + C \\
t=0, x = d, \dot{x} = kd^2: && kd^2 &= \frac12kd^2 + C \\
\Rightarrow && \dot{x} &= \frac12k(x^2+d^2) \\
\Rightarrow && \frac{\d x}{\d t} &= \frac12k(x^2+d^2) \\
\Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2+d^2)} \d x \\
&&&= \frac{2}{kd}\tan^{-1} \frac{x}{d} \\
\Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d} + C' \\
t = 0, x = d: && 0 &= \frac{\pi}{2kd} + C' \\
\Rightarrow && t &= \frac{2}{kd}\tan^{-1} \frac{x}{d}-\frac{\pi}{2kd}
\end{align*}
As \(x \to \infty\), \(t \to \frac{2}{kd} \frac{\pi}{2} - \frac{\pi}{2kd} = \frac{\pi}{2kd} \)
\(\,\) \begin{align*}
&& v &= \frac12kx^2 + C \\
t=0, x = d, \dot{x} = U && U &= \frac12kd^2 + C \\
\Rightarrow && \dot{x} &= \frac12k(x^2-d^2)+U \\
\Rightarrow && \frac{\d x}{\d t} &=\frac12k(x^2-d^2)+U \\
\Rightarrow && \int \d t &= \int \frac{1}{\frac12k(x^2-d^2)+U} \d x \\
&& &=\frac{2}{k} \int \frac{1}{x^2-d^2+\frac{2U}k} \d x \\
&&&= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \\
\Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} + C'' \\
t = 0, \dot{x} = d: && 0 &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \frac{ \sqrt{d^2-\frac{2U}k}-d}{d+\sqrt{d^2-\frac{2U}k}} + C'' \\
\Rightarrow && t &= \frac2{k} \frac{1}{2\sqrt{d^2-\frac{2U}k}} \ln \left ( \frac{ \sqrt{d^2-\frac{2U}k}-x}{x+\sqrt{d^2-\frac{2U}k}} \frac{d+\sqrt{d^2-\frac{2U}k}}{ \sqrt{d^2-\frac{2U}k}-d} \right )
\end{align*}
as \(t \to \infty\) the denominator needs to head to \(0\), ie \(x \to -\sqrt{d^2-\frac{2U}k}\)