$\textbf{The Question:}$
Consider a particle with initial velocity $v_0 > 0$, subject only to the retarding force $F = −kv|v|^{n-1}$ with $k,n > 0$. Find $v(t)$ and $x(t)$, and investigate the behavior of $v$ and $x$ as $t → +∞$.
There will be three cases:
(a) For small $n$, the particle comes to rest after a finite time, and thus has travelled a finite distance.
(b) For intermediate $n$, the particle comes to rest only asymptotically as $t → +∞$, but the distance it travels as $t → +∞$ is finite.
(c) For large $n$, the particle comes to rest only asymptotically as $t → +∞$, and it travels an infinite distance as $t → +∞$.
Prove this scenario, find the values of $n$ that form the dividing lines between these three cases, and put the dividing-line values of n into the correct cases.
$\textbf{My Attempt:}$
I have no real idea about how to do this question given that we have velocity with the power $n-1$ in the question.
I can see that taking $n=1$ gives $F=-kv$ which can then be solved to give $x(t) = A + Be^{\frac{-kt}{m}}$, so if $t \rightarrow\infty$ the distance travelled is A. Using initial conditions I get that $t \rightarrow \infty, x(t)=\frac{v_{0}m}{k}$ .
It seems to me if n is very large, then the particle will slow down very quickly but when $v<1$ the force will become very small so the particle will continue travelling for a long time, and vice versa if $n$ is close to $0$.
But I don't know how to find these 'dividing lines' of $n$. Any help would be appreciated, thanks.
To find the general solution of the differential equation $$m\frac{dv}{dt} = -kv|v|^{n-1}$$ Consider the substitution $u = |v|^{n-1}$ for the integration.
The solution of the above should be $$|v(t)|=\left( \frac{kt}{m} (n-1) + v_0^{1-n} \right) ^{1 \over 1-n}$$
To find $x(t)$, consider both cases of the absolute value.
You should find out that $n=2$ is also a dividing line from $x(t)$.