Consider a particle P of mass m which experiences air resistance of magnitude mkv where v is the speed and $k>0$ is a constant. Consider the situation where P is released from rest and from a fixed point O and falls under gravity.
I have previously worked out: $a=dv/dt=g-kv, v=-\frac{g}{k}e^{-kt}+\frac{g}{k}$
But then it asks; Using the expression $a = v\frac{dv}{ dx}$ where x is the displacement from O determine an expression for x in terms of t.
And I just can't work out how I need to do that. Any tips? hints?
Note that
$$a=dv/dt=g-kv\implies v=-\frac{g}{k}e^{-kt}+\frac{g}{k}$$
and from $a = v\frac{dv}{dx}$
$\implies dx=\frac v a dv=\frac v {g-kv} dv\implies x=-\frac{g\log(g-kv)+kv}{k^2}+c$
and since $x(0)=v(0)=0\implies c=-\frac{g\log g}{k^2}$