Since gravity for this problem is irrelevant I started from the following equation:
$$ma = -kv$$
From here I integrated both sides in order to find an expression of v as a function of t:
- V stands for initial velocity. $$-\frac{m}{k}*\int_V^v \frac1{v}dv = \int_0^tdt$$ which leads to: $$v=V*e^{-\frac{kt}m}$$ I integrated both sides again: $$V\int_0^te^{-\frac{kt}m}dt=\int_0^xdx ⇒ -\frac{m}ke^{-\frac{kt}m}+\frac{m}k=\frac{x}V⇒e^{\frac{kt}m}=\frac{mV}{mV-kx}$$ $$t=\frac{m}k*\ln{\frac{mV}{mV-kx}}$$ The problem with this solution is that it works only if $mV>kx$. I wonder if there is a solution that works for every $x>0$ or maybe I just missed something on the way.
Any help would be appriciated.
Your $t$ is the time to reach a given $x$. If $kx \gt mV$, you will never get there, which is what the equation is telling you.