Answer
$$
\begin{aligned}
& s=\frac{v_0}{k}\left(1-e^{-k u}\right) \\
& a=-k v_0 e^{-k u}
\end{aligned}
$$
Work Step by Step
Position:
$$
\begin{aligned}
(\rightarrow) \quad d t & =\frac{d s}{v} \\
\int_0^t d t & =\int_0^t \frac{d s}{v_0-k s} \\
t_0^t & =-\left.\frac{1}{k} \ln \left(v_0-k s\right)\right|_0 ^x \\
t & =\frac{1}{k} \ln \left(\frac{v_0}{v_0-k s}\right) \\
e^{k x} & =\frac{v_0}{v_0-k s} \\
s & =\frac{v_0}{k}\left(1-e^{-k x}\right)
\end{aligned}
$$
Velocity:
$$
\begin{aligned}
& v=\frac{d s}{d t}=\frac{d}{d t}\left[\frac{v_0}{k}\left(1-e^{-k x}\right)\right] \\
& v=v_0 e^{-k t}
\end{aligned}
$$
Acceleration:
$$
\begin{aligned}
& a=\frac{d v}{d t}=\frac{d}{d t}\left(v_0 e^{-k t}\right) \\
& a=-k v_0 e^{-k t}
\end{aligned}
$$
