Answer
Position = $s=\frac{2\left(2kt+\left(\frac{1}{v_{o^2}}\right)\right)^{\frac{1}{2}}}{2k}$
Velocity = $=\left(2kt+\left(\frac{1}{v_o^2}\right)\right)^{\frac{-1}{2}}$
Work Step by Step
$a=\frac{dv}{dt}=-kv^3$
$\int _{v0}^v\:v^3dv=\int _0^t\:-k\:dt$
$-\frac{1}{2}\left(v^{-2}-v_o^{-2}\right)=-kt$
Velocity
$=\left(2kt+\left(\frac{1}{v_o^2}\right)\right)^{\frac{-1}{2}}$
$ds\:=vdt$
$\int _0^sds=\int _0^t\:\:\frac{dt}{\left(2kt+\left(\frac{1}{v_0^2}\right)\right)^{\frac{1}{2}}}$
Position
$s=\frac{2\left(2kt+\left(\frac{1}{v_{o^2}}\right)\right)^{\frac{1}{2}}}{2k}$
Substituting the limits:
$s=\frac{2\left(2kt+\left(\frac{1}{v_{o^2}}\right)\right)^{\frac{1}{2}}}{2k}$
