A suppose a particle of mass $m$ is projected horizontally across two surfaces $S_1$ and $S_2$ with resistive forces $mkv$ and $mkv^2$ respectively where $k$ is a constant and $v$ is the velocity of the particle.
This means that the acceleration $a=\frac{F}{m}$ is given by $a_1=-kv_1$ and $a_2=-kv_2^2$. Using the chain rule, we derive that $a=\frac{dx}{dt}\times\frac{dv}{dx}=v\frac{dv}{dx}$.
Therefore, with the initial condition when $x\;(\text{the displacement})=0, v=u$, we get $v_1=-kx_1+u$ and $v_2=ue^{-kx_2}$.
Finally, with $v=\frac{dx}{dt}$ and initial conditions $x=0, t=0$ we get $x_1=\frac{u}{k}\left(1-e^{-kt}\right)$ and $x_2=\frac{1}{k}\ln(1+ukt)$.
I was wondering if someone could explain why $S_1$ has a limiting displacement of $\frac{u}{k}$ while $S_2$ does not have a limiting displacement. Intuitively I expected the particle which undergoes $mkv^2$ resistance to stop faster but with this model it does not stop at all. The only idea I have around why this might be is that for $v<1$, $mkv^2<mkv$ but this doesn't explain to me at all how there is no limiting motion. So if someone could clarify why my intuitive response is wrong or where the mathematical model might be at fault, that would be great.