Two-stage stopping under linear drag and dry friction ($t=\frac{M}{\lambda }\log \frac{U}{u_{\ast }}+\frac{u_{\ast }}{\mu g},d=\frac{M}{\lambda }(U-u_{\ast })+\frac{u_{\ast }^2}{2\mu g}$)

For a body of mass $M$ initially at speed $U$, suppose the total resistance is linear drag $\lambda u$ until speed $u_{\ast }>0$, followed by kinetic friction $\mu Mg$ until rest. The first stage solves $M\dot{u}=-\lambda u$, giving duration $(M/\lambda )\log (U/u_{\ast })$ and distance $(M/\lambda )(U-u_{\ast })$. Constant deceleration in the second stage gives duration $u_{\ast }/(\mu g)$ and distance $u_{\ast }^2/(2\mu g)$. The stopping distance is finite because the switch removes the infinite exponential stopping time of pure linear drag.