 Two-stage stopping under linear drag and dry friction

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.