= Two-stage stopping under linear drag and dry friction
{title2=$t=\frac M\lambda\log\frac U{u_*}+\frac{u_*}{\mu g},\quad d=\frac M\lambda(U-u_*)+\frac{u_*^2}{2\mu g}$}
For a body of mass $M$ initially at speed $U$, suppose the total resistance is <linear drag> $\lambda u$ until speed $u_*>0$, followed by <kinetic friction> $\mu Mg$ until rest. The first stage solves $M\dot u=-\lambda u$, giving duration $(M/\lambda)\log(U/u_*)$ and distance $(M/\lambda)(U-u_*)$. Constant deceleration in the second stage gives duration $u_*/(\mu g)$ and distance $u_*^2/(2\mu g)$. The stopping distance is finite because the switch removes the infinite exponential stopping time of pure linear drag.
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