 Solution

On the horizontal road the total normal reaction is $Mg$. The kinetic friction force has magnitude $\mu Mg$, so Newtonian mechanics gives $\dot{u}=-\mu g$ until rest. ThusFor the wet high-speed stage, add the four tyre drags: the total linear drag is $\lambda u$, not $\lambda u/4$. Hence $M\dot{u}=-\lambda u$ and $u(t)=U{e}^{-\lambda t/M}$. At the switching speed $u_{\ast }=U/4$, the elapsed time and distance areThe remaining stage has the same constant kinetic friction deceleration as on a dry road, but starts at $U/4$. It therefore contributes time $U/(4\mu g)$ and distance $U^2/(32\mu g)$. Adding both stages of the two-stage stopping under linear drag and dry friction givesThe exponential first stage reaches its positive switching speed in finite time; it need not reach zero under linear drag.