= 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. Thus
$$
\boxed{t_{{\rm dry}}=\frac U{\mu g},\qquad d_{{\rm dry}}=\frac{U^2}{2\mu g}.}
$$
For 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 $u(t)=Ue^{-\lambda t/M}$. At the switching speed $u_*=U/4$, the elapsed time and distance are
$$
t_*=\frac M\lambda\log4,\qquad
d_*=\int_0^{t_*}Ue^{-\lambda t/M}dt=\frac{3MU}{4\lambda}.
$$
The 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> gives
$$
\boxed{t_{{\rm wet}}=\frac M\lambda\log4+\frac U{4\mu g},\qquad
d_{{\rm wet}}=\frac{3MU}{4\lambda}+\frac{U^2}{32\mu g}.}
$$
The exponential first stage reaches its positive switching speed in finite time; it need not reach zero under <linear drag>.
Back to article page
