A car coasting at \(20 \mathrm{~m} / \mathrm{s}\) along a horizontal road has
its brakes suddenly applied and eventually comes to rest. What is the shortest
distance in which it can be stopped if the friction coefficient between tires
and road is \(0.90 ?\) Assume that all four wheels brake identically. If the
brakes don't lock, the car stops via static friction.
The friction force at one wheel, call it wheel 1 , is
$$
F_{\mathrm{fl}}=\mu_{5} F_{N 1}=\mu_{s} F_{W 1}
$$
where \(F_{W}\) is the weight carried by wheel 1 . We obtain the total friction
force \(F_{\mathrm{f}}\) by adding such terms for all four wheels:
$$
F_{\mathrm{f}}=\mu_{s} F_{W 1}+\mu_{s} F_{W_{2}}+\mu_{s} F_{W 3}+\mu_{s} F_{W
4}=\mu_{s}\left(F_{W_{1}}+F_{W 2}+F_{W 3}+F_{W_{4}}\right)=\mu_{s} F_{w}
$$
where \(F_{W}\) is the total weight of the car. (Notice that we are assuming
optimal braking at each wheel.) This friction force is the only unbalanced
force on the car (we neglect air friction). Writing \(F=m a\) for the car with
\(F\) replaced by \(-\mu_{s} F_{W}\) gives \(-\mu_{s} F_{W}=m a\), where \(m\) is the
car's mass and the positive direction is taken as the direction of motion.
However, \(F_{W}=m g\); so the car's acceleration is
$$
a=-\frac{\mu_{s} F_{W}}{m}=-\frac{\mu_{s} m g}{m}=-\mu_{s} g=(-0.90)\left(9.81
\mathrm{~m} / \mathrm{s}^{2}\right)=-8.829 \mathrm{~m} / \mathrm{s}^{2}
$$
We can determine how far the car went before stopping by solving the constant-
\(a\) motion problem. Knowing that \(v_{i}=20 \mathrm{~m} / \mathrm{s}, v_{f}=0\),
and \(a=-8.829 \mathrm{~m} / \mathrm{s}^{2}\), we find from
\(v_{f}^{2}-v_{i}^{2}=2 a x\) that
$$
x=\frac{(0-400) \mathrm{m}^{2} / \mathrm{s}^{2}}{-17.66 \mathrm{~m} /
\mathrm{s}^{2}}=22.65 \mathrm{~m} \quad \text { or } \quad 23 \mathrm{~m}
$$
If the four wheels had not all been braking optimally, the stopping distance
would have been longer.