# Is the car braking time formula $T = v / (\mu_s \, g)$ valid only for uniformly accelerated motion?

I'm wondering if the car braking time formula is valid only for uniformly accelerated motion.

$$T = \frac{v} {\mu_s \, g}$$

with $$v$$ average speed, $$\mu_s$$ static friction coefficient between the wheel and the ground, $$g$$ gravitational acceleration on the earth.

I derived it in this way ($$F_{s, max} = \mu_s \, N = \mu_s \, m \, g$$ maximum static friction force; $$N$$ normal force, $$m$$ car mass):

$$F_{s, max} = m \, a$$ $$\mu_s \, m \, g = m \, \frac{v} {T}$$ $$T = \frac{v} {\mu_s \, g}$$

where $$a$$ is the average acceleration of the car.

• Comments are not for extended discussion; this conversation has been moved to chat. Jan 13, 2020 at 14:25
• @JMac The same issue arises when we say static friction between the tire and road is responsible for accelerating a vehicle, because that is the only external force acting on the vehicle. But we don’t say the static friction does the work in accelerating the vehicle. We say the torque applied to the wheel by the engine is doing the work. Likewise, here we can say static friction is responsible for decelerating the vehicle since it is also the only external force acting on the vehicle. But the static friction does not do the work. It is the kinetic friction force of the brakes Jan 13, 2020 at 14:31
• @BobD You need the road to do work on, so that it can do work back on your car to actually provide movement. The car has mechanisms to do work on it's components to get them moving relative to each other, or prevent that movement; but without an external surface to act against and do work on the car, it won't actually gain any transnational motion. The road does work on the car, and the car does work on it. Consider a spacecraft. It does work against it's own exhaust because there is nothing else available.
– JMac
Jan 13, 2020 at 15:02

While the derivation you've used assumes uniform acceleration, it is also possible to show that the $$T$$ you have found is a lower bound on the stopping time of the car, even without assuming uniform acceleration. Roughly speaking, even if the acceleration varies with time, its magnitude can be no greater than $$\mu_s g$$, which implies that the stopping time can be no less than the $$T$$ you have found.

More formally: assume the frictional force and the acceleration vary with time. The magnitude of the frictional force $$F_\text{fr}(t)$$ is no greater that $$\mu_s$$ (the coefficient of static friction) times the normal force $$N$$: $$|F_\text{fr}(t)| \leq \mu_s N = \mu_s m g$$ assuming the car is on level ground. This means that the acceleration of the car is bounded by $$|a(t)| = |F_\text{fr}(t)/m| \leq \mu_s g.$$ If the car has a positive velocity $$v$$ initially, then as the car brakes we have $$a(t) < 0$$, and so $$a(t) > - \mu_s g$$. Using calculus, we then have \begin{align*} \Delta v &= \int_0^T a(t) \, dt \geq \int_0^T (- \mu_s g) \, dt \\ 0 - v &\geq - \mu_s g T \\ \mu_s g T &\geq v \\ T &\geq \frac{v}{\mu_s g}. \end{align*} Thus, no matter what the car does, it will not be able to stop more quickly (i.e., in less time) than the $$T$$ you have calculated assuming uniform acceleration.

• Hello @MichaelSeifert, thank you very much for your extremely clear answer; it is very helpful. Jan 13, 2020 at 16:49
• So the bottom line is you are simply saying T in the formula is a minimum because any slipping would result in a greater stopping time, correct? Jan 13, 2020 at 17:05
• @BobD: exactly. Any deceleration less than the maximum magnitude means you need more time to stop. Jan 13, 2020 at 17:22

Yes, your equation is only valid for uniformly accelerated motion. This is because you substituted $$a$$ as $$\frac vT$$ in your derivation and that is only valid during constant acceleration.

• In attempting to edit my answer I accidentally edited yours. Sorry. I assume you can simply decline the edit. I apologize for the inconvenience. Jan 14, 2020 at 12:52