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

Question

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.

Thank you in advance.

Answer 1

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.

Answer 2

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.