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.
Thank you in advance.
 A: 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.
A: 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.
