I already found the formula $d = v^2/2\mu g$ where '$v$' is for velocity, '$\mu$' is for the coefficient of friction between the tires and the road surface, and '$g$' is the acceleration of gravity. Are there any formulas similar to this one that incorporate the coefficient of friction, velocity, and deceleration?

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    $\begingroup$ But your formula already satisfies all the criteria that you listed. $\endgroup$
    – void_ptr
    Apr 25, 2019 at 19:35
  • $\begingroup$ @void_ptr it includes the acceleration of gravity but I'm creating a program where the user needs to input a negative deceleration value of a car coming to a complete stop. $\endgroup$ Apr 25, 2019 at 19:39
  • $\begingroup$ ug is the deceleration in this formula. Anyhow if you want one number a for deceleration, instead of measuring it relative to g, just use d = v^2/2a. $\endgroup$
    – void_ptr
    Apr 25, 2019 at 19:40
  • $\begingroup$ To think of it another way, besides tire friction, there aren't really other forces that can stop you (besides wind resistance, which would be very weak at low speeds). $\endgroup$
    – JMac
    Apr 25, 2019 at 19:44
  • $\begingroup$ @pythonstudent23, your formula accounts for the case where you decelerate the car at its maximum rate. Any higher deceleration would cause the tires to skid, and the stopping distance would increase. Do you want a formula that slows the car at a slower deceleration than this? If so, use $a=F/m$, which is Newton's 2nd law. $\endgroup$ Apr 25, 2019 at 21:05

1 Answer 1


What you have only works for a level road, and as mentioned in a comment, it is the minimum stopping distance without sliding. It is entirely possible to take a longer distance to stop.

If you consider motion along a road of constant slope, $\theta$ with respect to the horizontal, the situation changes and depends on whether you're traveling up the slope with initial speed $v$ or down the slope.

After drawing free-body diagrams and making standard assumptions of constant coefficient of friction and maximum braking without sliding, the results look like this:

For $v$ down the slope, $$d_{min}=\frac{v^2}{2g\left(\mu \cos\theta-\sin\theta \right)}$$

For $v$ up the slope, $$d_{min}=\frac{v^2}{2g\left(\mu \cos\theta+\sin\theta \right)}$$

Notice that if the slope angle is such that $\tan\theta > \mu$ the car moving up initially will slide down the slope after stopping, and car initially moving down will never stop.


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