How to find stopping distance of a car? I am trying to calculate the minimum stopping distance of a car once the brakes are applied. I know that $F = ma$, and the braking force is $F = \mu N = \mu m g$, so
$$a = \mu g.$$
Next, by applying the kinematics equation
$$v_f^2 - v_i^2 = 2 a x$$
I found
$$v^2 = 2 a d$$
where $a$ is the deceleration, $v$ is the initial velocity, and $d$ is the stopping distance. Then
$$d = \frac{v^2}{2a} = \frac{v^2}{2 \mu g}.$$
Is this reasoning right? 
 A: Your calculations are correct.  They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking.  If this is added to your calculations, then the two results should be similar.
A: So I'm assuming you're saying that the work done on the car in distance $d$ has to be equal to its kinetic energy $\frac{1}{2} m v^2$. Then, using $W = F d$:
$$
F d = \frac{1}{2} m v^2 \\
m a d = \frac{1}{2} m v^2 \\
d = \frac{v^2}{2 a}
$$
So, yes, this equation is correct. Your relation between the two forces is also correct. Since mass drops out, the coefficient of friction simply relates (vertical) acceleration due to gravity with (horizontal) acceleration due to friction. Assuming this is the acceleration the car is capable of throughout the process of stopping, we plug this in to get
$$
d = \frac{v^2}{2 \mu g}
$$
I can't address any differences between this and your model because I don't know what your model is.
A: When you're using the equation F=ma, the F is ALWAYS the total/resultant/net/unbalanced force, NOT one of the individual forces. It describes the effect (the acceleration) that happens due to the cause (the total force on an object). Here you happen to be right because (at least horizontally) there is only one force, the frictional force, so you should get a = mu*g, which is a constant, so the "SUVAT" equations for constant acceleration can be used.
