I'm sorry I missed that thread earlier.
If you are looking for a distance, you start with an equation that has a distance in it. You want to relate it to a force? We have a simple one that uses both, the definition of work:
$$\Delta E = F\Delta x$$
Where ##F## is an applied force, for a distance ##\Delta x##, that will require (or dissipate) an amount of energy ##\Delta E##. So:
$$ \Delta x = \frac{\Delta E}{F}$$
How can we relate this to a braking vehicle? Obviously, the force ##F## is the braking force. For the energy, we know the car will change speed during braking, thus changing its kinetic energy ##\frac{1}{2}mv^2##.
But what is the braking force? Well if we assume we are braking with the friction force created by all the tires supporting the car and that all tires have equal characteristics, then the braking force is ##\mu mg##. So:
$$\Delta x = \frac{\frac{1}{2}m(v_f^2 - v_0^2)}{\mu mg}$$
Or:
$$\Delta x = \frac{1}{2}\frac{v_f^2 - v_0^2}{\mu g}$$
That is the base in a very simplistic manner. The stopping distance required for the car speed to drop from ##v_f## to ##v_0## only depends on the tire-road coefficient of friction ##\mu##. We don't even know how the torque is applied to wheels, we just know that it creates a friction force; a friction force that has an upper limit set by ##\mu##.
But let's elaborate on the equation a little bit further.
First with the energy of the car. A car has also parts that rotate, like the wheels but also the braking system, driveshaft, transmission gears, etc. These parts will also reduce their RPM as the car slows down, which means more energy to dissipate. The good news is that it will be in proportion with the total mass of the car, thus introducing the concept of equivalent mass ##m_e## as shown in
post #92. I also suggest
this article for a deeper study on the subject. So, to resume, we get:
$$E = \frac{1}{2}m_ev^2$$
Where ##m_e = \lambda m##, and ##\lambda = 1 + \frac{\sum_x I_x g_{r\ x}}{mr_t}##.
##\lambda## depends on the summation of the product of the rotational inertia ##I_x## and gear ratio ##g_{r\ x}## with respect to the wheel for all rotating component ##x##, and the tire radius ##r_t##. (see the article for the derivation).
Second, let's look at the braking force. Yes, there is the tire-road friction force ##F_t##, but there is also the
drag ##F_D## and the
rolling resistance ##F_R##. So now ##F = F_t + F_D + F_R##.
Oh shoot! ##F_D## has the car speed in it (##\frac{1}{2}\rho C_D A v^2##). We now need to use the differential version and integrate:
$$F(v)dx = m_e vdv$$
Or:
$$\Delta x = \int_{v=v_0}^{v_f} \frac{m_e v}{F_t + \frac{1}{2}\rho C_D A v^2 + F_R} dv$$
(It's a fun one!)
Now, you can use the maximum tire-road friction force as ##F_t## to get the minimum braking distance required to slow down the car. But what if the braking system is unable to make enough torque to create that maximum friction force? Obviously, the distance will be greater. And now we're getting into the heart of your problem.
I'm going to let you digest this introduction and see where you want to go from here.
