If you include "reaction time/distance" in the formula, then yes.

You can derive the formula from first principles with a basic knowledge of high school kinematics. The formula is (and you can check its validity for yourself by dimensional analysis):

*s*_{S} = *v*_{0}∙*t*_{R} + ½∙*m*∙*v*_{0}^{2}/*F*_{B}where

*s*_{S} ― total stopping distance (metres);

*v*_{0} ― velocity of the vehicle before braking commences (metres per second);

*t*_{R} ― driver’s reaction time (seconds);

*m* ― vehicle’s mass (kg); and

*F*_{B} ― constant braking force (Newtons).

For example, a 1,500 kg vehicle travelling at 34 m/s (= 122 km/h) whose brakes and tyres are capable of exerting 10 kN (= 10,000 N) of braking force will have a total stopping distance of about 104 metres if the driver’s reaction time is half a second. At half that speed with everything else being equal, the stopping distance is a little over 30 metres. At twice the speed, the stopping distance is 381 metres.

Clearly, the dominant term on the right-hand side is the second one, which is quadratic in

*v*_{0}, and if you neglect the reaction time, the first term on the right-hand side falls away completely, making the formula a pure quadratic in

*v*_{0}.

'Luthon64