1) the braking distance s is equal to the sum of the reaction distance S 1 and the braking distance S2; (or through the distance S=S 1+L+S2+L0)

2) the reaction distance S 1 is proportional to the vehicle speed v, the reaction time is T 1, and the braking time is t2;

3) When braking, the comprehensive factor is U, and the work done by the maximum braking force (friction force) F is the change of vehicle kinetic energy.

Model structure

From assumption 2), it is concluded that: s1= t1v ... (1)

According to hypothesis 3), the work FS2 done by S2 under the force F changes the vehicle speed from V to 0, and the change of kinetic energy is MV*V/2, that is, FS2=MV*V/2, and F is directly proportional to M. According to Newton's second law, the acceleration a during braking is constant, so V * V = 2As2..(2).

Where: a=-ug (F=-uMg=Ma, g is the acceleration of gravity) ...

1), find the braking distance s:

Synthesize a, (1), (2) and (3) to get s = (t1+v/2ug) * v.

The vehicle speed from V to Vt=0 can be listed as VT = V+AT2...(4)

2), find the yellow light time t:

B, (3) and (4): T=T 1+V/ug (the reaction time can be obtained according to statistical data or empirical time). We only need to apply the model to practice, knowing the parameters T 1, v, u, the empirical estimated reaction time T 1 is 0.75 seconds, the length of the car body is 5 meters, the intersection is 20 meters, and the total friction coefficient and automobile braking performance coefficient are 0.4.