A car moving at a speed of 72 km / h was stopped as a result of braking. the coefficient of friction of wheels

A car moving at a speed of 72 km / h was stopped as a result of braking. the coefficient of friction of wheels on the road surface is 0.15. How far the car will travel from the start of braking to the end of the stop.

The friction force F, N, as a result of which the car stops, is determined by the formula

F = μ * N,

where μ is the coefficient of friction of the wheels on the road surface;

N is the reaction force of the support (road).

Since the road surface is horizontal, and besides the frictional force, no other forces act on the car, then

N = m * g,

where m is the mass of the vehicle;

g is the acceleration of gravity.

Then

F = μ * m * g.

In accordance with Newton’s second law, the acceleration a, m / s2, with which the car brakes, is equal to

a = F / m = (μ * m * g) / m = μ * g.

Equally slow motion of the car is described by the formula

v = v0 – a * t,

where v0 and v are the initial and final speeds, m / s;

t – movement time, s.

Since the car stopped at the end of the path, then v = 0. In this case, v0 = 72 km / h = 20 m / s. Then

0 = v0 – a * t,

whence the braking time of the car is

t = v0 / a = v0 / (μ * g);

t = 20 / (0.15 * 9.81) = 13.6 s.