The car is moving at a speed of 72 km / h. before the obstacle, the driver braked.

The car is moving at a speed of 72 km / h. before the obstacle, the driver braked. What distance the car will travel to a complete stop if the coefficient of friction is 0.2.

V0 = 72 km / h = 20 m / s.

V = 0 m / s.

g = 10 m / s2.

μ = 0.2.

S -?

We will assume that the vehicle was moving uniformly when braking. With uniformly accelerated braking, the path S traveled by the car is expressed by the formula:

S = (V02 – V2) / 2 * a, where V0 is its speed at the beginning of deceleration, V is the speed at the end of deceleration, a is the acceleration during deceleration.

Since at the end of its movement it stops V = 0 m / s, the formula will take the form: S = V02 / 2 * a.

m * a = Ftr + m * g + N.

ОХ: m * a = Ftr.

OU: 0 = N – m * g.

N = m * g.

The friction force Ffr is expressed by the formula: Ffr = μ * N = μ * m * g.

m * a = μ * m * g.

a = μ * g.

S = V02 / 2 * μ * g.

S = (20 m / s) 2/2 * 0.2 * 10 m / s2 = 100 m.

Answer: to a complete stop, the car will cover the path S = 100 m.