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. How far will the car 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 -?

When braking, the car moves with constant acceleration a. With uniformly accelerated movement, the path of the car S is expressed by the formula: S = (V0 ^ 2 – V ^ 2) / 2 * a.

Let’s find the acceleration of the car while braking. Let’s write Newton’s 2 law in vector form: m * a = m * g + N + Ftr, where m * g is the force of gravity, N is the reaction force of the road surface, Ftr is the friction force.

ОХ: m * a = Ftr.

OU: 0 = – m * g + N.

Ftr = m * a.

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 = (V0 ^ 2 – V ^ 2) / 2 * μ * g.

S = ((20 m / s) ^ 2 – (0 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.