On the section of the road where the road sign is installed (speed limit is 30 m), the driver applied emergency braking.

On the section of the road where the road sign is installed (speed limit is 30 m), the driver applied emergency braking. The traffic police inspector found on the track of the wheels that the braking distance was 12m. Has the driver violated the traffic rules if the coefficient of friction (rubber on dry asphalt) is 0.6?

Vmax = 30 km / h = 8.3 m / s.

S = 12 m.

V = 0 m / s

g = 10 m / s2.

μ = 0.6.

V0 -?

In order to establish whether the driver violated the rules of the road, it is necessary to find his speed at the moment when the car starts braking V0 and compare with the maximum allowable Vmax.

Let us write Newton’s 2 law for emergency braking of a car 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 sliding friction force of the wheels on the road.

ОХ: m * a = Ftr.

OU: 0 = – m * g + N.

N = m * g.

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

m * a = μ * m * g.

When braking, the car moves uniformly with acceleration a = μ * g.

a = (V0 ^ 2 – V ^ 2) / 2 * S.

So the car stopped, then V = 0 m / s.

a = V0 ^ 2/2 * S.

V0 = √ (2 * a * S) = √ (2 * μ * g * S)

V0 = √ (2 * 0.6 * 10 m / s2 * 12 m) = 12 m / s.

We see that V0> Vmax the driver violated the rules.

Answer: before the start of braking, the car was moving at a speed V0 = 12 m / s, which exceeds the maximum allowable value Vmax = 30 m / s. The driver violated the traffic rules.