The car moves on a horizontal road at a speed of 108 km / h (or 30 m / s). Determine its braking

| September 24, 2021 | education

The car moves on a horizontal road at a speed of 108 km / h (or 30 m / s). Determine its braking distance S, the coefficient of friction is 0.4.

V0 = 108 km / h = 30 m / s.

V = 0 m / s.

g = 10 m / s2.

μ = 0.4.

S -?

The braking distance of the car S is expressed by the formula: S = (V02 -V2) / 2 * a.

Since the car stops at the end of braking, then V = 0 m / s, S = V02 / 2 * a.

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

ОХ: m * a = Ftr.

OU: 0 = – m * g + N.

N = m * g.

a = Ftr / m.

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

a = μ * m * g / m = μ * g.

S = V02 / 2 * μ * g.

S = (30 m / s) 2/2 * 0.4 * 10 m / s2 = 112.5 m.

Answer: to a complete stop, the car will travel the path S = 112.5 m.



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