A car weighing 3.2 tons moves along a horizontal path at a speed of 54 km / h.

| February 15, 2021 | education

A car weighing 3.2 tons moves along a horizontal path at a speed of 54 km / h. at what distance will the car stop if the friction force is 45 kN during braking?

m = 3.2 t = 3200 kg.

V0 = 54 km / h = 15 m / s.

V = 0 m / s.

g = 9.8 m / s ^ 2.

Ftr = 45 kN = 45000 N.

S -?

With uniformly accelerated motion, the path of the body is determined by the formula: S = (V ^ 2 – V0 ^ 2) / 2 * a.

Since the car stopped, V = 0 m / s, the formula will take the form: S = -V0 ^ 2/2 * a.

The sign “-” means that the acceleration of the body is directed against the direction of motion, the body is inhibited.

Let’s write Newton’s 2 law: m * a = Ftr.

a = Ftr / m.

S = V0 ^ 2 * m / 2 * Ftr.

S = (15 m / s) ^ 2 * 3200 kg / 2 * 45000 N = 8 m.

Answer: the car will stop after a distance S = 8 m.



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