A car weighing 1.5 tons moves on a horizontal road at a speed of 54 km / h. find the braking time

A car weighing 1.5 tons moves on a horizontal road at a speed of 54 km / h. find the braking time if the braking force is 10 kN. find also the braking distance.

A car with a mass of m = 1.5 t = 1500 kg, moving along a horizontal road at a speed of v₀ = 54 km / h = 15 m / s, begins to stop under the action of a braking force F = 10 kN = 10000 N, that is, v = 0 m /with. According to Newton’s second law, this force: F = m ∙ а, where acceleration а = (v – v₀) / t or F = m ∙ (v – v₀) / t, then the braking time:

t = m ∙ (v – v₀) / F.

Since a = F / m, the braking distance will be:

S = (v ^ 2 – v₀ ^ 2) / (2 ∙ a) or S = (v ^ 2 – v₀ ^ 2) ∙ m / (2 ∙ F).

Substitute the values ​​of the quantities in the calculation formulas:

t = (1500 kg ∙ 15 m / s) / 10000 N; t = 2.25 s;

S = (15 m / s) ^ 2 ∙ 1500 kg / (2 ∙ 10000 N); S = 1.88 m.

Answer: The braking time is 2.25 s, the braking distance is 1.88 m.