Question

A 1300-kg car initially has a velocity of 22.2 m/s due south. It brakes to a stop over a 180 m distance.

(a) What is the magnitude of the car’s acceleration, in m/s 2
(b) What average net force magnitude was necessary to stop the car?
(c) Assuming the tires do not skid, what coefficient of static friction between tires and
pavement is needed?

Answer 1

(a.) The acceleration of the car can be calculated using the equation,
2ad = v²
Substituting,
2(a)(180 m) = (22.2 m/s)²
From the equation above, we achieve the value of a = 1.369 m/s²

(b) The force necessary is calculated below,
F = (1,300 kg)(1.369 m/s²)
F = 1,779.7 N

(c) Coefficient of friction,
1,779.7 N = (1,300)(9.81)(x)
The value of x from the generated equation,
x = 0.14.