A1300-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?

(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.