What's the 1) the acceleration 2) the distance travelled by the car before it comes to rest 3) what is the value of the friction while the car is skidding?

Question

A car of mass 1200kg is travelling along a straight horizontal road at a speed of 20m/s when it brakes sharply then skids. Friction brings the car to rest. The coefficient of friction between the tires and road is 0.8

Answer 1

m = 1200 kg; Weight = mg = 1200(9.8) = 11760 N = normal force (FN) because road is level

vo = 20 m/s; vf = 0

Friction force (Ff) = ?

µ = 0.8 = Ff/FN

3) Ff = 0.8FN = (0.8)(11760) = 9408 N

1) a = Fnet/m and here Fnet = Ff = 9408 so a = 9408/1200 = 7.84 m/s/s and it is negative because the car is slowing down

2) Use v²f = v²o + 2ad to find distance of the skid;

0 = 20² + 2(-7.84)d; d = (-400)/ (-15.68) = 25.5 m