A car is traveling at 52.4 km/h on a flat

highway.

The acceleration of gravity is 9.81 m/s2.

a) If the coefficient of kinetic friction be-

tween the road and the tires on a rainy day is

0.137, what is the minimum distance needed

for the car to stop?

Answer in units of m ... ?

Work is equal to a force over a distance and is equal to the change in kinetic energy, so

- Fd=ΔKE

-Fd=1/2m v22 - 1/2m v21

d = (1/2m v22 - 1/2m v21) / - F

We know that Ffric=kFnatural and Fnatural=mg so:

d = (1/2m v22 - 1/2m v21) / - (k∗mg) d = (-1/2 v21) / - (k∗g)

d = (-1/2 * (14.6m/s) 2) / - (0.137*9.8m/s)

d = 78.9m