Acar 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=1/2mv22−1/2mv21
d=(1/2mv22−1/2mv21)/−F

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

d=(1/2mv22−1/2mv21)/−(k∗mg)d=(−1/2v21)/−(k∗g)

d=(−1/2×(14.6m/s)2)/−(0.137×9.8m/s)

d = 78.9m

Yup no bull frogs here man