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

highway.

The acceleration of gravity is 9.81 m/s2 .

AIf 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 ...?

Verified Answer :

MissPhiladelphia

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

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