Posted by **jay** on Friday, December 24, 2010 at 8:54pm.

A car is traveling at 55.1 km/h on a ﬂat

highway.

The acceleration of gravity is 9.81 m/s

2. a) If the coeﬃcient of kinetic friction between the road and the tires on a rainy day is

0.116, what is the minimum distance needed

for the car to stop

b) What is the stopping distance when the

surface is dry and the coeﬃcient of kinetic

friction is 0.698

- physics -
**drwls**, Friday, December 24, 2010 at 9:50pm
First of all, convert 55.1 km/h to 15.3 m/s

2a) The stopping distance X is given by this equation that relates the kinital kinetic energy to the work done against friction:

(1/2)M V^2 = M*g*Uk*X

Notice that the mass M cancels out, which is good since they did not tell you the mass.

X = V^2/(2*g*Uk) = 103 m

2b) Use the same formula, but with the different value of Uk (the kinetic friction corefficient). The stopping distance is much less.

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