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How fast was a driver going if the car left skid marks that were 80 feet long on dry concrete? (The coefficient of friction is 1.02)?A)73 mphB)31 mphC)38 mphD)49 mph
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How fast was a driver going if the car left skid marks that were 80 feet long on dry concrete? (The coefficient of friction is 1.02)?A)73 mphB)31 mphC)38 mphD)49 mph

User Blackhole
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1 Answer

4 votes

In order to solve this question, we can use the formula


v=√(gdf)

Where g is gravity, d is distance, and f is frictional coefficient.

g = 30 ft/s^2

d = 80 feet

f = 1.02

v = sqrt(30 * 80 * 1.02) = 49.47 mph

The answer is D

User Mohsin Aljiwala
by
4.9k points
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