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13 votes
13 votes
How fast was a driver going if the car left skid marks that were 48 feet long on dry concrete? (The coefficient of friction is 1.02)?

A) 31 mph
B) 38 mph
C) 49 mph
D) 73 mph

User Viktor Sec
by
2.4k points

1 Answer

8 votes
8 votes

Answer:

B) 38 mph

Explanation:

s = √(30df)

s is the speed the car was traveling (in mph)

d is the distance the car skidded (in feet)

f is the coefficient of friction

s = √(30 x 48 x 1.02)

s = 38.32492661... = 38

User Jeff Saltfist
by
2.8k points