Question

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

Answer 1

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