How fast was a driver going if it left skid marks that were 63 feet long on wet concrete? (The coefficient of friction is 0.97) ?

Step-by-step explanation:

Let m be the mass of the vehicle

frictional force = μ mg

μ is coefficient of friction,

deceleration = μ g

=.97 x 32

a = - 31.04 ft / s

v² = u² - 2 a s

0 = u² - 2 a s

u² = 2 a s

= 2 x 31.04 x 63

= 3911.04

u = 62.54 ft / s