A car is traveling at 42.0 km/h on a flat highway. (a) If the coefficient of friction between road and tires on a rainy day is 0.107, what is the minimum distance in which the car will stop?

64.85 m

Explanation:

a)

μ = Coefficient of friction between road and tire on rainy day = 0.107

g = acceleration due to gravity = 9.8 m/s²

a = acceleration experienced by car due to friction = - μg = - (0.107) (9.8) = - 1.05 m/s²

v₀ = initial velocity of the car = 42 km/h = 11.67 m/s

v = final speed of the car = 0 m/s

d = minimum distance traveled before stopping

Using the equation

v² = v₀² + 2 a d

0² = 11.67² + 2 ( - 1.05) d

d = 64.85 m