Acar 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

Action forces is as a result of mass of your body times the acceleration of your motion.

Force = Mass × Acceleration (Newton's second law of motion)

Reaction Forces are as a result of friction force between you and the ground plus air, acting opposite to the direction you are heading. This is described in Newton's third law of motion which states that:

For every action, there must be an opposite and equal reaction.