Acar is traveling at 53.0 mi/h on a horizontal highway. (a) if the coefficient of static friction between road and tires on a rainy day is 0.104, what is the minimum distance in which the car will stop?

902 ft

Explanation:

First convert mi/h to ft/s.

53.0 mi/h × (5280 ft / mi) × (1 h / 3600 s) = 77.7 ft/s

Sum of the forces on the car in the y direction:

∑F = ma

N - W = 0

N = mg

Sum of the forces on the car in the x direction:

∑F = ma

-F = ma

-Nμ = ma

Substituting;

-mgμ = ma

-gμ = a

Acceleration is constant, so:

v² = v₀² + 2a(x - x₀)

(0 ft/s)² = (77.7 ft/s)² + 2(-32.2 ft/s² × 0.104)(x - 0)

x = 902 ft

The minimum stopping distance is 902 ft.

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