A chevrolet corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. what is its stopping distance on a roadway sloping downward at an angle of 17.0°?

By definition we have that the final speed is:

Vf² = Vo² + 2 * a * d

Where,

Vo: Final speed

a: acceleration

d: distance.

We cleared this expression the acceleration:

a = (Vf²-Vo²) / (2 * d)

Substituting the values:

a = ((0) ^ 2 - (60) ^ 2) / ((2) * (123) * (1/5280))

a = - 77268 mi / h ^ 2

its stopping distance on a roadway sloping downward at an angle of 17.0 ° is:

First you must make a free body diagram and see the acceleration of the car:

g = 32.2 feet / sec ^ 2

a = - 77268 (mi / h ^ 2) * (5280/1) (feet / mi) * (1/3600) ^ 2 (h / s) ^ 2

a = - 31.48 feet / sec ^ 2

A = a + g * sin (θ) = - 31.48 + 32.2 * sin17.0

A = - 22.07 feet / sec ^ 2

Clearing the braking distance:

Vf² = Vo² + 2 * a * d

d = (Vf²-Vo²) / (2 * a)

Substituting the values:

d = ((0) ^ 2 - (60 * (5280/3600)) ^ 2) / (2 * ( - 22.07))

d = 175.44 feet

answer:

its stopping distance on a roadway sloping downward at an angle of 17.0 ° is 175.44 feet