Question

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°?

Answer 1

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

Answer 2

Answer:

downward an angle of 17.0 ° is 175.44 feet

Hope this helps.

Explanation: