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11 November, 12:42

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°?

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  1. 11 November, 16:09
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    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
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