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9 November, 05:47

A ride-sharing car moving along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a speed of 28.0 m/s. Then the vehicle moves for 41.0 s at constant speed until the brakes are applied, stopping the vehicle in a uniform manner in an additional 5.00 s.

(a) How long is the ride-sharing car in motion (in s) ?

(b) What is the average velocity of the ride-sharing car for the motion described? (Enter the magnitude in m/s.)

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  1. 9 November, 06:46
    0
    Time taken to accelerate to 28 m / s

    = 28 / 2 = 14 s

    a) Total length of time in motion

    = 14 + 41 + 5

    = 60 s.

    b)

    Distance covered while accelerating

    s = ut + 1/2 at²

    = 0 +.5 x 2 x 14²

    = 196 m.

    Distance covered while moving in uniform motion

    = 28 x 41

    = 1148 m

    distance covered while decelerating

    v = u - at

    0 = 28 - a x 5

    a = 5.6 m / s²

    v² = u² - 2 a s

    0 = 28² - 2 x 5.6 x s

    s = 28² / 2 x 5.6

    = 70 m.

    Total distance covered

    = 196 + 1148 + 70

    = 1414 m

    total time taken = 60 s

    average velocity

    = 1414 / 60

    = 23.56 m / s.
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