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30 July, 21:54

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 31.0 m/s. Then the vehicle moves for 35.0 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?

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  1. 30 July, 22:20
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    A) total time = 55.5 seconds

    B) average velocity = 25.27 m/s

    Explanation:

    It starts from rest, so initial velocity, u = 0 m/s

    We are given;

    acceleration; a = 2 m/s²

    Final velocity; v = 31 m/s

    From Newton's first law of motion,

    v = u + at

    So, 31 = 0 + 2t

    t = 31/2

    t = 15.5 sec

    We are told that, after this time of 15.5 sec, the car travels 35 sec at a constant speed and after that it takes 5 sec additional time to stop. Thus;

    (a) Total time in which car is in motion = 15.5 + 35 + 5 = 55.5 seconds

    b) Total distance traveled during first 15.5 sec would be gotten from Newton's second equation of motion which is;

    S = ut + ½at²

    S1 = 0 + ½ (2 * 15.5²)

    S1 = 240.25 m

    Distance traveled in 35 sec with with velocity of 31 m/sec is;

    S2 = velocity x time

    S2 = 35 * 31 = 1085 m

    Now, for the final stage, final velocity (v) will now be 0 since the car comes to rest while initial velocity (u) will be 31 m/s.

    From the first equation of motion,

    a = (v - u) / t

    a = (0 - 31) / 5

    a = - 6.2 m/s²

    So, distance travelled is;

    S3 = ut + ½at²

    S3 = (31 * 5) + ½ (-6.2 * 5²)

    S3 = 155 - 77.5

    S3 = 77.5 m

    So overall total distance = S1 + S2 + S3

    Overall total distance = 240.25 + 1085 + 77.5 = 1402.75 m

    Average velocity = total distance/total time

    Average velocity = 1402.75/55.5 = 25.27 m/s
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