Alias robert

a car starting from rest is accelerated at 2m/s(square) for 10s, then moves at constant velocity for 20s and then decelerates at 1m/s(square), finally stops. how far does it travel during its trip? please help me! I need in solution for that problem

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  1. a = 2
    v = 2t + c, where t is in seconds
    when t=0, v = 0 , (it started from rest)
    0 = 0 + c, c = 0

    v = 2t

    s = t^2 + k
    when t = 0, s = 0 , (hasn't gone anywhere)

    after 10 seconds
    v = 20 m/s
    s = 100 m

    now goes for 20 s, at 20 m/s, covering 400 m
    so it can covered 500 m, and the speed is 20 m/s

    now repeat the process: starting again with a time of t=0, v = 20, s=0 (remembering we have already gone 500 m)
    a = -1
    v = -t + p
    when t = 0, v = 20
    20 = 0 + p
    so v = -t + 20
    s = (-1/2)t^2 + 20t + q
    q = 0
    s = (-1/2)t^2 + 20t

    when it came to a stop, v = 0
    0 = -t + 20
    t = 20, so it took 20 seconds to stop
    in those 20 seconds, the car went another
    (-1/2)(20)^2 + 20(20) m
    = 200 m

    so the whole "trip" was 700 m

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    posted by Reiny

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