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Math Help - Calc Word Problem

  1. #1
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    Calc Word Problem

    A particle moves along a line. The particle's position,  s, in meters at t seconds is modelled by s(t) = 2t^{3} - 15t^2 + 36t + 40, where t \geq 0.

    e) Determine the total distance travelled during the first five seconds



    Found the first derivative, used interval chart, found distances for time found and tried to find the total distance for the first 5 seconds:

    0 = 6t^2 - 30t + 36
    0 = 6(t - 3)(t - 2)
    t = 2s and t = 3s

    (0, 2) \rightarrow advancing
    (2, 3) \rightarrow retreating
    (3, \infty) \rightarrow advancing

    s(0) = 40m
    s(2) = 68m
    s(3) = 67m
    s(5) = 95m

    s(0) = 40m
    s(3) - s(2) = 1m
    s(5) - s(3) = 28m

    40m + 1m + 28m = 69m


    Textbook Answer: 57m
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  2. #2
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    Quote Originally Posted by Macleef View Post
    A particle moves along a line. The particle's position,  s, in meters at t seconds is modelled by s(t) = 2t^{3} - 15t^2 + 36t + 40, where t \geq 0.

    e) Determine the total distance travelled during the first five seconds


    Found the first derivative, used interval chart, found distances for time found and tried to find the total distance for the first 5 seconds:

    0 = 6t^2 - 30t + 36
    0 = 6(t - 3)(t - 2)
    t = 2s and t = 3s

    (0, 2) \rightarrow advancing
    (2, 3) \rightarrow retreating
    (3, \infty) \rightarrow advancing

    s(0) = 40m
    s(2) = 68m
    s(3) = 67m
    s(5) = 95m

    s(0) = 40m
    s(3) - s(2) = 1m
    s(5) - s(3) = 28m

    40m + 1m + 28m = 69m


    Textbook Answer: 57m
    s(0)=40m is where you started
    s(2)=68m is where it turns around
    so the first displacement should be s(2)-s(0)=28

    Then you get 28+1+28=57m

    I hope this clears it up.
    Good luck.
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