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23 December, 17:37

Given the position function, s of t equals t cubed divided by 3 minus 15 times t squared divided by 2 plus 50 times t, between t = 0 and t = 12, where s is given in feet and t is measured in seconds, find the interval in seconds where the particle is moving to the left.

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  1. 23 December, 21:19
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    s (t) = t^3/3 - 15t^2 / 2 + 50t between t = 0 and t = 12

    s' (t) = velocity = t^2 - 15t + 50

    The particle is moving to the left when the velocity is negative.

    factoring the velocity function we have

    v = (t - 10) (t - 5) so the velocity is 0 at t = 5 and t = 10 The graph of this is a parabola with a minimum between t = 5 and 10 (minimum because of the postive coefficient of t^2) so the require interval when the velocity is negative is between t = 5 and 10.

    Answer is its moving to the left for 5 seconds between t = 5 and t = 10 secs.
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