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Math Help - Rates of Change

  1. #1
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    Rates of Change

    1) s(t) = (t^3) - 8t + 1

    Find the displacement during the first 3 seconds.

    Do I find s(3) and s(1) and subtract the two? Or what?


    2) At what value or values of t does the particle change direction?

    Not sure where to start on this one
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  2. #2
    MHF Contributor
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    Quote Originally Posted by Dickson View Post
    1) s(t) = (t^3) - 8t + 1

    Find the displacement during the first 3 seconds.

    Do I find s(3) and s(1) and subtract the two?
    "The first three seconds" should, I believe, be from t = 0 to t = 3. But "the difference of the ending and starting positions" is the customary definition of "displacement". Check your book (or its worked examples) for specifics.

    Quote Originally Posted by Dickson View Post
    2) At what value or values of t does the particle change direction?
    Hint: If the velocity is positive, the particle is moving forward; if negative, then backward. And v(t) = s'(t).

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  3. #3
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    Hello, Dickson!

    1) s(t) \:=\: t^3 - 8t + 1

    Find the displacement during the first 3 seconds.

    Do I find s(3) and s(1) and subtract the two?
    Yes!


    2) At what value or values of t does the particle change direction?
    Big clue: In order to change the direction, the particle must come to a stop.

    The question becomes: When is the velocity equal to zero?

    Velocity is the derivative of the position function:
    . . . . v(t) \;=\;s'(t) \;=\;3t^2 - 8 \;=\;0

    We have: . t^2 \:=\:\tfrac{8}{3} \quad\Rightarrow\quad t \:=\:\pm\sqrt{\tfrac{8}{3}}\quad\Rightarrow\quad t \:=\:\pm\frac{2\sqrt{6}}{3}


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