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Speeding up?

  1. Nov 23, 2004 #1
    s(t)=t^3-3t

    which interval(s) of t is this object speeding up?

    s'(t)=3x^2-6
    0=3x^2-6
    x=sqrt(2) and

    ....+............-......-...........+
    inf...-sqrt(2).....0....sqrt(2) ...inf

    s''(t)=6x
    0=6x
    x=0

    ......-.......+.......
    inf.......0.........inf


    inc (-sqrt(2),0) V (sqrt(2),inf)

    correct?
     
  2. jcsd
  3. Nov 24, 2004 #2

    Galileo

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    Why did you change your t's to x's. :confused:

    The object is speeding up when s''(t) is positive.
    6t>0 iff t>0
     
  4. Nov 24, 2004 #3

    BobG

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    You made one mistake. Derivative of 3t is 3, not 6. That changes the interval where the velocity is positive or negative. Fortunately, that doesn't change your answer as to when acceleration is negative or positive (both 3 and 6 have a derivative of 0).

    What's the question actually asking. You asked when the object was speeding up (i.e. - acceleration was positive). Your final answer gave an incorrect answer as to when the object was moving forward (velocity positive).
     
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