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Calculus AB Problem

  1. Feb 17, 2015 #1
    1. The problem statement, all variables and given/known data
    The position of a particle moving along a line is given by s(t) = 2t^3 -24t^2 + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing?
    (a) 3 < t < 4 only
    (b) t > 4 only
    (c) t > 5 only
    (d) 0 < t < 3 and t > 5
    (e) 3 < t < 4 and t > 5
    2. Relevant equations
    d/dx [k*x^n] = kn*x(n-1) power rule
    d/dx [f(x) +- g(x)] = f'(x) +- g'(x)
    3. The attempt at a solution
    s(t) = 2t^3 -24t^2 + 90t + 7
    find the derivative:
    s'(t) = 6t^2 - 48t + 90
    find the second derivative:
    s''(t) = 12t - 48

    Since the problem is asking for acceleration "For what values of t is the speed of the particle increasing?", we find the point of inflection and find the intervals.
    s''(t) = 0
    12t - 48 = 0
    12(t - 4) = 0
    t=4
    Intervals: (0,4] and [4,infinity)
    We plug in a number for each interval:
    s''(1) = 12(1) -48
    =12 - 48
    =-36

    s''(5) = 12(5) - 48
    =60 - 48
    =12

    The values of t>4 are when the speed is increasing.

    The problem is that the solution to this problem is (e) 3 < t < 4 and t > 5 and I don't know why?
     
  2. jcsd
  3. Feb 17, 2015 #2

    DEvens

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    Hmm... Do you suppose the author of the question is thinking "magnitude of speed" rather than just speed?

    A positive speed getting smaller is not speed getting bigger. This is certain. But is a negative speed getting more negative, is the speed increasing?
     
  4. Feb 17, 2015 #3
    Is that what your teacher had in mind when he/she asked that kind of problem?
     
  5. Feb 17, 2015 #4

    DEvens

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    Oh, no. My physics teachers were perfect in every way. :wink:
     
  6. Feb 17, 2015 #5
    How about your calculus teacher? Did he/she ask that kind of problem?
     
  7. Feb 17, 2015 #6

    Ray Vickson

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    Standard nomenclature is that speed = absolute value of velocity. Your ##s'(t)## is velocity ##v(t)##, not speed. If you plot the speed ##|s'(t)|## you will see how the book's answer arises.

    If you want to do it without plotting, you need to be careful about when speed is increasing; this occurs if either (a) ##v(t) < 0## is decreasing, or (b) ##v(t) > 0## is increasing. Do you see why?
     
  8. Feb 17, 2015 #7
    So speed is different in that increasing speed depends only on the magnitude of the increase of speed, and not dependent on direction?
     
  9. Feb 17, 2015 #8

    Ray Vickson

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    No: increasing speed means increasing magnitude of velocity. If velocity is < 0 and is becoming < 0 at an even greater rate, speed increases; if velocity is > 0 and is becoming > 0 at an even greater rate, speed increases. Just think about velocity being > 0 if we are driving East and being < 0 if we are driving West, and think about what increasing speed means in both cases.
     
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