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Math Help - Motion of a Particle - Max Acceleration

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    Motion of a Particle - Max Acceleration

    Motion of a Particle - Max Acceleration-graph4.jpg

    Given: v(t) = t^{3} - 3t^{2} + 12t + 8

    \dfrac{dv}{dt} = a(t) = 3t^{2} - 6t + 12

    What is the max acceleration on the interval: 0 \le t \le 3? - Answer = 21

    Logic: When the graph has a maximum value, then it's derivative will be 0. So y' of a(t) = 0 at maximum points. You simply find which of these points lies within the given interval.

    So, \dfrac{da}{dt} = b(t) = 6t - 6 - Note  b(t) is random made up term.

    b(t) = 6t - 6 = 0 - Solve for t. See which t values like within 0 \le t \le 3

    t = 1 and 9 would be our acceleration value after plugging into a(t)

    Now test for max or min using 2nd derivative test:

    \dfrac{d^{2}a}{dt} = c(t) = 6

    That is positive so our answer of t = 1 is a minimum and the graph is concave up.

    In this case how to find the maximum?
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    MHF Contributor ebaines's Avatar
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    Re: Motion of a Particle - Max Acceleration

    Since you found the minimum, and given that a(t) is a parabola, you know that the function increases as you move away from t=1. So simply check the boundary points of t=0 and t=3.
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    Re: Motion of a Particle - Max Acceleration

    Quote Originally Posted by ebaines View Post
    Since you found the minimum, and given that a(t) is a parabola, you know that the function increases as you move away from t=1. So simply check the boundary points of t=0 and t=3.
    Thanks, makes sense.
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