Results 1 to 3 of 3

Thread: Motion of a Particle - Max Acceleration

  1. #1
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    Motion of a Particle - Max Acceleration

    Motion of a Particle - Max Acceleration-graph4.jpg

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

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

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

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

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

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

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

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

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

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

    In this case how to find the maximum?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,389
    Thanks
    420

    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.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor Jason76's Avatar
    Joined
    Oct 2012
    From
    USA
    Posts
    1,314
    Thanks
    21

    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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. particle motion
    Posted in the Calculus Forum
    Replies: 22
    Last Post: Dec 23rd 2017, 02:13 PM
  2. Particle Motion...
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 10th 2010, 09:04 PM
  3. Particle motion
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 25th 2009, 01:44 PM
  4. Replies: 5
    Last Post: Jun 1st 2009, 05:50 AM
  5. Particle Motion
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2009, 08:41 AM

Search Tags


/mathhelpforum @mathhelpforum