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Motion of a Particle - Max Acceleration

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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?

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.

Re: Motion of a Particle - Max Acceleration

Quote:

Originally Posted by

**ebaines** 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.