# Motion of a Particle - Max Acceleration

• Aug 21st 2013, 09:19 PM
Jason76
Motion of a Particle - Max Acceleration
Attachment 29044

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?
• Aug 22nd 2013, 03:44 AM
ebaines
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.
• Aug 26th 2013, 05:42 PM
Jason76
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.