How do I find the average acceleration and instantaneous acceleration using derivatives and limits?
I have completed parts $a$, $b$, and $c$ of the attached problem. However, I am having trouble with parts $d$ and $e$. Can anyone check to see if I am on the right track and how do I solve for parts $d$ and $e$?
1 Answer
$\begingroup$You are given the equation for position ($s = 2t^3 + 10$) and need to get equations for velocity and acceleration. Velocity is the derivative of position, which it looks like you found ($v = 6t^2$). The equation for acceleration is just another derivative ($a = 12t$).
Average acceleration is total change in velocity divided by total change in time. So for average acceleration, use the start time (0) and the end time (3). So you would evaluate the velocity equation at both points.
$$\frac{6(3)^2 - 6(0)^2}{3 - 0} = \frac{6\cdot 9 - 0}{3} = \frac{54}{3} = 18$$
For instantaneous acceleration, use the second derivative:
$$a = 12t = 12(3) = 36$$
$\endgroup$ 3
