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 $e$?
How do I find the average acceleration and instantaneous acceleration using derivatives and limits?
- $\begingroup$ Instantaneous velocity is $ds/dt=6t^2$. Instantaneous acceleration is $d^2s/dt^2=12t$. The acceleration at time $t=0$ is $d^2s/dt^2(0)=0$ and at time $t=3$ is $d^2s/dt^2(3)=36$. Therefore, the mean acceleration is $(36-0)/(3-0)=12$. $\endgroup$ Oct 25, 2019 at 0:27
- $\begingroup$ Check your work: in part b) you decided that the average velocity is 18 ly/second, but in part a) you say that while traveling at this average velocity for 3 seconds the saucer travels 64 ly. $\endgroup$– David KOct 25, 2019 at 1:09
- $\begingroup$ yes, I corrected that and I got 54 ly/ second for part a $\endgroup$ Oct 25, 2019 at 1:10
1 Answer
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:
$$a = 12t = 12(3) = 36$$
- $\begingroup$ Can you confirm that a, b, and c are correct? $\endgroup$ Oct 25, 2019 at 0:37
- $\begingroup$ (a) is wrong. The equation is for distance from, but the question is about distance travelled. To find that, subtract initial from final. (b) looks correct. (c) looks correct. $\endgroup$– johnnybOct 25, 2019 at 0:45
- $\begingroup$ So it would actually be 54? $\endgroup$ Oct 25, 2019 at 0:49