Question

question the height in feet of a rocket after t seconds is modeled by the function s(t) = 3t^3 - 2t. find the average velocity of the rocket over the following time intervals: 2, 2.1, 2, 2.01, 2, 2.001, and 2, 2.0001. use this to draw a conclusion about the instantaneous velocity of the rocket at t = 2 seconds. (round each answer to answer to 6 decimal places.) provide your answer below.

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula over the interval $[a,b]$ is $v_{avg}=\frac{s(b)-s(a)}{b - a}$, where $s(t)=3t^{2}-2t$.

Step2: For the interval $[2,2.1]$

First, find $s(2)$ and $s(2.1)$.
\[s(2)=3\times(2)^{2}-2\times2=3\times4 - 4=12 - 4 = 8\]
\[s(2.1)=3\times(2.1)^{2}-2\times2.1=3\times4.41-4.2 = 13.23-4.2=9.03\]
Then, $v_{avg}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{9.03 - 8}{0.1}=\frac{1.03}{0.1}=10.300000$.

Step3: For the interval $[2,2.01]$

Find $s(2.01)$.
\[s(2.01)=3\times(2.01)^{2}-2\times2.01=3\times4.0401-4.02=12.1203 - 4.02 = 8.1003\]
Then, $v_{avg}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{8.1003 - 8}{0.01}=\frac{0.1003}{0.01}=10.030000$.

Step4: For the interval $[2,2.001]$

Find $s(2.001)$.
\[s(2.001)=3\times(2.001)^{2}-2\times2.001=3\times4.004001-4.002=12.012003-4.002 = 8.010003\]
Then, $v_{avg}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{8.010003 - 8}{0.001}=\frac{0.010003}{0.001}=10.003000$.

Step5: For the interval $[2,2.0001]$

Find $s(2.0001)$.
\[s(2.0001)=3\times(2.0001)^{2}-2\times2.0001=3\times4.00040001-4.0002=12.00120003-4.0002 = 8.00100003\]
Then, $v_{avg}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{8.00100003 - 8}{0.0001}=\frac{0.00100003}{0.0001}=10.000300$.

As the intervals get smaller and closer to $t = 2$, the average velocity approaches $10$ feet per second. We can conclude that the instantaneous velocity of the rocket at $t = 2$ seconds is approximately $10.000000$ feet per second.

Answer:

For $[2,2.1]$: $10.300000$
For $[2,2.01]$: $10.030000$
For $[2,2.001]$: $10.003000$
For $[2,2.0001]$: $10.000300$