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QUESTION IMAGE

the height in feet of a rocket after t seconds is modeled by the functi…

Question

the height in feet of a rocket after t seconds is modeled by the function s(t)=3t² - 2. 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: t v_ave 2.1 10.3 2.01 10.03 2.001 10.003 2.0001 10.0003 v_ave≈

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{ave}$ over the interval $[a,b]$ for a position - function $s(t)$ is given by $v_{ave}=\frac{s(b)-s(a)}{b - a}$.

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

First, find $s(2)$ and $s(2.1)$.
\[s(2)=3\times(2)^{2}-2=3\times4 - 2=12 - 2 = 10\]
\[s(2.1)=3\times(2.1)^{2}-2=3\times4.41-2 = 13.23-2=11.23\]
\[v_{ave}=\frac{s(2.1)-s(2)}{2.1 - 2}=\frac{11.23 - 10}{0.1}=\frac{1.23}{0.1}=12.3\] (There seems to be an error in the provided answer for $t = 2.1$ in the image).

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

\[s(2.01)=3\times(2.01)^{2}-2=3\times4.0401-2=12.1203 - 2 = 10.1203\]
\[v_{ave}=\frac{s(2.01)-s(2)}{2.01 - 2}=\frac{10.1203 - 10}{0.01}=\frac{0.1203}{0.01}=12.03\]

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

\[s(2.001)=3\times(2.001)^{2}-2=3\times4.004001-2=12.012003 - 2 = 10.012003\]
\[v_{ave}=\frac{s(2.001)-s(2)}{2.001 - 2}=\frac{10.012003 - 10}{0.001}=\frac{0.012003}{0.001}=12.003\]

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

\[s(2.0001)=3\times(2.0001)^{2}-2=3\times4.00040001-2=12.00120003 - 2 = 10.00120003\]
\[v_{ave}=\frac{s(2.0001)-s(2)}{2.0001 - 2}=\frac{10.00120003 - 10}{0.0001}=\frac{0.00120003}{0.0001}=12.0003\]

Step6: Conclusion about instantaneous velocity

As the time - interval $[2,b]$ gets smaller and smaller (i.e., $b$ approaches $2$), the average velocity $v_{ave}$ approaches the instantaneous velocity at $t = 2$. The instantaneous velocity of the rocket at $t = 2$ seconds is $12$.

Answer:

The average velocities for the intervals $[2,2.1]$, $[2,2.01]$, $[2,2.001]$, $[2,2.0001]$ are approximately $12.3$, $12.03$, $12.003$, $12.0003$ respectively. The instantaneous velocity at $t = 2$ seconds is $12$.