Answer :
Let's consider the position function [tex]\( s(t) = 8 \sin(3t) \)[/tex], which describes a block bouncing vertically on a spring.
To find the average velocity over a given time interval [tex]\([t_0, t_1]\)[/tex], we use the formula:
[tex]\[ \text{Average Velocity} = \frac{s(t_1) - s(t_0)}{t_1 - t_0} \][/tex]
We need to calculate the average velocities over various time intervals and fill in the table. Let's list the intervals and then their corresponding average velocities.
The time intervals are:
1. [tex]\(\left[\frac{\pi}{2}, \pi\right]\)[/tex]
2. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\)[/tex]
3. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\)[/tex]
4. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\)[/tex]
5. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\)[/tex]
Using the given position function [tex]\( s(t) \)[/tex], let's calculate the average velocities for each interval:
- For the interval [tex]\( \left[\frac{\pi}{2}, \pi\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 5.092958178940653 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 3.5730808699515273 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.3599730008099652 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.03599997299997112 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.0035999999692397167 \][/tex]
We can now fill in the table with these results.
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Time \\ Interval \end{tabular} & \begin{tabular}{c} Average \\ Velocity \end{tabular} \\ \hline \(\left[\frac{\pi}{2}, \pi\right]\) & 5.092958178940653 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\) & 3.5730808699515273 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\) & 0.3599730008099652 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\) & 0.03599997299997112 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\) & 0.0035999999692397167 \\ \hline \end{tabular} \][/tex]
From the table, we observe that as the interval gets smaller and smaller, the average velocity approaches zero. Hence, we can make the conjecture that the instantaneous velocity at [tex]\( t = \frac{\pi}{2} \)[/tex] is approximately zero. This result is confirmed by the smallest average velocities in the list above, suggesting the instantaneous velocity value is [tex]\(-4.408728476930472 \times 10^{-15}\)[/tex], which is extremely close to zero.
To find the average velocity over a given time interval [tex]\([t_0, t_1]\)[/tex], we use the formula:
[tex]\[ \text{Average Velocity} = \frac{s(t_1) - s(t_0)}{t_1 - t_0} \][/tex]
We need to calculate the average velocities over various time intervals and fill in the table. Let's list the intervals and then their corresponding average velocities.
The time intervals are:
1. [tex]\(\left[\frac{\pi}{2}, \pi\right]\)[/tex]
2. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\)[/tex]
3. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\)[/tex]
4. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\)[/tex]
5. [tex]\(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\)[/tex]
Using the given position function [tex]\( s(t) \)[/tex], let's calculate the average velocities for each interval:
- For the interval [tex]\( \left[\frac{\pi}{2}, \pi\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 5.092958178940653 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 3.5730808699515273 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.3599730008099652 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.03599997299997112 \][/tex]
- For the interval [tex]\( \left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right] \)[/tex]:
[tex]\[ \text{Average Velocity} = 0.0035999999692397167 \][/tex]
We can now fill in the table with these results.
[tex]\[ \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Time \\ Interval \end{tabular} & \begin{tabular}{c} Average \\ Velocity \end{tabular} \\ \hline \(\left[\frac{\pi}{2}, \pi\right]\) & 5.092958178940653 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.1\right]\) & 3.5730808699515273 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.01\right]\) & 0.3599730008099652 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.001\right]\) & 0.03599997299997112 \\ \hline \(\left[\frac{\pi}{2}, \frac{\pi}{2} + 0.0001\right]\) & 0.0035999999692397167 \\ \hline \end{tabular} \][/tex]
From the table, we observe that as the interval gets smaller and smaller, the average velocity approaches zero. Hence, we can make the conjecture that the instantaneous velocity at [tex]\( t = \frac{\pi}{2} \)[/tex] is approximately zero. This result is confirmed by the smallest average velocities in the list above, suggesting the instantaneous velocity value is [tex]\(-4.408728476930472 \times 10^{-15}\)[/tex], which is extremely close to zero.
