At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
Sagot :
To fill in the table with the appropriate average velocities and then make a conjecture about the value of the instantaneous velocity at [tex]\( t=0 \)[/tex], we need to follow these steps:
1. Determine the position function: The position function given is [tex]\( s(t) = \frac{17}{t+1} \)[/tex].
2. Calculate the average velocity: The average velocity over a time interval [tex]\([t_1, t_2]\)[/tex] is given by the formula:
[tex]\[ v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Here, since we are always starting from [tex]\( t_1 = 0 \)[/tex], the formula simplifies to:
[tex]\[ v_{avg} = \frac{s(t) - s(0)}{t - 0} = \frac{s(t) - s(0)}{t} \][/tex]
3. Evaluate [tex]\( s(0) \)[/tex]:
[tex]\[ s(0) = \frac{17}{0+1} = 17 \][/tex]
4. Compute [tex]\( v_{avg} \)[/tex] for different [tex]\( t \)[/tex] values:
- For [tex]\( [0, 1] \)[/tex]:
[tex]\[ s(1) = \frac{17}{1+1} = \frac{17}{2} = 8.5 \][/tex]
[tex]\[ v_{avg} = \frac{s(1) - s(0)}{1 - 0} = \frac{8.5 - 17}{1} = -8.5 \][/tex]
- For [tex]\( [0, 0.5] \)[/tex]:
[tex]\[ s(0.5) = \frac{17}{0.5+1} = \frac{17}{1.5} \approx 11.333 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.5) - s(0)}{0.5 - 0} = \frac{11.333 - 17}{0.5} \approx -11.333 \][/tex]
- For [tex]\( [0, 0.1] \)[/tex]:
[tex]\[ s(0.1) = \frac{17}{0.1+1} \approx 15.4545 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.1) - s(0)}{0.1 - 0} = \frac{15.4545 - 17}{0.1} \approx -15.455 \][/tex]
- For [tex]\( [0, 0.01] \)[/tex]:
[tex]\[ s(0.01) = \frac{17}{0.01+1} \approx 16.8317 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.01) - s(0)}{0.01 - 0} = \frac{16.8317 - 17}{0.01} \approx -16.832 \][/tex]
- For [tex]\( [0, 0.001] \)[/tex]:
[tex]\[ s(0.001) = \frac{17}{0.001+1} \approx 16.983 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.001) - s(0)}{0.001 - 0} = \frac{16.983 - 17}{0.001} \approx -16.983 \][/tex]
5. Fill in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Time Interval} & [0,1] & [0,0.5] & [0,0.1] & [0,0.01] & [0,0.001] \\ \hline \text{Average Velocity} & -8.5 & -11.333 & -15.455 & -16.832 & -16.983 \\ \hline \end{array} \][/tex]
6. Conjecture about instantaneous velocity at [tex]\( t=0 \)[/tex]:
- As the time interval [tex]\([0, t]\)[/tex] gets smaller, the average velocity values get closer to approximately [tex]\(-17\)[/tex].
- Therefore, we can conjecture that the instantaneous velocity at [tex]\( t=0 \)[/tex] is:
[tex]\[ \lim_{t \to 0} \frac{17/(t+1) - 17}{t} \approx -17 \][/tex]
1. Determine the position function: The position function given is [tex]\( s(t) = \frac{17}{t+1} \)[/tex].
2. Calculate the average velocity: The average velocity over a time interval [tex]\([t_1, t_2]\)[/tex] is given by the formula:
[tex]\[ v_{avg} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Here, since we are always starting from [tex]\( t_1 = 0 \)[/tex], the formula simplifies to:
[tex]\[ v_{avg} = \frac{s(t) - s(0)}{t - 0} = \frac{s(t) - s(0)}{t} \][/tex]
3. Evaluate [tex]\( s(0) \)[/tex]:
[tex]\[ s(0) = \frac{17}{0+1} = 17 \][/tex]
4. Compute [tex]\( v_{avg} \)[/tex] for different [tex]\( t \)[/tex] values:
- For [tex]\( [0, 1] \)[/tex]:
[tex]\[ s(1) = \frac{17}{1+1} = \frac{17}{2} = 8.5 \][/tex]
[tex]\[ v_{avg} = \frac{s(1) - s(0)}{1 - 0} = \frac{8.5 - 17}{1} = -8.5 \][/tex]
- For [tex]\( [0, 0.5] \)[/tex]:
[tex]\[ s(0.5) = \frac{17}{0.5+1} = \frac{17}{1.5} \approx 11.333 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.5) - s(0)}{0.5 - 0} = \frac{11.333 - 17}{0.5} \approx -11.333 \][/tex]
- For [tex]\( [0, 0.1] \)[/tex]:
[tex]\[ s(0.1) = \frac{17}{0.1+1} \approx 15.4545 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.1) - s(0)}{0.1 - 0} = \frac{15.4545 - 17}{0.1} \approx -15.455 \][/tex]
- For [tex]\( [0, 0.01] \)[/tex]:
[tex]\[ s(0.01) = \frac{17}{0.01+1} \approx 16.8317 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.01) - s(0)}{0.01 - 0} = \frac{16.8317 - 17}{0.01} \approx -16.832 \][/tex]
- For [tex]\( [0, 0.001] \)[/tex]:
[tex]\[ s(0.001) = \frac{17}{0.001+1} \approx 16.983 \][/tex]
[tex]\[ v_{avg} = \frac{s(0.001) - s(0)}{0.001 - 0} = \frac{16.983 - 17}{0.001} \approx -16.983 \][/tex]
5. Fill in the table:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Time Interval} & [0,1] & [0,0.5] & [0,0.1] & [0,0.01] & [0,0.001] \\ \hline \text{Average Velocity} & -8.5 & -11.333 & -15.455 & -16.832 & -16.983 \\ \hline \end{array} \][/tex]
6. Conjecture about instantaneous velocity at [tex]\( t=0 \)[/tex]:
- As the time interval [tex]\([0, t]\)[/tex] gets smaller, the average velocity values get closer to approximately [tex]\(-17\)[/tex].
- Therefore, we can conjecture that the instantaneous velocity at [tex]\( t=0 \)[/tex] is:
[tex]\[ \lim_{t \to 0} \frac{17/(t+1) - 17}{t} \approx -17 \][/tex]
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
