Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A 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]
Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
