Answer :
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]
