Answer :
Let's start by calculating the average velocities over each of the given time intervals for the position function [tex]\(s(t) = -16 t^2 + 114 t\)[/tex].
For each time interval [tex]\([t_1, t_2]\)[/tex], the average velocity is given by:
[tex]\[ \text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
### Interval: [tex]\([1, 2]\)[/tex]
1. [tex]\(s(2) = -16(2)^2 + 114(2) = -64 + 228 = 164\)[/tex]
2. [tex]\(s(1) = -16(1)^2 + 114(1) = -16 + 114 = 98\)[/tex]
[tex]\[ \text{Average Velocity} = \frac{164 - 98}{2 - 1} = \frac{66}{1} = 66 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.5]\)[/tex]
1. [tex]\(s(1.5) = -16(1.5)^2 + 114(1.5) = -36 + 171 = 135\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{135 - 98}{1.5 - 1} = \frac{37}{0.5} = 74 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.1]\)[/tex]
1. [tex]\(s(1.1) = -16(1.1)^2 + 114(1.1) = -16(1.21) + 125.4 = -19.36 + 125.4 = 106.04\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{106.04 - 98}{1.1 - 1} = \frac{8.04}{0.1} = 80.4 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.01]\)[/tex]
1. [tex]\(s(1.01) = -16(1.01)^2 + 114(1.01) = -16(1.0201) + 115.14 = -16.3216 + 115.14 = 98.8184\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{98.8184 - 98}{1.01 - 1} = \frac{0.8184}{0.01} = 81.84 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.001]\)[/tex]
1. [tex]\(s(1.001) = -16(1.001)^2 + 114(1.001) = -16(1.002001) + 114.114 = -16.032016 + 114.114 = 98.081984\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{98.081984 - 98}{1.001 - 1} = \frac{0.081984}{0.001} = 81.984 \text{ ft/s} \][/tex]
Now we'll construct the table with the calculated values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Time Interval} & [1, 2] & [1, 1.5] & [1, 1.1] & [1, 1.01] & [1, 1.001] \\ \hline \text{Average Velocity} & 66 \text{ ft/s} & 74 \text{ ft/s} & 80.4 \text{ ft/s} & 81.84 \text{ ft/s} & 81.984 \text{ ft/s} \\ \hline \end{array} \][/tex]
To make a conjecture about the instantaneous velocity at [tex]\(t = 1\)[/tex], we observe how the average velocity gets closer to a specific value as the interval gets smaller. As the values approach, they seem to converge to approximately 82 ft/s.
Thus, the instantaneous velocity at [tex]\(t = 1\)[/tex] can be conjectured to be:
[tex]\[ \text{Instantaneous Velocity} = 82 \text{ ft/s} \][/tex]
For each time interval [tex]\([t_1, t_2]\)[/tex], the average velocity is given by:
[tex]\[ \text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
### Interval: [tex]\([1, 2]\)[/tex]
1. [tex]\(s(2) = -16(2)^2 + 114(2) = -64 + 228 = 164\)[/tex]
2. [tex]\(s(1) = -16(1)^2 + 114(1) = -16 + 114 = 98\)[/tex]
[tex]\[ \text{Average Velocity} = \frac{164 - 98}{2 - 1} = \frac{66}{1} = 66 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.5]\)[/tex]
1. [tex]\(s(1.5) = -16(1.5)^2 + 114(1.5) = -36 + 171 = 135\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{135 - 98}{1.5 - 1} = \frac{37}{0.5} = 74 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.1]\)[/tex]
1. [tex]\(s(1.1) = -16(1.1)^2 + 114(1.1) = -16(1.21) + 125.4 = -19.36 + 125.4 = 106.04\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{106.04 - 98}{1.1 - 1} = \frac{8.04}{0.1} = 80.4 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.01]\)[/tex]
1. [tex]\(s(1.01) = -16(1.01)^2 + 114(1.01) = -16(1.0201) + 115.14 = -16.3216 + 115.14 = 98.8184\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{98.8184 - 98}{1.01 - 1} = \frac{0.8184}{0.01} = 81.84 \text{ ft/s} \][/tex]
### Interval: [tex]\([1, 1.001]\)[/tex]
1. [tex]\(s(1.001) = -16(1.001)^2 + 114(1.001) = -16(1.002001) + 114.114 = -16.032016 + 114.114 = 98.081984\)[/tex]
2. [tex]\(s(1) = 98\)[/tex] (as calculated before)
[tex]\[ \text{Average Velocity} = \frac{98.081984 - 98}{1.001 - 1} = \frac{0.081984}{0.001} = 81.984 \text{ ft/s} \][/tex]
Now we'll construct the table with the calculated values:
[tex]\[ \begin{array}{|c|c|c|c|c|c|} \hline \text{Time Interval} & [1, 2] & [1, 1.5] & [1, 1.1] & [1, 1.01] & [1, 1.001] \\ \hline \text{Average Velocity} & 66 \text{ ft/s} & 74 \text{ ft/s} & 80.4 \text{ ft/s} & 81.84 \text{ ft/s} & 81.984 \text{ ft/s} \\ \hline \end{array} \][/tex]
To make a conjecture about the instantaneous velocity at [tex]\(t = 1\)[/tex], we observe how the average velocity gets closer to a specific value as the interval gets smaller. As the values approach, they seem to converge to approximately 82 ft/s.
Thus, the instantaneous velocity at [tex]\(t = 1\)[/tex] can be conjectured to be:
[tex]\[ \text{Instantaneous Velocity} = 82 \text{ ft/s} \][/tex]
