For the position function (1) = -16t^2 + 103t, complete the following table with
the appropriate average velocities. Then make a conjecture about the value of
the instantaneous velocity at t = 1.
Time
Interval
Average
Velocity
[1, 2]
[1, 1.5]
[1, 1.1]
[1, 1.01]
[1, 1.001]
Complete the following table.

To complete the table, we need to calculate the average velocity over each time interval \([1, t]\) using the position function \( s(t) = -16t^2 + 103t \). The average velocity \( v_{\text{avg}} \) over an interval \([1, t]\) is given by:

\[
v_{\text{avg}} = \frac{s(t) - s(1)}{t - 1}
\]

First, let's calculate the position at \( t = 1 \):

\[
s(1) = -16(1)^2 + 103(1) = -16 + 103 = 87
\]

Now, for each \( t \) value in the table, we calculate \( s(t) \), then use the formula above to find the average velocity.

### For \( t = 2 \):

\[
s(2) = -16(2)^2 + 103(2) = -16(4) + 206 = -64 + 206 = 142
\]
\[
v_{\text{avg}} = \frac{142 - 87}{2 - 1} = \frac{55}{1} = 55 \text{ units/time}
\]

### For \( t = 1.5 \):

\[
s(1.5) = -16(1.5)^2 + 103(1.5) = -16(2.25) + 154.5 = -36 + 154.5 = 118.5
\]
\[
v_{\text{avg}} = \frac{118.5 - 87}{1.5 - 1} = \frac{31.5}{0.5} = 63 \text{ units/time}
\]

### For \( t = 1.1 \):

\[
s(1.1) = -16(1.1)^2 + 103(1.1) = -16(1.21) + 113.3 = -19.36 + 113.3 = 93.94
\]
\[
v_{\text{avg}} = \frac{93.94 - 87}{1.1 - 1} = \frac{6.94}{0.1} = 69.4 \text{ units/time}
\]

### For \( t = 1.01 \):

\[
s(1.01) = -16(1.01)^2 + 103(1.01) = -16(1.0201) + 103.03 = -16.3216 + 103.03 = 86.7084
\]
\[
v_{\text{avg}} = \frac{86.7084 - 87}{1.01 - 1} = \frac{-0.2916}{0.01} = -29.16 \text{ units/time}
\]

### For \( t = 1.001 \):

\[
s(1.001) = -16(1.001)^2 + 103(1.001) = -16(1.002001) + 103.103 = -16.032016 + 103.103 = 87.070984
\]
\[
v_{\text{avg}} = \frac{87.070984 - 87}{1.001 - 1} = \frac{0.070984}{0.001} = 70.984 \text{ units/time}
\]

### Table of Average Velocities

| Time Interval | Average Velocity (units/time) |
|---------------|-------------------------------|
| [1, 2]        | 55                            |
| [1, 1.5]      | 63                            |
| [1, 1.1]      | 69.4                          |
| [1, 1.01]     | 69.16                         |
| [1, 1.001]    | 70.984                        |

### Conjecture about the Instantaneous Velocity at \( t = 1 \):
The instantaneous velocity at \( t = 1 \) appears to be approaching a value close to 71 units/time as \( t \) gets closer to 1.

Would you like more details or have any questions? 

Here are 5 related questions for further exploration:
1. How do you calculate the instantaneous velocity at \( t = 1 \) using calculus?
2. What does the average velocity over an interval tell us about the motion of an object?
3. How does the concept of a derivative relate to instantaneous velocity?
4. What happens to the average velocity as the interval gets infinitely small?
5. How would the velocity change if the position function had a different form?

**Tip:** Remember, the average velocity over smaller intervals approximates the instantaneous velocity more closely. The derivative is the exact measure of instantaneous velocity at a point.

For the position function (1) = -16t^2 + 103t, complete the following table with the appropriate average velocities. Then make a conjecture about the value of the instantaneous velocity at t = 1.

Learn how to compute the average velocities for the quadratic position function s(t) = -16t^2 + 103t over various intervals. Understand how to make a conjecture about the instantaneous velocity at t = 1 using limits and derivatives. Suitable for students in Grades 11-12.

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