The position function is given as $s(t) = -4.9t^2 + 31t + 22$.

Part 1: Calculating the Average Velocities

The average velocity over a time interval $[t_1, t_2]$ is given by:

$\text{Average Velocity} = \frac{s(t_2) - s(t_1)}{t_2 - t_1}$

We will use this formula to calculate the average velocities for each time interval.

1. For the interval $[4, 5]$:

$\text{Average Velocity} = \frac{s(5) - s(4)}{5 - 4}$ First, calculate $s(5)$ and $s(4)$: $s(5) = -4.9(5^2) + 31(5) + 22 = -4.9(25) + 155 + 22 = -122.5 + 177 = 54.5$ $s(4) = -4.9(4^2) + 31(4) + 22 = -4.9(16) + 124 + 22 = -78.4 + 124 + 22 = 67.6$ $\text{Average Velocity} = \frac{54.5 - 67.6}{1} = -13.1$

2. For the interval $[4, 4.1]$:

$\text{Average Velocity} = \frac{s(4.1) - s(4)}{4.1 - 4}$ First, calculate $s(4.1)$: $s(4.1) = -4.9(4.1^2) + 31(4.1) + 22 = -4.9(16.81) + 127.1 + 22 = -82.369 + 127.1 + 22 = 66.731$ $\text{Average Velocity} = \frac{66.731 - 67.6}{0.1} = -8.69$

3. For the interval $[4, 4.01]$:

$\text{Average Velocity} = \frac{s(4.01) - s(4)}{4.01 - 4}$ First, calculate $s(4.01)$: $s(4.01) = -4.9(4.01^2) + 31(4.01) + 22 = -4.9(16.0801) + 124.31 + 22 = -78.79249 + 124.31 + 22 = 67.51751$ $\text{Average Velocity} = \frac{67.51751 - 67.6}{0.01} = -8.249$

4. For the interval $[4, 4.001]$:

$\text{Average Velocity} = \frac{s(4.001) - s(4)}{4.001 - 4}$ First, calculate $s(4.001)$: $s(4.001) = -4.9(4.001^2) + 31(4.001) + 22 = -4.9(16.008001) + 124.031 + 22 = -78.4392049 + 124.031 + 22 = 67.5917951$ $\text{Average Velocity} = \frac{67.5917951 - 67.6}{0.001} = -8.2049$

5. For the interval $[4, 4.0001]$:

$\text{Average Velocity} = \frac{s(4.0001) - s(4)}{4.0001 - 4}$ First, calculate $s(4.0001)$: $s(4.0001) = -4.9(4.0001^2) + 31(4.0001) + 22 = -4.9(16.00080001) + 124.0031 + 22 = -78.403920049 + 124.0031 + 22 = 67.599179951$ $\text{Average Velocity} = \frac{67.599179951 - 67.6}{0.0001} = -8.2005$

Table of Average Velocities:

Part 2: Instantaneous Velocity at $t = 4$

To estimate the instantaneous velocity, we observe the trend in the average velocities as the time intervals get smaller. The values approach $-8.2$, suggesting that the instantaneous velocity at $t = 4$ is approximately $-8.2$ m/s.

The instantaneous velocity at $t = 4$ is $-8.2$ m/s (rounded to the nearest tenth).

Would you like more details or have any further questions? Here are 5 related questions to think about:

1. How would the instantaneous velocity change if we altered the coefficients in the position function?
2. What is the physical significance of a negative instantaneous velocity?
3. Can you determine the velocity at other points, such as $t = 2$ or $t = 6$?
4. How would you find the acceleration at $t = 4$?
5. How is the concept of instantaneous velocity related to the derivative of the position function?

Tip: Instantaneous velocity is the derivative of the position function, and you can find it by taking the first derivative of $s(t)$ with respect to $t$.