https://file.aiquickdraw.com/mathbot/file/17286025805241xs0ip6i.jpg

To solve the problem, we need to calculate the average velocity of the arrow over the given time intervals. The height function is given by:

\[
h(t) = 31t - 0.83t^2
\]

### Average velocity formula:
The average velocity \( v_{\text{avg}} \) over an interval \([t_1, t_2]\) is:

\[
v_{\text{avg}} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}
\]

We will apply this formula to the five time intervals provided.

#### 1. Interval [4, 5]:
\[
v_{\text{avg}} = \frac{h(5) - h(4)}{5 - 4}
\]

Let's compute the values of \( h(5) \) and \( h(4) \):
\[
h(5) = 31(5) - 0.83(5^2) = 155 - 20.75 = 134.25 \, \text{m}
\]
\[
h(4) = 31(4) - 0.83(4^2) = 124 - 13.28 = 110.72 \, \text{m}
\]

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{134.25 - 110.72}{5 - 4} = \frac{23.53}{1} = 23.53 \, \text{m/s}
\]

#### 2. Interval [4, 4.5]:
\[
v_{\text{avg}} = \frac{h(4.5) - h(4)}{4.5 - 4}
\]
Let's calculate \( h(4.5) \):
\[
h(4.5) = 31(4.5) - 0.83(4.5^2) = 139.5 - 16.798125 = 122.701875 \, \text{m}
\]

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{122.701875 - 110.72}{4.5 - 4} = \frac{11.981875}{0.5} = 23.96 \, \text{m/s}
\]

#### 3. Interval [4, 4.1]:
\[
v_{\text{avg}} = \frac{h(4.1) - h(4)}{4.1 - 4}
\]
Let's calculate \( h(4.1) \):
\[
h(4.1) = 31(4.1) - 0.83(4.1^2) = 127.1 - 14.00643 = 113.09357 \, \text{m}
\]

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{113.09357 - 110.72}{4.1 - 4} = \frac{2.37357}{0.1} = 23.74 \, \text{m/s}
\]

#### 4. Interval [4, 4.01]:
\[
v_{\text{avg}} = \frac{h(4.01) - h(4)}{4.01 - 4}
\]
Let's calculate \( h(4.01) \):
\[
h(4.01) = 31(4.01) - 0.83(4.01^2) = 124.31 - 13.315683 = 110.994317 \, \text{m}
\]

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{110.994317 - 110.72}{4.01 - 4} = \frac{0.274317}{0.01} = 27.43 \, \text{m/s}
\]

#### 5. Interval [4, 4.001]:
\[
v_{\text{avg}} = \frac{h(4.001) - h(4)}{4.001 - 4}
\]
Let's calculate \( h(4.001) \):
\[
h(4.001) = 31(4.001) - 0.83(4.001^2) = 124.031 - 13.28100683 = 110.7509932 \, \text{m}
\]

Now, compute the average velocity:
\[
v_{\text{avg}} = \frac{110.7509932 - 110.72}{4.001 - 4} = \frac{0.0309932}{0.001} = 30.99 \, \text{m/s}
\]

### Final answers:
1. [4, 5]: \( 23.53 \, \text{m/s} \)
2. [4, 4.5]: \( 23.96 \, \text{m/s} \)
3. [4, 4.1]: \( 23.74 \, \text{m/s} \)
4. [4, 4.01]: \( 27.43 \, \text{m/s} \)
5. [4, 4.001]: \( 30.99 \, \text{m/s} \)

Do you need further clarification or additional details?

Here are some related questions you can explore:
1. How would the height function change if gravity on the moon was different?
2. What is the significance of the quadratic term in the height equation?
3. Can we approximate the instantaneous velocity by using smaller intervals?
4. How would we compute the velocity if the function had higher powers of \( t \)?
5. What does the average velocity tell us in real-world terms?

**Tip:** The smaller the interval for average velocity, the closer it approximates the instantaneous velocity!

Suppose an arrow is shot straight up from the surface of the moon with velocity 31 m/s. The height after t seconds is given by h(t) = 31t - 0.83t^2. Find the average velocity over the following time intervals: [4, 5], [4, 4.5], [4, 4.1], [4, 4.01], [4, 4.001].

Learn how to calculate average velocity of an arrow on the moon using the height function h(t) = 31t - 0.83t^2. We will solve for five intervals, breaking down the steps for a clear understanding of average velocity in this quadratic motion scenario.

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