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### Problem 1a:
The function \( d(t) = 0.8t^2 \) represents the distance in meters that a rock dropped from a cliff on the Moon travels after \( t \) seconds. The question asks for the **average rate of change** in distance (i.e., velocity) of the rock between 5 seconds and 10 seconds after it was dropped.

#### Solution to 1a:
To find the **average rate of change** of \( d(t) \) over the interval \( t = 5 \) to \( t = 10 \), we use the formula for average rate of change:
\[
\text{Average rate of change} = \frac{d(t_2) - d(t_1)}{t_2 - t_1}
\]
where \( t_1 = 5 \) and \( t_2 = 10 \).

1. Find \( d(10) \):
\[
d(10) = 0.8(10)^2 = 0.8(100) = 80 \text{ meters}
\]

2. Find \( d(5) \):
\[
d(5) = 0.8(5)^2 = 0.8(25) = 20 \text{ meters}
\]

3. Calculate the average rate of change:
\[
\text{Average rate of change} = \frac{80 - 20}{10 - 5} = \frac{60}{5} = 12 \text{ meters per second}.
\]

Thus, the average rate of change in the rock's distance between 5 and 10 seconds is **12 meters per second**.

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### Problem 1b:
The second part of the question (1b) asks: **What does this rate of change mean as it relates to the rock falling?**

#### Explanation for 1b:
The average rate of change calculated in part 1a tells us that, between 5 and 10 seconds, the rock's velocity increased at an average of 12 meters per second. This means that, on average, the rock was falling faster and covering more distance in each second as time progressed. Since this is a uniform acceleration model, this is consistent with the rock accelerating due to the Moon's gravity.

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Do you want further details or have any questions?

Here are some follow-up questions you might explore:
1. What is the instantaneous velocity at \( t = 10 \) seconds?
2. How would the distance function change if the rock was dropped on Earth?
3. What is the total distance traveled by the rock after 15 seconds?
4. How would you interpret the average rate of change if \( t_1 = 0 \) and \( t_2 = 10 \)?
5. What is the acceleration implied by the function \( d(t) = 0.8t^2 \)?

**Tip**: The average rate of change over an interval is a useful tool to estimate velocity or growth, but the instantaneous rate of change (using derivatives) gives more precise insights.

An astronaut drops a rock off the edge of a cliff on the Moon. The distance, d(t), in meters, the rock travels after t seconds can be modeled by the function d(t) = 0.8t^2. What is the average rate of change, in meters per second, of the rock between 5 and 10 seconds after it was dropped? What does this rate of change mean as it relates to the rock falling?

Calculate the average rate of change for a rock dropped from a cliff on the Moon using the function d(t) = 0.8t^2. Learn how to find the rock's velocity between 5 and 10 seconds and interpret the result. Suitable for students in Grades 9-11.

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