https://file.aiquickdraw.com/mathbot/file/17303118491349ejgpen0.jpg

Let's go through each part of the worksheet based on the questions and graph data provided.

### Runner Analysis

1. **Which runner has a faster average speed for the first 9 seconds?**
   - We can determine this by finding the distance each runner covers in 9 seconds.
   - **Runner A**: Distance = 120 meters in 9 seconds
     - Average speed \( = \frac{120 \text{ m}}{9 \text{ s}} \approx 13.33 \text{ m/s} \)
   - **Runner B**: Distance = 120 meters in 9 seconds
     - Average speed \( = \frac{120 \text{ m}}{9 \text{ s}} \approx 13.33 \text{ m/s} \)
   - **Answer**: Both runners have the same average speed in the first 9 seconds.

2. **Which runner has a faster average speed from 9 to 20 seconds?**
   - To find this, we calculate the average speed for each runner between 9 and 20 seconds.
   - **Runner A**: Distance from 9s to 20s \( = 168 - 120 = 48 \text{ m}\)
     - Average speed \( = \frac{48 \text{ m}}{11 \text{ s}} \approx 4.36 \text{ m/s} \)
   - **Runner B**: Distance from 9s to 20s \( = 213 - 120 = 93 \text{ m}\)
     - Average speed \( = \frac{93 \text{ m}}{11 \text{ s}} \approx 8.45 \text{ m/s} \)
   - **Answer**: Runner B has a faster average speed from 9 to 20 seconds.

3. **Which runner has a faster average speed from 20 to 31 seconds?**
   - **Runner A**: Distance from 20s to 31s \( = 287 - 168 = 119 \text{ m}\)
     - Average speed \( = \frac{119 \text{ m}}{11 \text{ s}} \approx 10.82 \text{ m/s} \)
   - **Runner B**: Distance from 20s to 31s \( = 287 - 213 = 74 \text{ m}\)
     - Average speed \( = \frac{74 \text{ m}}{11 \text{ s}} \approx 6.73 \text{ m/s} \)
   - **Answer**: Runner A has a faster average speed from 20 to 31 seconds.

4. **Which runner has a faster average speed from 9 to 31 seconds?**
   - We calculate the total distance covered in this interval and divide by the time (22 seconds).
   - **Runner A**: Distance from 9s to 31s \( = 287 - 120 = 167 \text{ m}\)
     - Average speed \( = \frac{167 \text{ m}}{22 \text{ s}} \approx 7.59 \text{ m/s} \)
   - **Runner B**: Distance from 9s to 31s \( = 287 - 120 = 167 \text{ m}\)
     - Average speed \( = \frac{167 \text{ m}}{22 \text{ s}} \approx 7.59 \text{ m/s} \)
   - **Answer**: Both runners have the same average speed from 9 to 31 seconds.

5. **Which runner wins the race? How do you know?**
   - **Answer**: Both runners finish at 287 meters at 31 seconds, so it’s a tie.

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### Rate of Change for Given Graphs

To find the average rate of change, we use the formula:

\[
\text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a}
\]

6. **Graph from \([-1, 1]\)**
7. **Graph from \([0, 3]\)**
8. **Graph from \([1, 2]\)**
9. **Graph from \([0, 3]\)**
10. **Graph from \([-3, -1]\)**
11. **Graph from \([0, 1]\)**

If you want detailed steps for each graph or specific calculations, please let me know!

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### Related Questions

1. How do we determine instantaneous speed from a distance-time graph?
2. What other methods could we use to compare the runners' speeds?
3. How would the results change if the times were recorded in intervals other than 9, 20, and 31 seconds?
4. Can you explain how average rate of change relates to the slope of a graph?
5. Why is it important to consider both distance and time when comparing speeds?

---

**Tip:** To analyze speed and motion accurately, remember that distance-time graphs can be used to find both average and instantaneous speeds by interpreting slopes and intervals.

Analyze the distance-time data for two runners and determine the faster runner across different intervals. Calculate average rates of change for graphs over given intervals.

Analyze distance-time data to compare speeds of two runners over various intervals and calculate average rates of change for given graphs. This worksheet covers key concepts in average speed and rate of change suitable for Grades 8-10.

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