Question
The first point of he parabola is (0, 0), the second is at (2, 0.4), the third at (8, 1), and the fourth one at (11, 1.5).
Riley is training to run a 5-mile race. The graph represents the distance (miles) she ran with respect to time (minutes) in her first training run. Over which time interval did she run the fastest [0, 8] or [2, 11]?
The average rate of change over the interval [0, 8] is _ miles per minute.
The average rate of change over the interval [2, 11] is about _ miles per minute.
Riley ran the fastest over the time interval _
Riley is training to run a 5-mile race. The graph represents the distance (miles) she ran with respect to time (minutes) in her first training run. Over which time interval did she run the fastest [0, 8] or [2, 11]?
The average rate of change over the interval [0, 8] is _ miles per minute.
The average rate of change over the interval [2, 11] is about _ miles per minute.
Riley ran the fastest over the time interval _
Answers
GPT-4o mini
To determine which time interval Riley ran the fastest, we need to calculate the average rate of change over the two given intervals: [0, 8] and [2, 11].
**1. Calculate the average rate of change over the interval [0, 8]:**
- The starting point of the interval [0, 8] is (0, 0) and the endpoint is (8, 1).
- The average rate of change is calculated using the formula:
\[
\text{Average rate of change} = \frac{\text{Change in distance}}{\text{Change in time}} = \frac{d(8) - d(0)}{t(8) - t(0)} = \frac{1 - 0}{8 - 0} = \frac{1}{8} = 0.125 \text{ miles per minute}
\]
**2. Calculate the average rate of change over the interval [2, 11]:**
- The starting point of the interval [2, 11] is (2, 0.4) and the endpoint is (11, 1.5).
- Again, using the average rate of change formula:
\[
\text{Average rate of change} = \frac{d(11) - d(2)}{t(11) - t(2)} = \frac{1.5 - 0.4}{11 - 2} = \frac{1.1}{9} \approx 0.1222 \text{ miles per minute}
\]
**Summary of calculations:**
- Average rate of change over the interval [0, 8] is **0.125 miles per minute**.
- Average rate of change over the interval [2, 11] is approximately **0.1222 miles per minute**.
**3. Determine the fastest interval:**
Riley ran the fastest over the time interval **[0, 8]** because 0.125 miles per minute is greater than 0.1222 miles per minute.
Therefore, to summarize:
- The average rate of change over the interval [0, 8] is **0.125 miles per minute**.
- The average rate of change over the interval [2, 11] is about **0.1222 miles per minute**.
- Riley ran the fastest over the time interval **[0, 8]**.
**1. Calculate the average rate of change over the interval [0, 8]:**
- The starting point of the interval [0, 8] is (0, 0) and the endpoint is (8, 1).
- The average rate of change is calculated using the formula:
\[
\text{Average rate of change} = \frac{\text{Change in distance}}{\text{Change in time}} = \frac{d(8) - d(0)}{t(8) - t(0)} = \frac{1 - 0}{8 - 0} = \frac{1}{8} = 0.125 \text{ miles per minute}
\]
**2. Calculate the average rate of change over the interval [2, 11]:**
- The starting point of the interval [2, 11] is (2, 0.4) and the endpoint is (11, 1.5).
- Again, using the average rate of change formula:
\[
\text{Average rate of change} = \frac{d(11) - d(2)}{t(11) - t(2)} = \frac{1.5 - 0.4}{11 - 2} = \frac{1.1}{9} \approx 0.1222 \text{ miles per minute}
\]
**Summary of calculations:**
- Average rate of change over the interval [0, 8] is **0.125 miles per minute**.
- Average rate of change over the interval [2, 11] is approximately **0.1222 miles per minute**.
**3. Determine the fastest interval:**
Riley ran the fastest over the time interval **[0, 8]** because 0.125 miles per minute is greater than 0.1222 miles per minute.
Therefore, to summarize:
- The average rate of change over the interval [0, 8] is **0.125 miles per minute**.
- The average rate of change over the interval [2, 11] is about **0.1222 miles per minute**.
- Riley ran the fastest over the time interval **[0, 8]**.