Explain why if a runner completes a \( 6.2-\mathrm{mi} \) race in 38 min , then he must have been running at exactly \( 9 \mathrm{mi} / \mathrm{hr} \) at least twice in the race. Assume the runner's speed at the finish line is zero. Select the correct choice and fill in the answer boxes to complete your choice. (Round to one decimal place as needed.) A. The average speed is \( \square \mathrm{mi} / \mathrm{hr} \). By MVT, the speed was exactly \( \square \mathrm{mi} / \mathrm{hr} \) at least once. By the intermediate value theorem, all speeds between \( \square \) and \( \square \) mi/hr were reached. Because the initial and final speed was \( \square \) mi/hr, the speed of \( 9 \mathrm{mi} / \mathrm{hr} \) was reached at least twice in the race. B. The average speed is \( \square \mathrm{mi} / \mathrm{hr} \). By MVT, the speed was exactly \( \square \) mi/hr at least twice. By the intermediate value theorem, the speed between \( \square \) and \( \square \) mi/hr was constant. Therefore, the speed of \( 9 \mathrm{mi} / \mathrm{hr} \) was reached at least twice in the race. C. The average speed is \( \square \mathrm{mi} / \mathrm{hr} \). By the intermediate value theorem, the speed was exactly \( \square \mathrm{mi} / \mathrm{hr} \) at least twice. By MVT, all speeds between \( \square \) and \( \square \) mi/hr were reached. Because the initial and final speed was mi/hr, the speed of \( 9 \mathrm{mi} / \mathrm{hr} \) was reached at least twice in the race.

To solve this problem, we need to calculate the average speed of the runner and then apply the Mean Value Theorem (MVT) and the Intermediate Value Theorem (IVT) to explain why the runner must have been running at exactly \( 9 \mathrm{mi/hr} \) at least twice during the race. ### Step 1: Calculate the Average Speed The average speed \( v_{avg} \) can be calculated using the formula: \[ v_{avg} = \frac{\text{total distance}}{\text{total time}} \] Given: - Total distance = \( 6.2 \) miles - Total time = \( 38 \) minutes First, we need to convert the time from minutes to hours: \[ \text{Total time in hours} = \frac{38 \text{ min}}{60} = \frac{19}{30} \text{ hr} \] Now we can calculate the average speed: \[ v_{avg} = \frac{6.2 \text{ mi}}{\frac{19}{30} \text{ hr}} = 6.2 \times \frac{30}{19} \text{ mi/hr} \] Calculating this gives: \[ v_{avg} = \frac{186}{19} \approx 9.789 \text{ mi/hr} \] ### Step 2: Apply the Mean Value Theorem (MVT) According to the MVT, if a runner completes a race, there exists at least one point in time where the instantaneous speed \( v \) is equal to the average speed \( v_{avg} \). ### Step 3: Apply the Intermediate Value Theorem (IVT) Since the runner starts at some speed (let's denote it as \( v_0 \)) and finishes at \( 0 \) mi/hr, the IVT states that all speeds between \( v_0 \) and \( 0 \) must have been reached at least once during the race. ### Step 4: Fill in the Choices Now we can fill in the blanks in the provided options: 1. **Average Speed**: \( 9.8 \mathrm{mi/hr} \) (rounded to one decimal place). 2. **Speed at least once**: \( 9.8 \mathrm{mi/hr} \) (the average speed). 3. **Speeds between**: \( 0 \mathrm{mi/hr} \) and \( 9.8 \mathrm{mi/hr} \). 4. **Initial speed**: \( 9.8 \mathrm{mi/hr} \) (the average speed). ### Final Answer The correct choice is: **A. The average speed is \( 9.8 \mathrm{mi/hr} \). By MVT, the speed was exactly \( 9.8 \mathrm{mi/hr} \) at least once. By the intermediate value theorem, all speeds between \( 0 \) and \( 9.8 \) mi/hr were reached. Because the initial and final speed was \( 0 \) mi/hr, the speed of \( 9 \mathrm{mi/hr} \) was reached at least twice in the race.**