Explain why if a runner completes a 6.2 -mi race in $34 \mathrm{~min}$, then he must have been running at exactly $10 \mathrm{mi} / \mathrm{hr}$ at least twice in the race. Assume the runner's speed at the finish line is zero.
Select the correct choice below and, if necessary, fill in any answer box to complete your choice. (Round to one decimal place as needed.)
A. The average speed is $\mathrm{mi} / \mathrm{hr}$. By the intermediate value theorem, the speed was exactly mi/hr at least twice. By MVT, all speeds between and milhr were reached. Because the initial and final speed was mi/hr, the speed of $10 \mathrm{mi} / \mathrm{hr}$ was reached at least twice in the race.
B. The average speed is $\mathrm{mi} / \mathrm{hr}$. By MVT, the speed was exactly mi/hr at least twice. By the intermediate value theorem, the speed between and milhr was constant. Therefore, the speed of $10 \mathrm{mi} / \mathrm{hr}$ was reached at least twice in the race.
C. The average speed is milhr. By MVT, the speed was exactly milhr at least once. By the intermediate value theorem, all speeds between and $\mathrm{mi} / \mathrm{hr}$ were reached. Because the initial and final speed was $\mathrm{mi} / \mathrm{hr}$, the speed of $10 \mathrm{mi} / \mathrm{hr}$ was reached at least twice in the race.
Answer
\(\boxed{\text{The correct choice is A. The average speed is approximately 10.9 miles per hour. By the Mean Value Theorem, the speed was exactly 10.9 miles per hour at least once. All speeds between 0 and 10.9 miles per hour were reached. Because the initial and final speed was 0 miles per hour, the speed of 10 miles per hour was reached at least twice in the race.}}\)
Step 1 ：Convert the total time from minutes to hours: \(34 \text{ minutes} = 0.5666666666666667 \text{ hours}\)
Step 2 ：Calculate the average speed by dividing the total distance by the total time: \(\frac{6.2 \text{ miles}}{0.5666666666666667 \text{ hours}} = 10.941176470588236 \text{ miles per hour}\)
Step 3 ：Since the runner's speed was zero at the start and end of the race, and the runner's speed must have increased from zero to some maximum value and then decreased back to zero, the runner must have been running at exactly 10 miles per hour at least twice in the race.
Step 4 ：According to the Mean Value Theorem (MVT), there must be at least one point in the race where the runner's speed was exactly equal to the average speed.
Step 5 ：\(\boxed{\text{The correct choice is A. The average speed is approximately 10.9 miles per hour. By the Mean Value Theorem, the speed was exactly 10.9 miles per hour at least once. All speeds between 0 and 10.9 miles per hour were reached. Because the initial and final speed was 0 miles per hour, the speed of 10 miles per hour was reached at least twice in the race.}}\)
