Problem
The table below gives the height above the ground, $h$, of a passenger traveling on the Vegas High Roller, currently the largest Ferris wheel in the world with a 520-foot diameter. ${ }^{1}$ Assume passengers board the wheel at its lowest point ( 30 feet off the ground), time is measured in minutes since boarding, and a full ride takes 30 minutes. Write a difference quotient that represents the passenger's average vertical speed during the last two and a half minutes of her ride down, then calculate its value using data from the table. (Recall that heights on the way down are equal to heights on the way up.) \begin{tabular}{l|c|c|c|c|c|c|c} \hline$t$ & 0 & 2.5 & 5 & 7.5 & 10 & 12.5 & 15 \\ \hline$h=f(t)$ & 30 & 64.8 & 160 & 290 & 420 & 515.2 & 550 \\ \hline \end{tabular} Round your answer to two decimal places, if required.
Solution
Step 1 ：The average vertical speed can be calculated by the difference in height divided by the difference in time. In this case, we want to calculate the average speed during the last two and a half minutes of the ride. Looking at the table, we can see that the height at time \(t=12.5\) is 515.2 feet and at time \(t=15\) is 550 feet. So, the difference in height is \(550 - 515.2 = 34.8\) feet. The difference in time is \(15 - 12.5 = 2.5\) minutes. Therefore, the average speed is \(\frac{34.8}{2.5}\) feet per minute.
Step 2 ：The average vertical speed during the last two and a half minutes of the ride is approximately \(\boxed{13.92}\) feet per minute.
