We have two systems in this problem, namely:
System A. Mel slides down water slide A. So, let's name
the position of Mel at a given time, that represents her height. In this way, we know that:
![t=2s \rightarrow A(2)=50ft \\ \\ t=5s \rightarrow A(5)=35ft](https://tex.z-dn.net/?f=%20t%3D2s%20%5Crightarrow%20A%282%29%3D50ft%20%5C%5C%20%5C%5C%20t%3D5s%20%5Crightarrow%20A%285%29%3D35ft%20%20)
System B. Victor slides down water slide B. So, let's name
the position of Victor at a given time, that represents his height. Thus, we know that:
![t=1s \rightarrow B(1)=60ft \\ \\ t=4s \rightarrow B(4)=50ft](https://tex.z-dn.net/?f=%20t%3D1s%20%5Crightarrow%20B%281%29%3D60ft%20%5C%5C%20%5C%5C%20t%3D4s%20%5Crightarrow%20B%284%29%3D50ft%20)
So, we have the following questions:
1. Who was descending at a faster average rate?
For a nonlinear graph whose slope changes at each point, the average rate of change between any two points
is given by:
![ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}](https://tex.z-dn.net/?f=%20ARC%3D%5Cfrac%7Bf%28x_%7B2%7D%29-f%28x_%7B1%7D%29%7D%7Bx_%7B2%7D-x_%7B1%7D%7D%20)
![For \ Mel: \\ \\ ARC=\frac{35-50}{5-2}=-5ft/s \\ \\ \\ For \ Victor: \\ \\ ARC=\frac{50-60}{4-1}=-3.33ft/s](https://tex.z-dn.net/?f=%20For%20%5C%20Mel%3A%20%5C%5C%20%5C%5C%20ARC%3D%5Cfrac%7B35-50%7D%7B5-2%7D%3D-5ft%2Fs%20%5C%5C%20%5C%5C%20%5C%5C%20For%20%5C%20Victor%3A%20%5C%5C%20%5C%5C%20ARC%3D%5Cfrac%7B50-60%7D%7B4-1%7D%3D-3.33ft%2Fs%20)
<em>So Mel was descending at a faster average rate.</em>
2. Ordered pairs relating Mel’s positions at a given time.
We can write these ordered pairs as follows:
![\boxed{Mel: (2, 50) \ and \ (5, 35)}](https://tex.z-dn.net/?f=%20%5Cboxed%7BMel%3A%20%282%2C%2050%29%20%5C%20and%20%5C%20%285%2C%2035%29%7D%20)
That is, after 2 seconds, Mel was 50 feet in the air, and after 5 seconds, she was 35 feet in the air.
3. Ordered pairs relating Victor’s positions at a given time.
We can write these ordered pairs as follows:
![\boxed{Victor: (1,60) \ and \ (4,50)}](https://tex.z-dn.net/?f=%20%5Cboxed%7BVictor%3A%20%281%2C60%29%20%5C%20and%20%5C%20%284%2C50%29%7D%20)
That is, after 1 second Victor was 60 feet in the air, and after 4 seconds, he was 50 feet in the air.