We have two systems in this problem, namely:
System A. Mel slides down water slide A. So, let's name [tex]A(t)[/tex] the position of Mel at a given time, that represents her height. In this way, we know that:
[tex] t=2s \rightarrow A(2)=50ft \\ \\ t=5s \rightarrow A(5)=35ft [/tex]
System B. Victor slides down water slide B. So, let's name [tex]B(t)[/tex] the position of Victor at a given time, that represents his height. Thus, we know that:
[tex] t=1s \rightarrow B(1)=60ft \\ \\ t=4s \rightarrow B(4)=50ft [/tex]
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 [tex](x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})[/tex] is given by:
[tex] ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}} [/tex]
[tex] For \ Mel: \\ \\ ARC=\frac{35-50}{5-2}=-5ft/s \\ \\ \\ For \ Victor: \\ \\ ARC=\frac{50-60}{4-1}=-3.33ft/s [/tex]
So Mel was descending at a faster average rate.
2. Ordered pairs relating Mel’s positions at a given time.
We can write these ordered pairs as follows:
[tex] \boxed{Mel: (2, 50) \ and \ (5, 35)} [/tex]
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:
[tex] \boxed{Victor: (1,60) \ and \ (4,50)} [/tex]
That is, after 1 second Victor was 60 feet in the air, and after 4 seconds, he was 50 feet in the air.
