Question

Mel slides down waterslide A, and Victor slides down waterslide B. After 2 seconds, Mel was 50 feet in the air, and after 5 seconds, she was 35 feet in the air. After 1 second, Victor was 60 feet in the air, and after 4 seconds, he was 50 feet in the air. Who was descending at a faster average rate?
Write ordered pairs relating Mel’s and Victor’s positions at a given time (i.e., (time, height)).

Mel: (2, ) and (5, )

Victor: (1, ) and (, )

Answer 1

We have two systems in this problem, namely:

System A. Mel slides down water slide A. So, let's name $A(t)$ the position of Mel at a given time, that represents her height. In this way, we know that:

$t=2s \rightarrow A(2)=50ft \\ \\ t=5s \rightarrow A(5)=35ft$

System B. Victor slides down water slide B. So, let's name $B(t)$ the position of Victor at a given time, that represents his height. Thus, we know that:

$t=1s \rightarrow B(1)=60ft \\ \\ t=4s \rightarrow B(4)=50ft$

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 $(x_{1},f(x_{1}) \ and \ (x_{2},f(x_{2})$ is given by:

$ARC=\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}$

$For \ Mel: \\ \\ ARC=\frac{35-50}{5-2}=-5ft/s \\ \\ \\ For \ Victor: \\ \\ ARC=\frac{50-60}{4-1}=-3.33ft/s$

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:

$\boxed{Mel: (2, 50) \ and \ (5, 35)}$

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:

$\boxed{Victor: (1,60) \ and \ (4,50)}$

That is, after 1 second Victor was 60 feet in the air, and after 4 seconds, he was 50 feet in the air.

Answer 2

Mel (2, 50) and (5, 35)
Victor (1, 60) and (4, 50)