Question

An elevator began at an elevation of 85.5 feet and ascended at a rate of 2.75 feet per second. Which expression represents the height of the elevator after s seconds?

Answer 1

We know that an elevator began at an elevation of 85.5 feet, and ascended at a rate of 2.75 feet per second.

This means that at 0 seconds, the elevator is at an elevation of 85.5feet.

`\begin{gathered} \text{At 1 second: }85.5+1\cdot2.75 \\ \text{At 2 seconds: }85.5+2\cdot2.75 \\ \text{At 3 seconds: }85.5+3\cdot2.75 \end{gathered}`

and so on. This means that if s is the number of seconds, the height will have the expression:

`\begin{gathered} h=85.5+s\cdot2.75 \\ =85.5+2.75s \end{gathered}`