Question

A swimming pool was filling with water at a constant rate of 150 gallons per hour. The pool had 75 gallons before the timer started. How do you write an equation in slope-intercept form to model the situation?

Answer 1

`y = 15/6t + 75`

Explanation:

The problem notes that the water is being added to the pool at a constant rate which implies a linear relationship between the time and the volume of water in the pool. That is, the relationship can be expressed as the graph of a line.

The slope point form of a line is `y = mx + b` where m is the slope and b is the y-intercept.

Let t = time in seconds after the timer started and y = volume of water in gallons.

It follows that `y = mt + b` where t is the independent variable and y the dependent variable. Visualize this as the the graph of a line with t on the horizontal axis and y the vertical axis. The faster the water is being pumped into the pool, the steeper the line will be.

Now, it's just a matter of finding m and b to complete the equation to model the situation.

To find b, note that when the timer started, t = 0 and y = 75, Substituting:

`y = mt + b`
`75 = m0 + b`
`b = 75`

To find m, note that m, the slope, or steepness, can be interpreted as the "rise" divided by the "run" of the graph of line representing `y = mt + b`.

Recall that when t = 0, the pool contains 75 gallons of water; when t = 60, the pool has 150 more gallons.

Visualizing the graph of the line, it follows that over a run of 60, there is a rise of 150.

So, rise/run = m = `150/60 = 15/6`.

Substituting our m and b into `y = mt + b` yields:

`y = 15/6t + 75`