• Use the Distance, Rate, and Time formula
  • Solve a formula for a specific variable

Assignment

Use the Distance, Rate, and Time Formula

A rate is how something changes over time, and a common rate you likely see every day is miles per hour, which relates distance and time through division (the per). Our rate of miles per hour is defined as the distance divided by time.

\[\begin{align} r = \frac{d}{t} \end{align}\]

This formula can be used to solve various problems involving rate, distance and time. The book has a number of these examples, so I’ll just put one here and we’ll do a few more in class.

Jamal’s Bike

Jamal rides his bike at a uniform rate of $12$ miles per hour for $3.5$ hours. What distance has he traveled?

They give us a rate and a time, while looking for a distance so the formula is a perfect fit. Make sure you units all match up before plugging anything in.

\[\begin{align} 12 \text{ mph}= \frac{d}{3.5\text{ hours}} \end{align}\]

We can solve for $d$ by multiplying by $3.5 \text{ hours}$ on both sides.

\[\begin{align} 3.5\text{ hours} \cdot 12 \text{ mph} &= \frac{d}{3.5\text{ hours}} \cdot 3.5\text{ hours} \\ 42 \text{ miles} &= d \end{align}\]

A good way to make sure everything was done correctly—aside from plugging back in and checking–is to look at your units. Miles per hour is literally miles divided by hours, or $\frac{\text{miles}}{\text{hour}}$, so multiplying by $\text{hours}$ leaves just $\text{miles}$. We were looking for a distance, so it checks out.

Solve a Formula for a Specific Variable

We can solve our distance equation for a different variable, like $d$. To do that, we multiply both sides by $t$.

\[\begin{align} r &= \frac{d}{t} \\ t \cdot r &= \frac{d}{t} \cdot t \\ rt &= d \end{align}\]

And for completion’s sake, you can solve for $t$.

\[\begin{align} rt &= d \\ \frac{rt}{r} &= \frac{d}{r} \\ t &= \frac{d}{r} \end{align}\]

To solve a formula for a specific variable means to isolate it and move every other variable and constant to the other side. Your strategies from before still work, with the only real change coming from the fact you won’t have to do as much arithmetic.

Here’s the area of a triangle solved for $h$. I’m going to start leaving some of the steps out. Make sure you are able to fill in the blanks.

\[\begin{align} A &= \frac{1}{2}bh \\ 2A &= bh \\ \frac{2A}{b} & = h \end{align}\]