# Distance, Rate, and Time

One of the most important parts of mathematics is knowing how to apply the skills and formulas that you learn to real-world applications. Formulas are simply mathematical structures that define relationships between variables and constants. In this video, we will explore a formula that relates distance, rate of speed, and time. It may help if you have a pencil and paper handy.

As mentioned, the formula, distance equals rate times time \(d = r ⋅ t\) relates the variables that represent distance, rate of speed, and time. Conceptually, it makes sense, as the rate of speed, \(r\), is a ratio of distance over some unit of time. This formula is used intuitively all the time, as you will see in the following examples:

Suppose you are driving for 2.5 hours at an average speed of 50 miles per hour mph. How many miles did you travel?

The distance formula, *d* equals *r* times *t*, will look like this:

Algebraically, the unit of hours cancels from the right side of the equation, and you are left with a total distance traveled of \(50 ⋅ (2.5) = 125 \:miles\).

Of course, sometimes the word problems that we are given are not as straightforward. There are instances when the units of measure, distance or time, must be converted before using the formula.

**Suppose you are driving at an average speed of 22 miles per hour mph for 45 minutes. How many miles did you travel?**

Because the rate of speed is measured in miles per hour and the time is given in minutes, you must calculate the fraction of one hour that 45 minutes represents. There are 60 minutes in one hour, so 45 minutes represents 0.75, or three-fourths \(\frac{3}{4}\), of one hour:

\(\frac{45 \:minutes}{60\: minutes}=.75\: hours\)Once the units of time are the same, the distance can be determined, like this:

\(d=\frac{22\: miles}{1\: hour}\times (.75\: hours)\)

This results in a total distance traveled of 22 times 0.75 which equals 16.5 miles. \(22(.75)=16.5 \:miles\).

Now let’s have you try one. You are on your way to a new job interview, and you want to arrive early to make a good impression. Unfortunately, you did not factor in the heavy traffic that you encounter in your commuting time, and now you are wondering if you are going to be late! Your average speed is only 35 miles per hour. How far have you traveled after 25 minutes? If you’d like to try this for yourself, go ahead and pause the video.

The answer is 14.7 miles. Here’s how we got that.

First, convert 25 minutes to hours by dividing 25 over 60, which gives you 0.42 hours. \(\frac{25}{60}=.42 \:hours\)

Then, use the distance formula: \(d=35(.4)=14.7\: miles\)

After driving for 25 minutes at an average rate of speed of 35 miles per hour, 14 miles have been traveled.

Once variable relationships are defined by formulas, algebraic rules allow us to solve for any of the variables. Here are a few other word problems that use the distance formula to determine rate of speed or time traveled.

**A passenger plane is flying for 4 hours and has traveled 1,640 miles. What was the rate of speed of the plane?**

In this example, we know the time and the distance traveled. The units of hours and miles will result in the rate of the plane, as expressed in miles per hour. The distance formula can be rearranged algebraically to solve for the variable, r, as shown:

\(d=rt\)

So you’re going to divide by t on both sides

\(\frac{d}{t}=r\)

(So you get d divided by t is equal to r) Now we can just substitute the information and solve for the rate of speed, r. So we have 1640 miles divided by 4 hours gives you 410 miles per hour.

Okay, here’s another one for you to try. Use the following information to determine the rate of speed that you would have had to maintain to get to your job interview on time: The interview location is approximately 30 miles from your home, and you gave yourself 50 minutes to drive to the interview. Pause the video now and see if you can figure it out!

The answer is 36 miles per hour. Let’s look at how we got that.

\(\frac{d}{t}=r\)

First we’re going to solve the equation for the rate of speed.

\(t=\frac{50}{60}\)

Then we’re going to convert 50 minutes to a fraction of one hour. 50 minutes over 60 minutes gives us 0.83 hours. Now we’re going to plug it into our equation. So 30 divided by 0.83 gives us 36 miles per hour.

In order to get to the interview on time, an average rate of speed of 36 miles per hour must be maintained.

The next example allows us to answer questions regarding time in a distance word problem.

**How long does the train from Boston to New York take if the distance between train stations is approximately 306 kilometers and the train travels at an average rate of speed of 106 kilometers per hour?**

Again, this problem requires you to algebraically rearrange the distance formula. By dividing both sides of the equation by the rate of speed, r, we are able to solve for time, t, as shown. Distance equals rate times time, you divide by r on both sides. You get distance over rate equals time. Now we’re going to substitute the given information and solve for the time, t

\(d=rt\)

Isolate the variable, t.

\(\frac{d}{r}=t\)

Substitute the given information to solve for the time, t.

In this example, the distance unit of kilometers cancels out, leaving the number of hours the trip will take.

Okay, now that you are familiar with this formula and how to apply it, you are well on your way to understanding the relation of distance, rate of speed, and time.

Thanks for watching, and happy studying!

## Distance, Rate, and Time Practice Problems

**Question #1:**

If you travel at 65 miles per hour for 15 minutes, how far did you travel?

15.5 miles

16.25 miles

17.75 miles

18 miles

**Answer:**

The correct answer is 16.25 miles. To solve this problem, use this formula:

\(d=rt\)

The rate (*r*) is 65 mph and the time (*t*) is \(\frac{1}{4}\) hr (since time is measured in hours and 15 minutes is \(\frac{1}{4}\) of an hour).

\(d=(65\text{ mph})(\frac{1}{4}\text{ hr})=16\frac{1}{4}=16.25\text{ miles}\)

The distance traveled is 16.25 miles.

**Question #2:**

Amber is making a 3.5-hour weekend trip and travels at an average speed of 55 miles per hour. How many miles did she travel?

15.7 miles

192.5 miles

82.5 miles

112.7 miles

**Answer:**

The correct answer is 192.5 miles. To solve this problem, use the formula:

\(d=rt\)

The rate (*r*) is 55 mph and the time is 3.5 hrs.

\(d=(55\text{ mph})(3.5\text{ hrs})=192.5\text{ miles}\)

Amber traveled 192.5 miles.

**Question #3:**

If Daniel drives 120 miles at an average speed of 60 miles per hour, how long will he be driving?

1 hour

1.5 hours

2 hours

2.5 hours

**Answer:**

The correct answer is 2 hours. First, we need to manipulate our \(d=rt\) formula so we can solve for time (*t*). We do this by dividing by the rate on both sides. Our formula now looks like this:

\(t=\frac{d}{r}\)

Our distance (*d*) is 120 mi and our rate (*r*) is 60 mph.

\(t=\frac{120\text{ mi}}{60\text{ mph}}=2\text{ hrs}\)

Daniel will be driving for 2 hours.

**Question #4:**

Sarah gets to her friend’s house 15 miles away in 12 minutes. If she drove at a constant speed the whole time, how fast was she driving?

75 mph

65 mph

55 mph

45 mph

**Answer:**

The correct answer is 75 mph. First, we need to manipulate our \(d=rt\) formula so we can solve for the rate (*r*). We do this by dividing by time on both sides. Our formula now looks like this:

\(r=\frac{d}{t}\)

The distance is 15 mi and the time is 12 min, which needs to be converted to hrs.

\(\frac{12\text{ mins}}{60\text{ mins in an hour}}=\frac{1}{5}\text{ hr}\)

The time (*t*) is \(\frac{1}{5}\text{ hr}\).

\(r=\frac{15}{\frac{1}{5}}=15\times5=75\text{ mph}\)

Sarah was driving 75 miles per hour.

**Question #5:**

Jonathan traveled at 50 miles per hour for an hour and 45 minutes. What is the distance he traveled?

38 miles

49.5 miles

76 miles

87.5 miles

**Answer:**

The correct answer is 87.5 miles. To solve this problem, use the formula:

\(d=rt\)

The rate (*r*) is 50 mph and the time (*t*) is 1 hour and 45 minutes, which equals \(1\frac{3}{4}\) hours, or \(\frac{7}{4}\) hours.

\(d=(50\text{ mph})(\frac{7}{4}\text{ hrs})=87\frac{1}{2}=87.5\text{ mi}\)

Jonathan traveled 87.5 miles.