In algebra distance word problems are a struggle for some. Use this free lesson to help you learn how to solve distance word problems.

**Distance word problems** are a common type of algebra word problems. They involve a scenario in which you need to figure out how **fast**, how **far**, or how **long** one or more objects have traveled. These are often called **train problems** because one of the most famous types of distance problems involves finding out when two trains heading toward each other cross paths.

In this lesson, you'll learn how to solve train problems and a few other common types of distance problems. But first, let's look at some basic principles that apply to **any** distance problem.

There are three basic aspects to movement and travel: **distance**, **rate**, and **time**. To understand the difference among these, think about the last time you drove somewhere.

The **distance** is how **far** you traveled. The **rate** is how **fast** you traveled. The **time** is how **long** the trip took.

The relationship among these things can be described by this formula:

distance = rate x time

d = rt

In other words, the **distance** you drove is equal to the **rate** at which you drove times the amount of **time** you drove. For an example of how this would work in real life, just imagine your last trip was like this:

- You drove 25 miles—that's the
**distance**. - You drove an average of 50 mph—that's the
**rate**. - The drive took you 30 minutes, or 0.5 hours—that's the
**time**.

According to the formula, if we multiply the **rate** and **time**, the product should be our distance.

And it is! We drove 50 mph for 0.5 hours—and **50 ⋅ 0.5** equals 25, which is our distance.

What if we drove 60 mph instead of 50? How far could we drive in 30 minutes? We could use the same formula to figure this out.

60 **⋅** 0.5 is 30, so our distance would be **30** miles.

When you solve any distance problem, you'll have to do what we just did—use the formula to find **distance**, **rate**, or **time**. Let's try another simple problem.

On his day off, Lee took a trip to the zoo. He drove an average speed of 65 mph, and it took him two-and-a-half hours to get from his house to the zoo. How far is the zoo from his house?

First, we should identify the information we know. Remember, we're looking for any information about distance, rate, or time. According to the problem:

- The
**rate**is 65 mph. - The
**time**is two-and-a-half hours, or 2.5 hours. - The
**distance**is unknown—it's what we're trying to find.

You could picture Lee's trip with a diagram like this:

This diagram is a start to understanding this problem, but we still have to figure out what to do with the numbers for **distance**, **rate**, and **time**. To keep track of the information in the problem, we'll set up a table. (This might seem excessive now, but it's a good habit for even simple problems and can make solving complicated problems much easier.) Here's what our table looks like:

distance | rate | time |
---|---|---|

d | 65 | 2.5 |

We can put this information into our formula: **distance = rate ⋅ time**.

We can use the **distance = rate ⋅ time** formula to find the distance Lee traveled.

d = rt

The formula *d = rt* looks like this when we plug in the numbers from the problem. The unknown **distance** is represented with the variable *d*.

d = 65 ⋅ 2.5

To find *d*, all we have to do is multiply 65 and 2.5. **65 ⋅ 2.5** equals 162.5.

d = 162.5

We have an answer to our problem: *d* = 162.5. In other words, the distance Lee drove from his house to the zoo is 162.5 miles.

Be careful to use the same **units of measurement** for rate and time. It's possible to multiply 65 miles per **hour** by 2.5 **hours** because they use the same unit: an **hour**. However, what if the time had been written in a different unit, like in **minutes**? In that case, you'd have to convert the time into hours so it would use the same unit as the rate.

In the problem we just solved we calculated for **distance**, but you can use the *d = rt* formula to solve for **rate** and **time** too. For example, take a look at this problem:

After work, Janae walked in her neighborhood for a half hour. She walked a mile-and-a-half total. What was her average speed in miles per hour?

We can picture Janae's walk as something like this:

And we can set up the information from the problem we know like this:

distance | rate | time |
---|---|---|

1.5 | r | 0.5 |

The table is repeating the facts we already know from the problem. Janae walked **one-and-a-half miles** or 1.5 miles in a half hour, or 0.5 hours.

As always, we start with our formula. Next, we'll fill in the formula with the information from our table.

d = rt

The rate is represented by *r* because we don't yet know how fast Janae was walking. Since we're solving for *r*, we'll have to get it alone on one side of the equation.

1.5 = r ⋅ 0.5

Our equation calls for *r* to be **multiplied** by 0.5, so we can get *r* alone on one side of the equation by **dividing** both sides by 0.5: **1.5 / 0.5 =** 3.

3 = r

*r* = 3, so 3 is the answer to our problem. Janae walked **3** miles per hour.

In the problems on this page, we solved for **distance** and **rate** of travel, but you can also use the travel equation to solve for **time**. You can even use it to solve certain problems where you're trying to figure out the distance, rate, or time of two or more moving objects. We'll look at problems like this on the next few pages.