Algebra Topics
Distance Word Problems
What are 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.
The basics of distance problems
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.
Solving distance problems
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.
Solving for rate and time
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.
Two-part and round-trip problems
Do you know how to solve this problem?
Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?
This problem is a classic two-part trip problem because it's asking you to find information about one part of a two-part trip. This problem might seem complicated, but don't be intimidated!
You can solve it using the same tools we used to solve the simpler problems on the first page:
- The travel equation d = rt
- A table to keep track of important information
Let's start with the table. Take another look at the problem. This time, the information relating to distance, rate, and time has been underlined.
Bill took a trip to see a friend. His friend lives 225 miles away. He drove in town at an average of 30 mph, then he drove on the interstate at an average of 70 mph. The trip took three-and-a-half hours total. How far did Bill drive on the interstate?
If you tried to fill in the table the way we did on the last page, you might have noticed a problem: There's too much information. For instance, the problem contains two rates—30 mph and 70 mph. To include all of this information, let's create a table with an extra row. The top row of numbers and variables will be labeled in town, and the bottom row will be labeled interstate.
| distance | rate | time | |
|---|---|---|---|
| in town | 30 | ||
| interstate | 70 |
We filled in the rates, but what about the distance and time? If you look back at the problem, you'll see that these are the total figures, meaning they include both the time in town and on the interstate. So the total distance is 225. This means this is true:
Interstate distance + in-town distance = Total distance
Together, the interstate distance and in-town distance are equal to the total distance. See?
In any case, we're trying to find out how far Bill drove on the interstate, so let's represent this number with d. If the interstate distance is d, it means the in-town distance is a number that equals the total, 225, when added to d. In other words, it's equal to 225 - d.
We can fill in our chart like this:
| distance | rate | time | |
|---|---|---|---|
| in town | 225 - d | 30 | |
| interstate | d | 70 |
We can use the same technique to fill in the time column. The total time is 3.5 hours. If we say the time on the interstate is t, then the remaining time in town is equal to 3.5 - t. We can fill in the rest of our chart.
| distance | rate | time | |
|---|---|---|---|
| in town | 225 - d | 30 | 3.5 - t |
| interstate | d | 70 | t |
Now we can work on solving the problem. The main difference between the problems on the first page and this problem is that this problem involves two equations. Here's the one for in-town travel:
225 - d = 30 ⋅ (3.5 - t)
And here's the one for interstate travel:
d = 70t
If you tried to solve either of these on its own, you might have found it impossible: since each equation contains two unknown variables, they can't be solved on their own. Try for yourself. If you work either equation on its own, you won't be able to find a numerical value for d. In order to find the value of d, we'll also have to know the value of t.
We can find the value of t in both problems by combining them. Let's take another look at our travel equation for interstate travel.
While we don't know the numerical value of d, this equation does tell us that d is equal to 70t.
d = 70t
Since 70t and d are equal, we can replace d with 70t. Substituting 70t for d in our equation for interstate travel won't help us find the value of t—all it tells us is that 70t is equal to itself, which we already knew.
70t = 70t
But what about our other equation, the one for in-town travel?
225 - d = 30 ⋅ (3.5 - t)
When we replace the d in that equation with 70t, the equation suddenly gets much easier to solve.
225 - 70t = 30 ⋅ (3.5 - t)
Our new equation might look more complicated, but it's actually something we can solve. This is because it only has one variable: t. Once we find t, we can use it to calculate the value of d—and find the answer to our problem.
To simplify this equation and find the value of t, we'll have to get the t alone on one side of the equals sign. We'll also have to simplify the right side as much as possible.
225 - 70t = 30 ⋅ (3.5 - t)
Let's start with the right side: 30 times (3.5 - t) is 105 - 30t.
225 - 70t = 105 - 30t
Next, let's cancel out the 225 next to 70t. To do this, we'll subtract 225 from both sides. On the right side, it means subtracting 225 from 105. 105 - 225 is -120.
- 70t = -120 - 30t
Our next step is to group like terms—remember, our eventual goal is to have t on the left side of the equals sign and a number on the right. We'll cancel out the -30t on the right side by adding 30t to both sides. On the right side, we'll add it to -70t. -70t + 30t is -40t.
- 40t = -120
Finally, to get t on its own, we'll divide each side by its coefficient: -40. -120 / - 40 is 3.
t = 3
So t is equal to 3. In other words, the time Bill traveled on the interstate is equal to 3 hours. Remember, we're ultimately trying to find the distance Bill traveled on the interstate. Let's look at the interstate row of our chart again and see if we have enough information to find out.
| distance | rate | time | |
|---|---|---|---|
| interstate | d | 70 | 3 |
It looks like we do. Now that we're only missing one variable, we should be able to find its value pretty quickly.
To find the distance, we'll use the travel formula distance = rate ⋅ time.
d = rt
We now know that Bill traveled on the interstate for 3 hours at 70 mph, so we can fill in this information.
d = 3 ⋅ 70
Finally, we finished simplifying the right side of the equation. 3 ⋅ 70 is 210.
d = 210
So d = 210. We have the answer to our problem! The distance is 210. In other words, Bill drove 210 miles on the interstate.
Solving a round-trip problem
It might have seemed like it took a long time to solve the first problem. The more practice you get with these problems, the quicker they'll go. Let's try a similar problem. This one is called a round-trip problem because it describes a round trip—a trip that includes a return journey. Even though the trip described in this problem is slightly different from the one in our first problem, you should be able to solve it the same way. Let's take a look:
Eva drove to work at an average speed of 36 mph. On the way home, she hit traffic and only drove an average of 27 mph. Her total time in the car was 1 hour and 45 minutes, or 1.75 hours. How far does Eva live from work?
If you're having trouble understanding this problem, you might want to visualize Eva's commute like this:
As always, let's start by filling in a table with the important information. We'll make a row with information about her trip to work and from work.
1.75 - t to describe the trip from work. (Remember, the total travel time is 1.75 hours, so the time to work and from work should equal 1.75.)
From our table, we can write two equations:
- The trip to work can be represented as d = 36t.
- The trip from work can be represented as d = 27 (1.75 - t).
In both equations, d represents the total distance. From the diagram, you can see that these two equations are equal to each other—after all, Eva drives the same distance to and from work.
Just like with the last problem we solved, we can solve this one by combining the two equations.
We'll start with our equation for the trip from work.
d = 27 (1.75 - t)
Next, we'll substitute in the value of d from our to work equation, d = 36t. Since the value of d is 36t, we can replace any occurrence of d with 36t.
36t = 27 (1.75 - t)
Now, let's simplify the right side. 27 ⋅(1.75 - t) is 47.25.
36t = 47.25 - 27t
Next, we'll cancel out -27t by adding 27t to both sides of the equation. 36t + 27t is 63t.
63t = 47.25
Finally, we can get t on its own by dividing both sides by its coefficient: 63. 47.25 / 63 is .75.
t = .75
t is equal to .75. In other words, the time it took Eva to drive to work is .75 hours. Now that we know the value of t, we'll be able to can find the distance to Eva's work.
If you guessed that we were going to use the travel equation again, you were right. We now know the value of two out of the three variables, which means we know enough to solve our problem.
d = rt
First, let's fill in the values we know. We'll work with the numbers for the trip to work. We already knew the rate: 36. And we just learned the time: .75.
d = 36 ⋅ .75
Now all we have to do is simplify the equation: 36 ⋅ .75 = 27.
d = 27
d is equal to 27. In other words, the distance to Eva's work is 27 miles. Our problem is solved.
Intersecting distance problems
An intersecting distance problem is one where two things are moving toward each other. Here's a typical problem:
Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading to Springfield at the same time a train leaves Springfield heading to Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?
This problem is asking you to calculate how long it will take these two trains moving toward each other to cross paths. This might seem confusing at first. Even though it's a real-world situation, it can be difficult to imagine distance and motion abstractly. This diagram might help you get a sense of what this situation looks like:
If you're still confused, don't worry! You can solve this problem the same way you solved the two-part problems on the last page. You'll just need a chart and the travel formula.
Pawnee and Springfield are 420 miles apart. A train leaves Pawnee heading toward Springfield at the same time a train leaves Springfield heading toward Pawnee. One train is moving at a speed of 45 mph, and the other is moving 60 mph. How long will they travel before they meet?
Let's start by filling in our chart. Here's the problem again, this time with the important information underlined. We can start by filling in the most obvious information: rate. The problem gives us the speed of each train. We'll label them fast train and slow train. The fast train goes 60 mph. The slow train goes only 45 mph.
We can also put this information into a table:
| distance | rate | time | |
|---|---|---|---|
| fast train | 60 | ||
| slow train | 45 |
We don't know the distance each train travels to meet the other yet—we just know the total distance. In order to meet, the trains will cover a combined distance equal to the total distance. As you can see in this diagram, this is true no matter how far each train travels.
This means that—just like last time—we'll represent the distance of one with d and the distance of the other with the total minus d. So the distance for the fast train will be d, and the distance for the slow train will be 420 - d.
| distance | rate | time | |
|---|---|---|---|
| fast train | d | 60 | |
| slow train | 420 - d | 45 |
Because we're looking for the time both trains travel before they meet, the time will be the same for both trains. We can represent it with t.
| distance | rate | time | |
|---|---|---|---|
| fast train | d | 60 | t |
| slow train | 420 - d | 45 | t |
The table gives us two equations: d = 60t and 420 - d = 45t. Just like we did with the two-part problems, we can combine these two equations.
The equation for the fast train isn't solvable on its own, but it does tell us that d is equal to 60t.
d = 60t
The other equation, which describes the slow train, can't be solved alone either. However, we can replace the d with its value from the first equation.
420 - d = 45t
Because we know that d is equal to 60t, we can replace the d in this equation with 60t. Now we have an equation we can solve.
420 - 60t = 45t
To solve this equation, we'll need to get t and its coefficients on one side of the equals sign and any other numbers on the other. We can start by canceling out the -60t on the left by adding 60t to both sides. 45t + 60t is 105t.
420 = 105t
Now we just need to get rid of the coefficient next to t. We can do this by dividing both sides by 105. 420 / 105 is 4.
4 = t
t = 4. In other words, the time it takes the trains to meet is 4 hours. Our problem is solved!
If you want to be sure of your answer, you can check it by using the distance equation with t equal to 4. For our fast train, the equation would be d = 60 ⋅ 4. 60 ⋅ 4 is 240, so the distance our fast train traveled would be 240 miles. For our slow train, the equation would be d = 45 ⋅ 4. 45 ⋅ 4 is 180, so the distance traveled by the slow train is 180 miles.
Remember how we said the distance the slow train and fast train travel should equal the total distance? 240 miles + 180 miles equals 420 miles, which is the total distance from our problem. Our answer is correct.
Practice problem 1
Here's another intersecting distance problem. It's similar to the one we just solved. See if you can solve it on your own. When you're finished, scroll down to see the answer and an explanation.
Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove an average of 70 mph. How long did they drive before they met up?
Problem 1 answer
Here's practice problem 1:
Jon and Dani live 270 miles apart. One day, they decided to drive toward each other and hang out wherever they met. Jon drove an average of 65 mph, and Dani drove 70 mph. How long did they drive before they met up?
Answer: 2 hours.
Let's solve this problem like we solved the others. First, try making the chart. It should look like this:
| distance | rate | time | |
|---|---|---|---|
| Jon | d | 65 | t |
| Dani | 270 - d | 70 | t |
Here's how we filled in the chart:
- Distance: Together, Dani and Jon will have covered the total distance between them by the time they meet up. That's 270. Jon's distance is represented by d, so Dani's distance is 270 - d.
- Rate: The problem tells us Dani and Jon's speeds. Dani drives 65 mph, and Jon drives 70 mph.
- Time: Because Jon and Dani drive the same amount of time before they meet up, both of their travel times are represented by t.
Now we have two equations. The equation for Jon's travel is d = 65t. The equation for Dani's travel is 270 - d = 70t. To solve this problem, we'll need to combine them.
The equation for Jon tells us that d is equal to 65t. This means we can combine the two equations by replacing the d in Dani's equation with 65t.
270 - 65t = 70t
Let's get t on one side of the equation and a number on the other. The first step to doing this is to get rid of -65t on the left side. We'll cancel it out by adding 65t to both sides: 70t + 65t is 135t.
270 = 135t
All that's left to do is to get rid of the 135 next to the t. We can do this by dividing both sides by 135: 270 / 135 is 2.
2 = t
That's it. t is equal to 2. We have the answer to our problem: Dani and Jon drove 2 hours before they met up.
Overtaking distance problems
The final type of distance problem we'll discuss in this lesson is a problem in which one moving object overtakes—or passes—another. Here's a typical overtaking problem:
The Hill family and the Platter family are going on a road trip. The Hills left 3 hours before the Platters, but the Platters drive an average of 15 mph faster. If it takes the Platter family 13 hours to catch up with the Hill family, how fast are the Hills driving?
You can picture the moment the Platter family left for the road trip a little like this:
The problem tells us that the Platter family will catch up with the Hill family in 13 hours and asks us to use this information to find the Hill family's rate. Like some of the other problems we've solved in this lesson, it might not seem like we have enough information to solve this problem—but we do. Let's start making our chart. The distance can be d for both the Hills and the Platters—when the Platters catch up with the Hills, both families will have driven the exact same distance.
| distance | rate | time | |
|---|---|---|---|
| the Hills | d | ||
| the Platters | d |
Filling in the rate and time will require a little more thought. We don't know the rate for either family—remember, that's what we're trying to find out. However, we do know that the Platters drove 15 mph faster than the Hills. This means if the Hill family's rate is r, the Platter family's rate would be r + 15.
| distance | rate | time | |
|---|---|---|---|
| the Hills | d | r | |
| the Platters | d | r + 15 |
Now all that's left is the time. We know it took the Platters 13 hours to catch up with the Hills. However, remember that the Hills left 3 hours earlier than the Platters—which means when the Platters caught up, they'd been driving 3 hours more than the Platters. 13 + 3 is 16, so we know the Hills had been driving 16 hours by the time the Platters caught up with them.
| distance | rate | time | |
|---|---|---|---|
| the Hills | d | r | 16 |
| the Platters | d | r + 15 | 13 |
Our chart gives us two equations. The Hill family's trip can be described by d = r ⋅ 16. The equation for the Platter family's trip is d = (r + 15) ⋅ 13. Just like with our other problems, we can combine these equations by replacing a variable in one of them.
The Hill family equation already has the value of d equal to r ⋅ 16. So we'll replace the d in the Platter equation with r ⋅ 16. This way, it will be an equation we can solve.
r ⋅ 16 = (r + 15) ⋅ 13
First, let's simplify the right side: r ⋅ 16 is 16r.
16r = (r + 15) ⋅ 13
Next, we'll simplify the right side and multiply (r + 15) by 13.
16r = 13r + 195
We can get both r and their coefficients on the left side by subtracting 13r from 16r : 16r - 13r is 3r.
3r = 195
Now all that's left to do is get rid of the 3 next to the r. To do this, we'll divide both sides by 3: 195 / 3 is 65.
r = 65
So there's our answer: r = 65. The Hill family drove an average of 65 mph.
You can solve any overtaking problem the same way we solved this one. Just remember to pay special attention when you're setting up your chart. Just like the Hill family did in this problem, the person or vehicle who started moving first will always have a greater travel time.
Practice problem 2
Try solving this problem. It's similar to the problem we just solved. When you're finished, scroll down to see the answer and an explanation.
A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?
Problem 2 answer
Here's practice problem 2:
A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?
Answer: 4 p.m.
To solve this problem, start by making a chart. Here's how it should look:
| distance | rate | time | |
|---|---|---|---|
| fast train | d | 80 | t |
| slow train | d | 60 | t + 1 |
Here's an explanation of the chart:
- Distance: Both trains will have traveled the same distance by the time the fast train catches up with the slow one, so the distance for both is d.
- Rate: The problem tells us how fast each train was going. The fast train has a rate of 80 mph, and the slow train has a rate of 60 mph.
- Time: We'll use t to represent the fast train's travel time before it catches up. Because the slow train started an hour before the fast one, it will have been traveling one hour more by the time the fast train catches up. It's t + 1.
Now we have two equations. The equation for the fast train is d = 80t. The equation for the slow train is d = 60 (t + 1). To solve this problem, we'll need to combine the equations.
The equation for the fast train says d is equal to 80t. This means we can combine the two equations by replacing the d in the slow train's equation with 80t.
80t = 60 (t + 1)
First, let's simplify the right side of the equation: 60 ⋅ (t + 1) is 60t + 60.
80t = 60t + 60
To solve the equation, we'll have to get t on one side of the equals sign and a number on the other. We can get rid of 60t on the right side by subtracting 60t from both sides: 80t - 60t is 20t.
20t = 60
Finally, we can get rid of the 20 next to t by dividing both sides by 20. 60 divided by 20 is 3.
t = 3
So t is equal to 3. The fast train traveled for 3 hours. However, it's not the answer to our problem. Let's look at the original problem again. Pay attention to the last sentence, which is the question we're trying to answer.
A train moving 60 mph leaves the station at noon. An hour later, a train moving 80 mph leaves heading the same direction on a parallel track. What time does the second train catch up to the first?
Our problem doesn't ask how long either of the trains traveled. It asks what time the second train catches up with the first.
The problem tells us that the slow train left at noon and the fast one left an hour later. This means the fast train left at 1 p.m. From our equations, we know the fast train traveled 3 hours. 1 + 3 is 4, so the fast train caught up with the slow one at 4 p.m. The answer to the problem is 4 p.m.
Assessment
Want even more practice? Try out a short assessment to test your skills by clicking the link below: