Distance Word Problems

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

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.

distanceratetime
the Hillsd
the Plattersd

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.

distanceratetime
the Hillsdr
the Plattersdr + 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.

distanceratetime
the Hillsdr16
the Plattersdr + 1513

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:

distanceratetime
fast traind80t
slow traind60t + 1

Here's an explanation of the chart:

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

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