Word problems distance problems

Related Pages
Rate Distance Time Word Problems
Distance Problems
Average Speed Problems

What Are Distance Word Problems Or Distance Rate Time Problems?

Distance problems are word problems that involve the distance
an object will travel at a certain average rate for a given
period of time.

The formula for distance problems is:
distance = rate × time or
d = r × t

Things to watch out for:
Make sure that you change the units when necessary. For example, if the rate is given in miles
per hour and the time is given in minutes then change the units appropriately.

It would be helpful to use a table to organize the information for distance problems. A table
helps you to think about one number at a time instead being confused by the question.

The following diagrams give the steps to solve Distance-Rate-Time Problems. Scroll down the page
for examples and solutions.

Distance Problems: Traveling At Different Rates

Example:
A bus traveling at an average rate of 50 kilometers per hour made the trip to town in 6 hours.
If it had traveled at 45 kilometers per hour, how many more minutes would it have taken to make
the trip?

Solution:
Step 1: Set up a rtd table.

	r	t	d
Case 1
Case 2

Step 2: Fill in the table with information given in the question.

A bus traveling at an average rate of 50 kilometers per hour
made the trip to town in 6 hours. If it had traveled at
45 kilometers per hour, how many more minutes would it
have taken to make the trip?

Let t = time to make the trip in Case 2.

	r	t	d
Case 1	50	6
Case 2	45	t

Step 3: Fill in the values for d using the formula d = rt

	r	t	d
Case 1	50	6	50 × 6 = 300
Case 2	45	t	45t

Step 4: Since the distances traveled in both cases are the same,
we get the equation:

45t = 300
Isolate variable t

Step 5: Beware — the
question asked for “how many more minutes would it
have taken to make the trip”, so we need to deduct the original 6 hours taken.

Answer: The time taken would have been 40 minutes longer.

How to solve different types of distance, rate, time problems?

The following examples illustrate three types of problems involving distance, rate and time: opposite directions, same direction and roundtrip.
Example 1:
Lea left home and drove toward the ferry office at an average speed of 22 km.h. Trevon left at the same time and drove in the opposite direction with an average speed of 43 km/h. How long does Trevon need to drive before they are 65 km apart?

Example 2:
Maria left home and traveled toward the capital at an average speed of 80 km/h. Some time later, Imani left traveling in the opposite direction with an average speed of 45 km/h. After Maria has traveled for six hours they were 705 km apart. Find the number of hours Imani traveled.

Example 3:
A passenger train traveled to the repair yards and back. On the trip there, it traveled 68 mph and on the return trip it went 85 mph. How long did the trip there take if the return trip took four hours?

Example 4:
A submarine traveled to St. Vincent and back. It took three hours less time to get there than it did to get back. The average speed on the trip there was 30 km/h. The average speed on the way back was 20 km/h. How many hours did the trip there take?

Example 5:
Micaela left the hardware store and traveled toward her friend’s house at an average speed of 25 km/h. Nicole left some time later traveling in the same direction at the average speed of 30 km/h. After traveling for five hours Nicole caught up with Micaela. How long did Micaela travel before Nicole caught up?

Example 6:
A passenger plane left New York and flew east. A jet left one hour later flying at 280 km/h in a effort to catch up to the passenger plane. After flying for seven hours the jet finally caught up. Find the passenger plane’s average speed.

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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!

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.

Before you get into distance, time and speed word problems, take a few minutes to read this first and understand: How to build your credit score in USA as an international student.

Problems involving Time, Distance and Speed are solved based on one simple formula.

Distance = Speed * Time

Which implies →

Speed = Distance / Time   and

Time = Distance / Speed

Solution

Solution

Solution

Distance covered = 4*40 = 160 miles

Example 4. If a person walks at 4 mph, he covers a certain distance. If he walks at 9 mph, he covers 7.5 miles more. How much distance did he actually cover?

Now we can see that the direct application of our usual formula Distance = Speed * Time or its variations cannot be done in this case and we need to put in extra effort to calculate the given parameters.

Let us see how this question can be solved.

Solution

For these kinds of questions, a table like this might make it easier to solve.

Distance	Speed	Time
d	4	t
d+7.5	9	t

 
Let the distance covered by that person be ‘d’.

Walking at 4 mph and covering a distance ‘d’ is done in a time of ‘d/4’

IF he walks at 9 mph, he covers 7.5 miles more than the actual distance d, which is ‘d+7.5’.

He does this in a time of (d+7.5)/9.

Since the time is same in both the cases →

d/4 = (d+7.5)/9            →        9d = 4(d+7.5)   →        9d=4d+30        →        d = 6.

Solution

Here, we see that the distance is same.

Let us assume that its usual speed is ‘s’ and time is ‘t’, then

Distance	Speed	Time
d	s	t min
d	S+1/3	t+30 min

 
s*t = (1/3)s*(t+30)      →        t = t/3 + 10      →        t = 15.

So the actual time taken to cover the distance is 15 minutes.

Note: Note the time is expressed in terms of ‘minutes’. When we express distance in terms of miles or kilometers, time is expressed in terms of hours and has to be converted into appropriate units of measurement.

Solved Questions on Trains

Example 1. X and Y are two stations which are 320 miles apart. A train starts at a certain time from X and travels towards Y at 70 mph. After 2 hours, another train starts from Y and travels towards X at 20 mph. At what time do they meet?

Solution

Let the time after which they meet be ‘t’ hours.

Then the time travelled by second train becomes ‘t-2’.

Now,

Distance covered by first train+Distance covered by second train = 320 miles

70t+20(t-2) = 320

Solving this gives t = 4.

Solution

Let the speed of the first train be ‘s’.

Distance covered by the first train in (2+4) hours = Distance covered by second train in 4 hours

Therefore, 6s = 60*4

Solving which gives s=40.

So the slower train is moving at the rate of 40 mph.
 

Questions on Boats/Airplanes

For problems with boats and streams,

Speed of the boat upstream (against the current) = Speed of the boat in still water – speed of the stream

[As the stream obstructs the speed of the boat in still water, its speed has to be subtracted from the usual speed of the boat]

Speed of the boat downstream (along with the current) = Speed of the boat in still water + speed of the stream

[As the stream pushes the boat and makes it easier for the boat to reach the destination faster, speed of the stream has to be added]

Similarly, for airplanes travelling with/against the wind,

Speed of the plane with the wind = speed of the plane + speed of the wind

Speed of the plane against the wind = speed of the plane – speed of the wind

Let us look at some examples.

Example 1. A man travels at 3 mph in still water. If the current’s velocity is 1 mph, it takes 3 hours to row to a place and come back. How far is the place?

Solution

Let the distance be ‘d’ miles.

Time taken to cover the distance upstream + Time taken to cover the distance downstream = 3

Speed upstream = 3-1 = 2 mph

Speed downstream = 3+1 = 4 mph

Solution

Let the speed of the plane be ‘a’ and that of the wind be ‘w’.

Our table looks like this:
 

	Distance	Speed	Time
With the wind	2400	a+w	4
Against the wind	2400	a-w	6

 
4(a+w) = 2400 and 6(a-w) = 2400

Expressing one unknown variable in terms of the other makes it easier to solve, which means

a+w = 600 → w=600-a

Substituting the value of w in the second equation,

a-w = 400

a-(600-a) = 400 → a = 500

The speed of the plane is 500 kmph and that of the wind is 100 kmph.
 

More solved examples on Speed, Distance and Time

Example 1. A boy travelled by train which moved at the speed of 30 mph. He then boarded a bus which moved at the speed of 40 mph and reached his destination. The entire distance covered was 100 miles and the entire duration of the journey was 3 hours. Find the distance he travelled by bus.

Solution

	Distance	Speed	Time
Train	d	30	t
Bus	100-d	40	3-t

Let the time taken by the train be ‘t’. Then that of bus is ‘3-t’.

The entire distance covered was 100 miles

So, 30t + 40(3-t) = 100

Solving which gives t=2.

Substituting the value of t in 40(3-t), we get the distance travelled by bus is 40 miles.

Alternatively, we can add the time and equate it to 3 hours, which directly gives the distance.

d/30 + (100-d)/40 = 3

Solution

Our table looks like this.

	Distance	Speed	Time
1st part of journey	d	100	t
2nd part of journey	630-d	110	6-t

Assuming the distance covered in the 1^st part of journey to be ‘d’, the distance covered in the second half becomes ‘630-d’.

Assuming the time taken for the first part of the journey to be ‘t’, the time taken for the second half becomes ‘6-t’.

From the first equation, d=100t

The second equation is 630-d = 110(6-t).

Substituting the value of d from the first equation, we get

630-100t = 110(6-t)

Solving this gives t=3.

Solution

	Distance	Speed	Time
First person	d	3	t
Second person	20-d	4	t

 
Assuming the distance travelled by the first person to be ‘d’, the distance travelled by the second person is ’20-d’.

The time is ‘t’ for both of them because when they meet, they would have walked for the same time.

Since time is same, we can equate as

d/3 = (20-d)/4

Solving this gives d=60/7 miles (8.5 miles approximately)

Then t = 20/7 hours

So the two persons meet after 2 6/7 hours.
 

Practice Questions for you to solve

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Problem 1: Click here

Answer 1: Click here

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Problem 2: Click here

Answer 2: Click here

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Chapter 8: Rational Expressions

8.8 Rate Word Problems: Speed, Distance and Time

Distance, rate and time problems are a standard application of linear equations. When solving these problems, use the relationship rate (speed or velocity) times time equals distance.

[latex]rcdot t=d[/latex]

For example, suppose a person were to travel 30 km/h for 4 h. To find the total distance, multiply rate times time or (30km/h)(4h) = 120 km.

The problems to be solved here will have a few more steps than described above. So to keep the information in the problem organized, use a table. An example of the basic structure of the table is below:

Example of a Distance, Rate and Time Chart

Who or What	Rate	Time	Distance

The third column, distance, will always be filled in by multiplying the rate and time columns together. If given a total distance of both persons or trips, put this information in the distance column. Now use this table to set up and solve the following examples.

Joey and Natasha start from the same point and walk in opposite directions. Joey walks 2 km/h faster than Natasha. After 3 hours, they are 30 kilometres apart. How fast did each walk?

Who or What	Rate	Time	Distance
Natasha	[latex]r[/latex]	[latex]text{3 h}[/latex]	[latex]text{3 h}(r)[/latex]
Joey	[latex]r + 2[/latex]	[latex]text{3 h}[/latex]	[latex]text{3 h}(r + 2)[/latex]

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]begin{array}{rrrrrrl} 3r&+&3(r&+&2)&=&30 \ 3r&+&3r&+&6&=&30 \ &&&-&6&&-6 \ hline &&&&dfrac{6r}{6}&=&dfrac{24}{6} \ \ &&&&r&=&4 text{ km/h} end{array}[/latex]

This means that Natasha walks at 4 km/h and Joey walks at 6 km/h.

Nick and Chloe left their campsite by canoe and paddled downstream at an average speed of 12 km/h. They turned around and paddled back upstream at an average rate of 4 km/h. The total trip took 1 hour. After how much time did the campers turn around downstream?

Who or What	Rate	Time	Distance
Downstream	[latex]text{12 km/h}[/latex]	[latex]t[/latex]	[latex]text{12 km/h } (t)[/latex]
Upstream	[latex]text{4 km/h}[/latex]	[latex](1 — t)[/latex]	[latex]text{4 km/h } (1 — t)[/latex]

The distance travelled downstream is the same distance that they travelled upstream. Therefore, the equation to be solved is:

[latex]begin{array}{rrlll} 12(t)&=&4(1&-&t) \ 12t&=&4&-&4t \ +4t&&&+&4t \ hline dfrac{16t}{16}&=&dfrac{4}{16}&& \ \ t&=&0.25&& end{array}[/latex]

This means the campers paddled downstream for 0.25 h and spent 0.75 h paddling back.

Terry leaves his house riding a bike at 20 km/h. Sally leaves 6 h later on a scooter to catch up with him travelling at 80 km/h. How long will it take her to catch up with him?

Who or What	Rate	Time	Distance
Terry	[latex]text{20 km/h}[/latex]	[latex]t[/latex]	[latex]text{20 km/h }(t)[/latex]
Sally	[latex]text{80 km/h}[/latex]	[latex](t — text{6 h})[/latex]	[latex]text{80 km/h }(t — text {6 h})[/latex]

The distance travelled by both is the same. Therefore, the equation to be solved is:

[latex]begin{array}{rrrrr} 20(t)&=&80(t&-&6) \ 20t&=&80t&-&480 \ -80t&&-80t&& \ hline dfrac{-60t}{-60}&=&dfrac{-480}{-60}&& \ \ t&=&8&& end{array}[/latex]

This means that Terry travels for 8 h and Sally only needs 2 h to catch up to him.

On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took 2.5 h. For how long did the car travel 40 km/h?

Who or What	Rate	Time	Distance
Fifty-five	[latex]text{55 km/h}[/latex]	[latex]t[/latex]	[latex]text{55 km/h }(t)[/latex]
Forty	[latex]text{40 km/h}[/latex]	[latex](text{2.5 h}-t)[/latex]	[latex]text{40 km/h }(text{2.5 h}-t)[/latex]

The distance travelled by both is 30 km. Therefore, the equation to be solved is:

[latex]begin{array}{rrrrrrr} 55(t)&+&40(2.5&-&t)&=&130 \ 55t&+&100&-&40t&=&130 \ &-&100&&&&-100 \ hline &&&&dfrac{15t}{15}&=&dfrac{30}{15} \ \ &&&&t&=&2 end{array}[/latex]

This means that the time spent travelling at 40 km/h was 0.5 h.

Distance, time and rate problems have a few variations that mix the unknowns between distance, rate and time. They generally involve solving a problem that uses the combined distance travelled to equal some distance or a problem in which the distances travelled by both parties is the same. These distance, rate and time problems will be revisited later on in this textbook where quadratic solutions are required to solve them.

Questions

For Questions 1 to 8, find the equations needed to solve the problems. Do not solve.

1. A is 60 kilometres from B. An automobile at A starts for B at the rate of 20 km/h at the same time that an automobile at B starts for A at the rate of 25 km/h. How long will it be before the automobiles meet?
2. Two automobiles are 276 kilometres apart and start to travel toward each other at the same time. They travel at rates differing by 5 km/h. If they meet after 6 h, find the rate of each.
3. Two trains starting at the same station head in opposite directions. They travel at the rates of 25 and 40 km/h, respectively. If they start at the same time, how soon will they be 195 kilometres apart?
4. Two bike messengers, Jerry and Susan, ride in opposite directions. If Jerry rides at the rate of 20 km/h, at what rate must Susan ride if they are 150 kilometres apart in 5 hours?
5. A passenger and a freight train start toward each other at the same time from two points 300 kilometres apart. If the rate of the passenger train exceeds the rate of the freight train by 15 km/h, and they meet after 4 hours, what must the rate of each be?
6. Two automobiles started travelling in opposite directions at the same time from the same point. Their rates were 25 and 35 km/h, respectively. After how many hours were they 180 kilometres apart?
7. A man having ten hours at his disposal made an excursion by bike, riding out at the rate of 10 km/h and returning on foot at the rate of 3 km/h. Find the distance he rode.
8. A man walks at the rate of 4 km/h. How far can he walk into the country and ride back on a trolley that travels at the rate of 20 km/h, if he must be back home 3 hours from the time he started?

Solve Questions 9 to 22.

1. A boy rides away from home in an automobile at the rate of 28 km/h and walks back at the rate of 4 km/h. The round trip requires 2 hours. How far does he ride?
2. A motorboat leaves a harbour and travels at an average speed of 15 km/h toward an island. The average speed on the return trip was 10 km/h. How far was the island from the harbour if the trip took a total of 5 hours?
3. A family drove to a resort at an average speed of 30 km/h and later returned over the same road at an average speed of 50 km/h. Find the distance to the resort if the total driving time was 8 hours.
4. As part of his flight training, a student pilot was required to fly to an airport and then return. The average speed to the airport was 90 km/h, and the average speed returning was 120 km/h. Find the distance between the two airports if the total flying time was 7 hours.
5. Sam starts travelling at 4 km/h from a campsite 2 hours ahead of Sue, who travels 6 km/h in the same direction. How many hours will it take for Sue to catch up to Sam?
6. A man travels 5 km/h. After travelling for 6 hours, another man starts at the same place as the first man did, following at the rate of 8 km/h. When will the second man overtake the first?
7. A motorboat leaves a harbour and travels at an average speed of 8 km/h toward a small island. Two hours later, a cabin cruiser leaves the same harbour and travels at an average speed of 16 km/h toward the same island. How many hours after the cabin cruiser leaves will it be alongside the motorboat?
8. A long distance runner started on a course, running at an average speed of 6 km/h. One hour later, a second runner began the same course at an average speed of 8 km/h. How long after the second runner started will they overtake the first runner?
9. Two men are travelling in opposite directions at the rate of 20 and 30 km/h at the same time and from the same place. In how many hours will they be 300 kilometres apart?
10. Two trains start at the same time from the same place and travel in opposite directions. If the rate of one is 6 km/h more than the rate of the other and they are 168 kilometres apart at the end of 4 hours, what is the rate of each?
11. Two cyclists start from the same point and ride in opposite directions. One cyclist rides twice as fast as the other. In three hours, they are 72 kilometres apart. Find the rate of each cyclist.
12. Two small planes start from the same point and fly in opposite directions. The first plane is flying 25 km/h slower than the second plane. In two hours, the planes are 430 kilometres apart. Find the rate of each plane.
13. On a 130-kilometre trip, a car travelled at an average speed of 55 km/h and then reduced its speed to 40 km/h for the remainder of the trip. The trip took a total of 2.5 hours. For how long did the car travel at 40 km/h?
14. Running at an average rate of 8 m/s, a sprinter ran to the end of a track and then jogged back to the starting point at an average of 3 m/s. The sprinter took 55 s to run to the end of the track and jog back. Find the length of the track.

Answer Key 8.8