Solve Uniform Motion Applications

When planning a road trip, it often helps to know how long it will take to reach the destination or how far to travel each day. We would use the distance, rate, and time formula, $D = r t,$ which we have already seen.

In this section, we will use this formula in situations that require a little more algebra to solve than the ones we saw earlier. Generally, we will be looking at comparing two scenarios, such as two vehicles travelling at different rates or in opposite directions. When the speed of each vehicle is constant, we call applications like this uniform motion problems.

Our problem-solving strategies will still apply here, but we will add to the first step. The first step will include drawing a diagram that shows what is happening in the example. Drawing the diagram helps us understand what is happening so that we will write an appropriate equation. Then we will make a table to organize the information, like we did for the money applications.

The steps are listed here for easy reference:

Use a Problem-Solving Strategy in Distance, Rate, and Time Applications.

Read the problem. Make sure all the words and ideas are understood.
- Draw a diagram to illustrate what it happening.
- Create a table to organize the information.
- Label the columns rate, time, distance.
- List the two scenarios.
- Write in the information you know.
Identify what we are looking for.
Name what we are looking for. Choose a variable to represent that quantity.
- Complete the chart.
- Use variable expressions to represent that quantity in each row.
- Multiply the rate times the time to get the distance.
Translate into an equation.
- Restate the problem in one sentence with all the important information.
- Then, translate the sentence into an equation.
Solve the equation using good algebra techniques.
Check the answer in the problem and make sure it makes sense.
Answer the question with a complete sentence.

An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in 4 hours and the local train takes 5 hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains.

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening. Shown below is a sketch of what is happening in the example.
Create a table to organize the information.
Label the columns “Rate,” “Time,” and “Distance.”
List the two scenarios.
Write in the information you know.

Step 2. Identify what we are looking for.

We are asked to find the speed of both trains.
Notice that the distance formula uses the word “rate,” but it is more common to use “speed” when we talk about vehicles in everyday English.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Complete the chart
Use variable expressions to represent that quantity in each row.
We are looking for the speed of the trains. Let’s let r represent the speed of the local train. Since the speed of the express train is 12 mph faster, we represent that as $r + 12 .$

$\begin{matrix} r & = & speed of the local train \\ r + 12 & = & speed of the express train \end{matrix}$

Fill in the speeds into the chart.

A table with three rows and four columns. The first row is a header row and reads from left to right _____, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have Express and then Local. Below the Rate header cell, we have r plus 12 and then r. Below the Time header cell, we have 4 and then 5. The rest of the cells are blank.

Multiply the rate times the time to get the distance.

Step 4. Translate into an equation.

Restate the problem in one sentence with all the important information.
Then, translate the sentence into an equation.
The equation to model this situation will come from the relation between the distances. Look at the diagram we drew above. How is the distance travelled by the express train related to the distance travelled by the local train?
Since both trains leave from Pittsburgh and travel to Washington, D.C. they travel the same distance. So we write:

The sentence, “The distance traveled by the express train equals the distance traveled by the local train,” can be translated to an equation. Translate “distance traveled by the express train” to 4 times the quantity r plus 12, and translate “distance traveled by the local train” to 5r. The full equation is 4 times the quantity r plus 12 equals 5r.

Step 5. Solve the equation using good algebra techniques.

Now solve this equation.	So the speed of the local train is 48 mph.
Find the speed of the express train.	The speed of the express train is 60 mph.

Step 6. Check the answer in the problem and make sure it makes sense.

Comparison of express and local train travel, detailing speed, time, and how both achieve a 240-mile distance.
express train	$60 mph (4 hours) = 240 miles$
local train	$48 mph (5 hours) = 240 miles ✓$

Step 7. Answer the question with a complete sentence.

The speed of the local train is 48 mph and the speed of the express train is 60 mph.

In Example 1, the last example, we had two trains traveling the same distance. The diagram and the chart helped us write the equation we solved. Let’s see how this works in another case.

Christopher and his parents live 115 miles apart. They met at a restaurant between their homes to celebrate his mother’s birthday. Christopher drove 1.5 hours while his parents drove 1 hour to get to the restaurant. Christopher’s average speed was 10 miles per hour faster than his parents’ average speed. What were the average speeds of Christopher and of his parents as they drove to the restaurant?

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening. Below shows a sketch of what is happening in the example.
Create a table to organize the information.
Label the columns rate, time, distance.
List the two scenarios.
Write in the information you know.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have Christopher and Parents. Below the time header cell, we have 1.5 and 1. The extra cell contains 115. The rest of the cells are blank.

Step 2. Identify what we are looking for.

We are asked to find the average speeds of Christopher and his parents.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

Complete the chart.
Use variable expressions to represent that quantity in each row.
We are looking for their average speeds. Let’s let r represent the average speed of the parents. Since the Christopher’s speed is 10 mph faster, we represent that as $r + 10 .$

Fill in the speeds into the chart.

Multiply the rate times the time to get the distance.

Step 4. Translate into an equation.

Restate the problem in one sentence with all the important information.
Then, translate the sentence into an equation.
Again, we need to identify a relationship between the distances in order to write an equation. Look at the diagram we created above and notice the relationship between the distance Christopher traveled and the distance his parents traveled.

The distance Christopher travelled plus the distance his parents travel must add up to 115 miles. So we write:

The sentence, “The distance traveled by Christopher plus the distance traveled by his parents equals 115 miles,” can be translated to an equation. Translate “distance traveled by Christopher” to 1.5 times the quantity r plus 10, and translate “distance traveled by his parents” to r. The full equation is 1.5 times the quantity r plus 10, plus r equals 115.

Step 5. Solve the equation using good algebra techniques.

Step-by-step solution for a word problem calculating parent's and Christopher's speeds using an algebraic equation.
Now solve this equation.	$\begin{matrix} 1.5 (r + 10) + r & = & 115 \\ 1.5 r + 15 + r & = & 115 \\ 2.5 r + 15 & = & 115 \\ 2.5 r & = & 100 \\ r & = & 40 \end{matrix}$
	So the parents' speed was 40 mph.
Christopher's speed is $r + 10$ .	$\begin{matrix} r + 10 \\ 40 + 10 \\ 50 \end{matrix}$
	Christopher's speed was 50 mph.

Step 6. Check the answer in the problem and make sure it makes sense.

Calculated distances traveled by Christopher and his parents, summing up to a total journey distance.
Christopher drove	$\begin{matrix} 50 mph (1.5 hours) & = & 75 miles \end{matrix}$
His parents drove	$\begin{matrix} 40 mph (1 hours) & = & \underset{_______}{40 miles} \\ 115 miles \end{matrix}$

This table outlines a problem-solving step, specifically providing the calculated speeds for Christopher and his parents.
Step 7. Answer the question with a complete sentence.	Christopher's speed was 50 mph. His parents' speed was 40 mph.

As you read the next example, think about the relationship of the distances traveled. Which of the previous two examples is more similar to this situation?

Two truck drivers leave a rest area on the interstate at the same time. One truck travels east and the other one travels west. The truck traveling west travels at 70 mph and the truck traveling east has an average speed of 60 mph. How long will they travel before they are 325 miles apart?

Solution

Step 1. Read the problem. Make sure all the words and ideas are understood.

Draw a diagram to illustrate what it happening.
Create a table to organize the information.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have West and East. Below the rate header cell, we have 70 and 60. The extra cell contains 325. The rest of the cells are blank.

Step 2. Identify what we are looking for.

We are asked to find the amount of time the trucks will travel until they are 325 miles apart.

Step 3. Name what we are looking for. Choose a variable to represent that quantity.

We are looking for the time travelled. Both trucks will travel the same amount of time. Let’s call the time t. Since their speeds are different, they will travel different distances.
Complete the chart.

Step 4. Translate into an equation.

We need to find a relation between the distances in order to write an equation. Looking at the diagram, what is the relationship between the distance each of the trucks will travel?
The distance traveled by the truck going west plus the distance travelled by the truck going east must add up to 325 miles. So we write:

Distance traveled by westbound truck plus distance traveled by eastbound truck equals 325. The first part corresponds to 70t and the second part corresponds to 60.

Step 5. Solve the equation using good algebra techniques.

Step-by-step solution for the algebraic equation 70t + 60t = 325, demonstrating the process to find the value of 't'.
Now solve this equation.	$\begin{matrix} 70 t + 60 t & = & 325 \\ 130 t & = & 325 \\ t & = & 2.5 \end{matrix}$

So it will take the trucks 2.5 hours to be 325 miles apart.

Step 6. Check the answer in the problem and make sure it makes sense.

Distances covered by two trucks traveling in opposite directions over 2.5 hours, including total distance.
Truck going West	$\begin{matrix} 70 mph (2.5 hours) & = & 175 miles \end{matrix}$
Truck going East	$\begin{matrix} 60 mph (2.5 hours) & = & \underset{________}{150 miles} \\ 325 miles \end{matrix}$

This table presents a problem-solving step and its corresponding answer.
Step 7. Answer the question with a complete sentence.	It will take the trucks 2.5 hours to be 325 miles apart.

When Katie Mae walks to school, it takes her 30 minutes. If she rides her bike, it takes her 15 minutes. Her speed is three miles per hour faster when she rides her bike than when she walks. What are her walking speed and her speed riding her bike?

Solution

First, we draw a diagram that represents the situation to help us see what is happening.

A house and a school are represented by two separate lines. There is a line marked walking from the house to the school that takes 30 minutes. There is a line marked biking from the house to the school that take 15 minutes and is 3 mph faster. The space between the house and school is marked distance.

We are asked to find her speed walking and riding her bike. Let’s call her walking speed r. Since her biking speed is three miles per hour faster, we will call that speed $r + 3 .$ We write the speeds in the chart.

The speed is in miles per hour, so we need to express the times in hours, too, in order for the units to be the same. Remember, one hour is 60 minutes. So:

\begin{matrix} 30 minutes is \frac{30}{60} or \frac{1}{2} hour \\ 15 minutes is \frac{15}{60} or \frac{1}{4} hour \end{matrix}

Next, we multiply rate times time to fill in the distance column.

A table with three rows and four columns. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have walk and bike. Below the rate header cell, we have r and r plus 3. Below the time header cell, we have 1/2 and 1/4. Below the distance cell we have 1/2 times r and 1/4 times the quantity (r plus 3).

The equation will come from the fact that the distance from Katie Mae’s home to her school is the same whether she is walking or riding her bike.

So we say:


Translate into an equation.
Solve this equation.
Clear the fractions by multiplying by the LCD of all the fractions in the equation.
Simplify.	6 mph (Katie Mae's biking speed)
Let's check if this works. Walk 3 mph (0.5 hour) = 1.5 miles Bike 6 mph (0.25 hour) = 1.5 miles
Yes, either way Katie Mae travels 1.5 miles to school.	Katie Mae’s walking speed is 3 mph. Her speed riding her bike is 6 mph.

In the distance, rate, and time formula, time represents the actual amount of elapsed time (in hours, minutes, etc.). If a problem gives us starting and ending times as clock times, we must find the elapsed time in order to use the formula.

Hamilton loves to travel to Las Vegas, 255 miles from his home in Orange County. On his last trip, he left his house at 2:00 pm. The first part of his trip was on congested city freeways. At 4:00 pm, the traffic cleared and he was able to drive through the desert at a speed 1.75 times as fast as when he drove in the congested area. He arrived in Las Vegas at 6:30 pm. How fast was he driving during each part of his trip?

Solution

A diagram will help us model this trip.

Home (2:00 pm) and Las Vegas (6:30 pm) are represented by two separate lines. The space between home and Las Vegas is marked 255 miles. There is an arrow marked city driving from Home/2:00 pm to 4:00 pm. Then there is an arrow marked desert driving from the tip of the previous one at 4:00 pm to Las Vegas/6:30 pm.

Next, we create a table to organize the information.

We know the total distance is 255 miles. We are looking for the rate of speed for each part of the trip. The rate in the desert is 1.75 times the rate in the city. If we let $r =$ the rate in the city, then the rate in the desert is $1.75 r .$

The times here are given as clock times. Hamilton started from home at 2:00 pm and entered the desert at 4:30 pm. So he spent two hours driving the congested freeways in the city. Then he drove faster from 4:00 pm until 6:30 pm in the desert. So he drove 2.5 hours in the desert.

Now, we multiply the rates by the times.

A table with three rows and four columns and an extra cell at the bottom of the fourth column. The first row is a header row and reads from left to right blank, Rate (mph), Time (hrs), and Distance (miles). Below the blank header cell, we have city and desert. Below the rate header cell, we have r and 1.75r. Below the time head cell, we have 2 and 2.5. Below the Distance header cell we have 2r, 2.5 times 1.75r, and 255.

By looking at the diagram below, we can see that the sum of the distance driven in the city and the distance driven in the desert is 255 miles.


Translate into an equation.
Solve this equation.
Check.
	Hamilton drove 40 mph in the city and 70 mph in the desert.

Key Concepts

Distance, Rate, and Time
- D = rt where D = distance, r = rate, t = time
Problem-Solving Strategy—Distance, Rate, and Time Applications
1. Read the problem. Make sure all the words and ideas are understood.
  Draw a diagram to illustrate what it happening.
  Create a table to organize the information: Label the columns rate, time, distance. List the two scenarios. Write in the information you know.
2. Identify what we are looking for.
3. Name what we are looking for. Choose a variable to represent that quantity.
  Complete the chart.
  Use variable expressions to represent that quantity in each row.
  Multiply the rate times the time to get the distance.
4. Translate into an equation.
  Restate the problem in one sentence with all the important information.
  Then, translate the sentence into an equation.
5. Solve the equation using good algebra techniques.
6. Check the answer in the problem and make sure it makes sense.
7. Answer the question with a complete sentence.

Practice Makes Perfect

Solve Uniform Motion Applications

In the following exercises, solve.

Lilah is moving from Portland to Seattle. It takes her three hours to go by train. Mason leaves the train station in Portland and drives to the train station in Seattle with all Lilah’s boxes in his car. It takes him 2.4 hours to get to Seattle, driving at 15 miles per hour faster than the speed of the train. Find Mason’s speed and the speed of the train.

Solution

Mason 75 mph, train 60 mph

Kathy and Cheryl are walking in a fundraiser. Kathy completes the course in 4.8 hours and Cheryl completes the course in 8 hours. Kathy walks two miles per hour faster than Cheryl. Find Kathy’s speed and Cheryl’s speed.

Two busses go from Sacramento for San Diego. The express bus makes the trip in 6.8 hours and the local bus takes 10.2 hours for the trip. The speed of the express bus is 25 mph faster than the speed of the local bus. Find the speed of both busses.

Solution

express bus 75mph, local 50 mph

A commercial jet and a private airplane fly from Denver to Phoenix. It takes the commercial jet 1.1 hours for the flight, and it takes the private airplane 1.8 hours. The speed of the commercial jet is 210 miles per hour faster than the speed of the private airplane. Find the speed of both airplanes.

Saul drove his truck 3 hours from Dallas towards Kansas City and stopped at a truck stop to get dinner. At the truck stop he met Erwin, who had driven 4 hours from Kansas City towards Dallas. The distance between Dallas and Kansas City is 542 miles, and Erwin’s speed was eight miles per hour slower than Saul’s speed. Find the speed of the two truckers.

Solution

Saul 82 mph, Erwin 74 mph

Charlie and Violet met for lunch at a restaurant between Memphis and New Orleans. Charlie had left Memphis and drove 4.8 hours towards New Orleans. Violet had left New Orleans and drove 2 hours towards Memphis, at a speed 10 miles per hour faster than Charlie’s speed. The distance between Memphis and New Orleans is 394 miles. Find the speed of the two drivers.

Sisters Helen and Anne live 332 miles apart. For Thanksgiving, they met at their other sister’s house partway between their homes. Helen drove 3.2 hours and Anne drove 2.8 hours. Helen’s average speed was four miles per hour faster than Anne’s. Find Helen’s average speed and Anne’s average speed.

Solution

Helen 57.2 mph, Anne 53.2 mph

Ethan and Leo start riding their bikes at the opposite ends of a 65-mile bike path. After Ethan has ridden 1.5 hours and Leo has ridden 2 hours, they meet on the path. Ethan’s speed is six miles per hour faster than Leo’s speed. Find the speed of the two bikers.

Elvira and Aletheia live 3.1 miles apart on the same street. They are in a study group that meets at a coffee shop between their houses. It took Elvira half an hour and Aletheia two-thirds of an hour to walk to the coffee shop. Aletheia’s speed is 0.6 miles per hour slower than Elvira’s speed. Find both women’s walking speeds.

Solution

Aletheia 2.4 mph, Elvira 3 mph

DaMarcus and Fabian live 23 miles apart and play soccer at a park between their homes. DaMarcus rode his bike for three-quarters of an hour and Fabian rode his bike for half an hour to get to the park. Fabian’s speed was six miles per hour faster than DaMarcus’ speed. Find the speed of both soccer players.

Cindy and Richard leave their dorm in Charleston at the same time. Cindy rides her bicycle north at a speed of 18 miles per hour. Richard rides his bicycle south at a speed of 14 miles per hour. How long will it take them to be 96 miles apart?

Solution

3 hours

Matt and Chris leave their uncle’s house in Phoenix at the same time. Matt drives west on I-60 at a speed of 76 miles per hour. Chris drives east on I-60 at a speed of 82 miles per hour. How many hours will it take them to be 632 miles apart?

Two busses leave Billings at the same time. The Seattle bus heads west on I-90 at a speed of 73 miles per hour while the Chicago bus heads east at a speed of 79 miles an hour. How many hours will it take them to be 532 miles apart?

Solution

3.5 hours

Two boats leave the same dock in Cairo at the same time. One heads north on the Mississippi River while the other heads south. The northbound boat travels four miles per hour. The southbound boat goes eight miles per hour. How long will it take them to be 54 miles apart?

Lorena walks the path around the park in 30 minutes. If she jogs, it takes her 20 minutes. Her jogging speed is 1.5 miles per hour faster than her walking speed. Find Lorena’s walking speed and jogging speed.

Solution

walking 3 mph, jogging 4.5 mph

Julian rides his bike uphill for 45 minutes, then turns around and rides back downhill. It takes him 15 minutes to get back to where he started. His uphill speed is 3.2 miles per hour slower than his downhill speed. Find Julian’s uphill and downhill speed.

Cassius drives his boat upstream for 45 minutes. It takes him 30 minutes to return downstream. His speed going upstream is three miles per hour slower than his speed going downstream. Find his upstream and downstream speeds.

Solution

upstream 6 mph, downstream 9 mph

It takes Darline 20 minutes to drive to work in light traffic. To come home, when there is heavy traffic, it takes her 36 minutes. Her speed in light traffic is 24 miles per hour faster than her speed in heavy traffic. Find her speed in light traffic and in heavy traffic.

At 1:30 Marlon left his house to go to the beach, a distance of 7.6 miles. He rode his skateboard until 2:15, then walked the rest of the way. He arrived at the beach at 3:00. Marlon’s speed on his skateboard is 2.5 times his walking speed. Find his speed when skateboarding and when walking.

Solution

skateboarding 7.2 mph, walking 2.9 mph

Aaron left at 9:15 to drive to his mountain cabin 108 miles away. He drove on the freeway until 10:45, and then he drove on the mountain road. He arrived at 11:05. His speed on the freeway was three times his speed on the mountain road. Find Aaron’s speed on the freeway and on the mountain road.

Marisol left Los Angeles at 2:30 to drive to Santa Barbara, a distance of 95 miles. The traffic was heavy until 3:20. She drove the rest of the way in very light traffic and arrived at 4:20. Her speed in heavy traffic was 40 miles per hour slower than her speed in light traffic. Find her speed in heavy traffic and in light traffic.

Solution

heavy traffic 30 mph, light traffic 70 mph

Lizette is training for a marathon. At 7:00 she left her house and ran until 8:15, then she walked until 11:15. She covered a total distance of 19 miles. Her running speed was five miles per hour faster than her walking speed. Find her running and walking speeds.

Everyday Math

John left his house in Irvine at 8:35 am to drive to a meeting in Los Angeles, 45 miles away. He arrived at the meeting at 9:50. At 3:30 pm, he left the meeting and drove home. He arrived home at 5:18.

ⓐ What was his average speed on the drive from Irvine to Los Angeles?
ⓑ What was his average speed on the drive from Los Angeles to Irvine?
ⓓ John drove a total of 90 miles roundtrip. Find his average speed. (Round to the nearest tenth.)

Solution

Sarah wants to arrive at her friend’s wedding at 3:00. The distance from Sarah’s house to the wedding is 95 miles. Based on usual traffic patterns, Sarah predicts she can drive the first 15 miles at 60 miles per hour, the next 10 miles at 30 miles per hour, and the remainder of the drive at 70 miles per hour.

ⓐ How long will it take Sarah to drive the first 15 miles?
ⓑ How long will it take Sarah to drive the next 10 miles?
ⓓ What time should Sarah leave her house?

Writing Exercises

When solving a uniform motion problem, how does drawing a diagram of the situation help you?

Solution

Answers will vary.

When solving a uniform motion problem, how does creating a table help you?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

A table with two rows and four columns. The first row reads: 'I can...', 'Confidently', 'With Some Help', and 'No-I don't get it!' The second row reads: 'solve uniform motion applications' with the rest of the cells blank for students to fill out.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?