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Solving Uniform Motion Applications

Solving 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, \(D=rt,\) 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.

  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.
    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 _____, Rate, Time, and Distance. The rest of the cells are blank.
  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.

Example

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.
    Pittsburgh and Washington, DC, are represented by two separate lines. There is a line marked Express Train from Pittsburgh to Washington that is 12 mph faster and 4 hours long. There is a line marked Local Train from Pittsburgh to Washington that take 5 hours. The space between Pittsburgh and Washington is marked distance.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 Time header cell, we have 4 and then 5. The rest of the cells are blank.
  • 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 \(r+12.\)

\(\begin{array}{ccc}\hfill r& =\hfill & \text{speed of the local train}\hfill \\ \hfill r+12& =\hfill & \text{speed of the express train}\hfill \end{array}\)

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.

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. Below the Distance header cell, we have 4 times the quantity (r plus 12) and then 5r.

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.

\(\begin{array}{cccc}\phantom{\rule{2.7em}{0ex}}\text{express train}\hfill & & & \text{60 mph (4 hours) = 240 miles}\hfill \\ \phantom{\rule{2.7em}{0ex}}\text{local train}\hfill & & & \text{48 mph (5 hours) = 240 miles ✓}\hfill \end{array}\)

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 the example above, 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.

Example

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.Christopher and Parents are represented by two separate lines. The distance between these two lines is marked 115 miles. Lunch is also located between Christopher and Parents. There is an arrow from Christopher that is marked 10 mph faster and 1.5 hours. There is an arrow from Parents marked 1 hour. These two arrows meet somewhere between Christopher and Parents.
  • 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.
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 rate header cell, we have r plus 10 and r. Below the time header cell, we have 1.5 and 1. Below the distance header cell, we have 1.5 times the quantity (r plus 10), r, and 115.

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.

\(\begin{array}{cccc}\text{Now solve this equation.}\hfill & & & \hfill \begin{array}{ccc}\hfill 1.5\left(r+10\right)+r& =\hfill & 115\hfill \\ \hfill 1.5r+15+r& =\hfill & 115\hfill \\ \hfill 2.5r+15& =\hfill & 115\hfill \\ \hfill 2.5r& =\hfill & 100\hfill \\ \hfill r& =\hfill & 40\hfill \end{array}\hfill \\ & & & \text{So the parents’ speed was 40 mph.}\hfill \\ \text{Christopher’s speed is}\phantom{\rule{0.2em}{0ex}}r+10.\hfill & & & \hfill \begin{array}{c}\hfill r+10\hfill \\ \hfill 40+10\hfill \\ \hfill 50\hfill \end{array}\hfill \\ & & & \text{Christopher’s speed was 50 mph.}\hfill \end{array}\)

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

\(\begin{array}{cccccc}\phantom{\rule{2.7em}{0ex}}\text{Christopher drove}\hfill & & & \text{50 mph (1.5 hours)}\hfill & =\hfill & \text{75 miles}\hfill \\ \phantom{\rule{2.7em}{0ex}}\text{His parents drove}\hfill & & & \text{40 mph (1 hours)}\hfill & =\hfill & \underset{\text{_______}}{\text{40 miles}}\hfill \\ & & & & & \text{115 miles}\hfill \end{array}\)

\(\begin{array}{cccc}\begin{array}{c}\mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with a complete sentence.}\hfill \\ \end{array}\hfill & & & \begin{array}{c}\text{Christopher’s speed was 50 mph.}\hfill \\ \text{His parents’ speed was 40 mph.}\hfill \end{array}\hfill \end{array}\)

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?

Example

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.
    West and East are represented by two separate lines. The distance between these two lines is marked 325 miles. Rest stop is also located between West and East. There is an arrow from Rest stop heading toward West that is marked 70 mph. There is an arrow from Rest stop heading toward East that is marked 60 mph.
  • 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.
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. Below the time head cell, we have t and t. Below the Distance header cell we have 70t, 60t, and 325.

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.

\(\phantom{\rule{2.7em}{0ex}}\begin{array}{cccccc}\text{Now solve this equation.}\hfill & & & \hfill 70t+60t& =\hfill & 325\hfill \\ & & & \hfill 130t& =\hfill & 325\hfill \\ & & & \hfill t& =\hfill & 2.5\hfill \end{array}\)

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.

\(\begin{array}{cccccc}\text{Truck going West}\hfill & & & \text{70 mph (2.5 hours)}\hfill & =\hfill & \text{175 miles}\hfill \\ \text{Truck going East}\hfill & & & \text{60 mph (2.5 hours)}\hfill & =\hfill & \underset{\text{________}}{\text{150 miles}}\hfill \\ & & & & & \text{325 miles}\hfill \end{array}\)

\(\begin{array}{cccc}\mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question with a complete sentence.}\hfill & & & \text{It will take the trucks 2.5 hours to be 325 miles apart.}\hfill \end{array}\)

Matching Units in Problems

It is important to make sure the units match when we use the distance rate and time formula. For instance, if the rate is in miles per hour, then the time must be in hours.

Example

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:

\(\begin{array}{}\\ \text{30 minutes is}\phantom{\rule{0.4em}{0ex}}\frac{30}{60}\phantom{\rule{0.4em}{0ex}}\text{or}\phantom{\rule{0.4em}{0ex}}\frac{1}{2}\phantom{\rule{0.4em}{0ex}}\text{hour}\hfill \\ \text{15 minutes is}\phantom{\rule{0.4em}{0ex}}\frac{15}{60}\phantom{\rule{0.4em}{0ex}}\text{or}\phantom{\rule{0.4em}{0ex}}\frac{1}{4}\phantom{\rule{0.4em}{0ex}}\text{hour}\hfill \end{array}\)

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.

Example

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 faster than 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 \(1.75r.\)

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.

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

This lesson is part of:

Math Models and Geometry II

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