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By the end of this section, you will be able to:
  • Solve uniform motion applications
  • Solve work applications

Before you get started, take this readiness quiz.

If you miss a problem, go back to the section listed and review the material.

  1. An express train and a local bus leave Chicago to travel to Champaign. The express bus can make the trip in 2 hours and the local bus takes 5 hours for the trip. The speed of the express bus is 42 miles per hour faster than the speed of the local bus. Find the speed of the local bus.
    If you missed this problem, review [link] .
  2. Solve 1 3 x + 1 4 x = 5 6 .
    If you missed this problem, review [link] .
  3. Solve: 18 t 2 30 = −33 t .
    If you missed this problem, review [link] .

Solve uniform motion applications

We have solved uniform motion problems using the formula D = r t in previous chapters. We used a table like the one below to organize the information and lead us to the equation.

The above image is a table with 4 columns and three rows. The first row is the header row. The second column in the header row has the word “rate”. The third column has the word, “Time”. The fourth column says “Distance”. The rest of the spaces are blank.

The formula D = r t assumes we know r and t and use them to find D. If we know D and r and need to find t , we would solve the equation for t and get the formula t = D r .

We have also explained how flying with or against a current affects the speed of a vehicle. We will revisit that idea in the next example.

An airplane can fly 200 miles into a 30 mph headwind in the same amount of time it takes to fly 300 miles with a 30 mph tailwind. What is the speed of the airplane?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

The above image has two parallel arrows. The first arrow has its tip pointing to the right The words,”300 miles with the wind r plus 30” above the arrow tip. Below that is a squiggly line. To the left of the squiggly line it says, “Wind 30 miles per hour”. Below that is an arrow with its tip pointing to the left. Below that are the words, “200 miles against the wind r minus 30”.

We fill in the chart to organize the information.

We are looking for the speed of the airplane. Let r = the speed of the airplane.
When the plane flies with the wind, the wind increases its speed and the rate is r + 30 .
When the plane flies against the wind, the wind decreases its speed and the rate is r 30 .
Write in the rates.
Write in the distances.
Since D = r t , we solve for t and get D r .
We divide the distance by the rate in each row, and place the expression in the time column.
.
We know the times are equal and so we write our equation. 200 r 30 = 300 r + 30
We multiply both sides by the LCD.
200 ( r + 30 ) = 300 ( r 30 )
( r + 30 ) ( r 30 ) ( 200 r 30 ) = ( r + 30 ) ( r 30 ) ( 300 r + 30 )
Simplify. ( r + 30 ) ( 200 ) = ( r 30 ) ( 300 )
200 r + 6000 = 300 r 9000
Solve. 15000 = 100 r
150 = r
Check.
Is 150 mph a reasonable speed for an airplane? Yes. If the plane is traveling 150 mph and the wind is 30 mph:
Tailwind 150 + 30 = 180 mph 300 180 = 5 3 hours
Headwind 150 30 = 120 mph 200 120 = 5 3 hours
The times are equal, so it checks. The plane was traveling 150 mph.

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Link can ride his bike 20 miles into a 3 mph headwind in the same amount of time he can ride 30 miles with a 3 mph tailwind. What is Link’s biking speed?

15 mph

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Judy can sail her boat 5 miles into a 7 mph headwind in the same amount of time she can sail 12 miles with a 7 mph tailwind. What is the speed of Judy’s boat without a wind?

17 mph

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In the next example, we will know the total time resulting from travelling different distances at different speeds.

Jazmine trained for 3 hours on Saturday. She ran 8 miles and then biked 24 miles. Her biking speed is 4 mph faster than her running speed. What is her running speed?

Solution

This is a uniform motion situation. A diagram will help us visualize the situation.

The above image is a straight line with two arrow heads pointing to the right. At the far left, above the line, it reads, “run” and further down, slightly past the first arrow head, it reads, “bike”. Slightly below the line, to the far left, before the first arrow head, it reads, “8 miles” and to the far right, after the first arrow head it reads, “12 miles”. The region from the far left to the far right of the arrow is grouped to indicate the entire length of the line is 3 hours.

We fill in the chart to organize the information.

We are looking for Jazmine’s running speed. Let r = Jazmine’s running speed.
Her biking speed is 4 miles faster than her running speed. r + 4 = her biking speed
The distances are given, enter them into the chart.
Since D = r t , we solve for t and get t = D r .
We divide the distance by the rate in each row, and place the expression in the time column.
.
Write a word sentence. Her time plus the time biking is 3 hours.
Translate the sentence to get the equation. 8 r + 24 r + 4 = 3
Solve. r ( r + 4 ) ( 8 r + 24 r + 4 ) = 3 r ( r + 4 ) 8 ( r + 4 ) + 24 r = 3 r ( r + 4 ) 8 r + 32 + 24 r = 3 r 2 + 12 r 32 + 32 r = 3 r 2 + 12 r 0 = 3 r 2 20 r 32 0 = ( 3 r + 4 ) ( r 8 )
( 3 r + 4 ) = 0 ( r 8 ) = 0
r = 4 3 r = 8
Check. r = 4 3 r = 8
A negative speed does not make sense in this problem, so r = 8 is the solution.
Is 8 mph a reasonable running speed? Yes.
Run 8 mph 8 miles 8 mph = 1 hour Bike 12 mph 24 miles 12 mph = 2 hours Total 3 hours Jazmine’s running speed is 8 mph.

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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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