Solve Uniform Motion and Work Applications

By the end of this section, you will be able to: solve uniform motion applications, and solve work applications.

Solve uniform motion applications

We have solved uniform motion problems using the formula $D = rt$ in previous chapters. We used a table to organize the information and lead us to the equation.

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

t = \frac{D}{r}.

In each row of the rate table we divide the distance by the rate to fill in the time column. When two legs of a trip take equal times, we set the two time expressions equal; when their times differ by a known amount, or add to a known total, we build the equation from that relationship.

We have also explained how flying with or against a current affects the speed of a vehicle — a tailwind adds to the speed and a headwind subtracts from it.

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?

Let $r =$ the speed of the airplane in still air. Flying into the headwind the rate is $r - 30$ ; flying with the tailwind the rate is $r + 30$ . We divide each distance by its rate to get the time, and record everything in the chart.

	Rate	Time	Distance
Headwind	$r - 30$	$\tfrac{200}{r-30}$	$200$
Tailwind	$r + 30$	$\tfrac{300}{r+30}$	$300$

The two times are equal, so we write and solve the equation. Multiply both sides by the LCD $(r-30)(r+30)$ :

\begin{array}{lrcl} & \tfrac{200}{r-30} &=& \tfrac{300}{r+30} \\[4pt] \text{Clear the fractions.} & 200(r+30) &=& 300(r-30) \\[4pt] \text{Distribute.} & 200r + 6000 &=& 300r - 9000 \\[4pt] \text{Solve.} & 15000 &=& 100r \\[4pt] & 150 &=& r \end{array}

Check. Is $150$ mph reasonable for an airplane? Yes. At $150$ mph the tailwind speed is $180$ mph and $\tfrac{300}{180} = \tfrac{5}{3}$ hours; the headwind speed is $120$ mph and $\tfrac{200}{120} = \tfrac{5}{3}$ hours. The times are equal, so it checks. The plane was traveling $150$ mph.

An airplane can fly $200$ miles into a $30$ mph headwind in the same time it takes to fly $300$ miles with a $30$ mph tailwind. Find the speed of the airplane in still air, in mph (enter the number).

Mary takes a helicopter tour that flies $450$ miles against a $35$ mph headwind in the same time it flies $702$ miles with a $35$ mph tailwind. Find the speed of the helicopter in still air, in mph (enter the number).

A total time from two legs

Sometimes we know the total time for a trip made up of two legs traveled at different speeds. We still divide each distance by its rate to get each leg’s time, then add the two times to get the total.

Example. 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?

Let $r =$ Jazmine’s running speed, so $r + 4 =$ her biking speed.

	Rate	Time	Distance
Run	$r$	$\tfrac{8}{r}$	$8$
Bike	$r + 4$	$\tfrac{24}{r+4}$	$24$

Her running time plus her biking time is $3$ hours. Multiply both sides by the LCD $r(r+4)$ :

\begin{array}{lrcl} & \tfrac{8}{r} + \tfrac{24}{r+4} &=& 3 \\[4pt] \text{Clear the fractions.} & 8(r+4) + 24r &=& 3r(r+4) \\[4pt] \text{Distribute.} & 8r + 32 + 24r &=& 3r^2 + 12r \\[4pt] \text{Write in standard form.} & 0 &=& 3r^2 - 20r - 32 \\[4pt] \text{Factor.} & 0 &=& (3r + 4)(r - 8) \end{array}

So $r = -\tfrac{4}{3}$ or $r = 8$ . A negative speed does not make sense here, so $r = 8$ . Check: at $8$ mph running takes $\tfrac{8}{8} = 1$ hour and at $12$ mph biking takes $\tfrac{24}{12} = 2$ hours, a total of $3$ hours. Jazmine’s running speed is $8$ mph.

Jazmine trained for $3$ hours total. She ran $8$ miles and biked $24$ miles, biking $4$ mph faster than she ran. Find her running speed, in mph (enter the number).

One leg takes longer than the other

When a trip out and back covers the same distance but the two directions take different times, we express one time as more than the other.

Example. Hamilton rode his bike downhill $12$ miles from his house to the ocean and then rode uphill to return home. His uphill speed was $8$ mph slower than his downhill speed. It took him $2$ hours longer to get home than it took to get to the ocean. Find Hamilton’s downhill speed.

Let $r =$ Hamilton’s downhill speed, so $r - 8 =$ his uphill speed. The distance is $12$ miles each way.

	Rate	Time	Distance
Downhill	$r$	$\tfrac{12}{r}$	$12$
Uphill	$r - 8$	$\tfrac{12}{r-8}$	$12$

The uphill time is $2$ more than the downhill time. Multiply both sides by the LCD $r(r-8)$ :

\begin{array}{lrcl} & \tfrac{12}{r-8} &=& \tfrac{12}{r} + 2 \\[4pt] \text{Clear the fractions.} & 12r &=& 12(r-8) + 2r(r-8) \\[4pt] \text{Distribute.} & 12r &=& 12r - 96 + 2r^2 - 16r \\[4pt] \text{Write in standard form.} & 0 &=& 2r^2 - 16r - 96 \\[4pt] \text{Factor out } 2 \text{ and factor.} & 0 &=& 2(r - 12)(r + 4) \end{array}

So $r = 12$ or $r = -4$ . A negative speed makes no sense, so $r = 12$ . Check: downhill at $12$ mph takes $\tfrac{12}{12} = 1$ hour; uphill at $12 - 8 = 4$ mph takes $\tfrac{12}{4} = 3$ hours, which is $2$ hours more. Hamilton’s downhill speed is $12$ mph.

Hamilton rode $12$ miles downhill to the ocean, then $12$ miles uphill home at a speed $8$ mph slower. The uphill trip took $2$ hours longer. Find his downhill speed, in mph (enter the number).

Solve work applications

Suppose Pete can paint a room in $10$ hours. Working at a steady pace, in $1$ hour he paints $\tfrac{1}{10}$ of the room. If Alicia would take $8$ hours to paint the same room, then in $1$ hour she paints $\tfrac{1}{8}$ of the room. How long would it take them to paint the room working together?

A work application has three quantities: the time each person takes alone, and the time it takes them together. Let $t$ be the number of hours it takes them together. Then in $1$ hour working together they complete $\tfrac{1}{t}$ of the job. The key model is:

\text{part done by first} + \text{part done by second} = \text{part done together}

So Pete’s part plus Alicia’s part equals the whole rate:

\begin{array}{lrcl} & \tfrac{1}{10} + \tfrac{1}{8} &=& \tfrac{1}{t} \\[4pt] \text{Multiply by the LCD } 40t. & 4t + 5t &=& 40 \\[4pt] \text{Simplify and solve.} & 9t &=& 40 \\[4pt] & t &=& \tfrac{40}{9} \end{array}

Written as a mixed number, $t = 4\tfrac{4}{9}$ hours. Since $\tfrac{4}{9}$ of $60$ minutes is about $27$ minutes, it would take Pete and Alicia about $4$ hours and $27$ minutes to paint the room together. Notice it takes less time together than either person alone, as it should.

Pete paints a room in $10$ hours and Alicia in $8$ hours. Working together, they finish in $t$ hours, where $\tfrac{1}{10} + \tfrac{1}{8} = \tfrac{1}{t}$ . Find $t$ (enter it as a fraction of hours).

Solve a work application.

Read the problem and let $t$ be the time it takes to do the job together.
Find each worker’s rate: if a job takes $a$ hours alone, that worker does $\tfrac{1}{a}$ of the job per hour.
Set the sum of the individual per-hour parts equal to the together rate $\tfrac{1}{t}$ .
Clear the fractions with the LCD, solve, and check the answer is reasonable.

Example. Press #1 takes $6$ hours to print a magazine and Press #2 takes $12$ hours. How long will it take to print the magazine with both presses running together?

Let $t =$ the hours to finish together. Each row records the hours to complete the whole job and the part completed per hour.

	Hours to complete	Part per hour
Press #1	$6$	$\tfrac{1}{6}$
Press #2	$12$	$\tfrac{1}{12}$
Together	$t$	$\tfrac{1}{t}$

The part completed by Press #1 plus the part by Press #2 equals the amount completed together. Multiply both sides by the LCD $12t$ :

\begin{array}{lrcl} & \tfrac{1}{6} + \tfrac{1}{12} &=& \tfrac{1}{t} \\[4pt] \text{Clear the fractions.} & 2t + t &=& 12 \\[4pt] \text{Simplify and solve.} & 3t &=& 12 \\[4pt] & t &=& 4 \end{array}

When both presses run together it takes $4$ hours to do the job.

Press #1 prints a magazine in $6$ hours and Press #2 in $12$ hours. Running together they finish in $t$ hours, where $\tfrac{1}{6} + \tfrac{1}{12} = \tfrac{1}{t}$ . Find $t$ , in hours (enter the number).

Some work problems give the together time and one worker’s time, and ask for the other worker’s time alone.

Example. Corey can shovel all the snow from the sidewalk and driveway in $4$ hours. If he and his twin Casey work together, they finish in $2$ hours. How many hours would it take Casey to do the job alone?

Let $t =$ the hours Casey needs alone.

	Hours to complete	Part per hour
Corey	$4$	$\tfrac{1}{4}$
Casey	$t$	$\tfrac{1}{t}$
Together	$2$	$\tfrac{1}{2}$

Corey’s part plus Casey’s part equals the together part. Multiply both sides by the LCD $4t$ :

\begin{array}{lrcl} & \tfrac{1}{4} + \tfrac{1}{t} &=& \tfrac{1}{2} \\[4pt] \text{Clear the fractions.} & t + 4 &=& 2t \\[4pt] \text{Solve.} & 4 &=& t \end{array}

It would take Casey $4$ hours to do the job alone.

Corey shovels the snow in $4$ hours; with his twin Casey they finish in $2$ hours. Casey alone needs $t$ hours, where $\tfrac{1}{4} + \tfrac{1}{t} = \tfrac{1}{2}$ . Find $t$ , in hours (enter the number).

Key terms

uniform motion — motion at a constant rate, modeled by $D = rt$ ; solving for time gives $t = \tfrac{D}{r}$ , which fills the time column of a rate table. headwind / tailwind — a wind (or current) that decreases or increases a vehicle’s speed, changing the rate by a fixed amount. work application — a problem where two or more workers share a job; each does $\tfrac{1}{a}$ of the job per hour when alone, and their per-hour parts sum to $\tfrac{1}{t}$ , the together rate.

This section is adapted from Elementary Algebra 2e, Section 8.8: Solve Uniform Motion and Work Applications by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: condensed the worked examples into rate/work tables and aligned step tables, recast the “How To” procedure as a callout, and grouped the uniform motion examples by the relationship used (equal times, total time, and a time difference); omitted the Be Prepared quiz, the diagrams, media links, and end-of-section exercises; and converted the practice problems (“Try Its”) and representative exercises into interactive exercises with instant feedback.

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