Lesson 8.8: Solve Uniform Motion and Work Applications

New Concept

This lesson applies algebra to solve "how fast?" and "how long?" questions. You'll use the distance formula, $D=rt$, and work-rate principles to build and solve rational equations for real-world motion and work scenarios.

What’s next

Next, you'll walk through interactive examples for both motion and work problems, followed by a series of practice cards to master the setup.

Property

When solving uniform motion problems where the time is the same for two different trips, we use the formula $t = \frac{D}{r}$ for each trip and set them equal. Let $t_1$ and $t_2$ be the times for two trips. If $t_1 = t_2$, then:

This is common in problems involving a headwind (speed is $r-w$) and a tailwind (speed is $r+w$), where $r$ is the vehicle's speed and $w$ is the wind or current speed.

Examples

• An airplane flies 480 miles with a 40 mph tailwind in the same time it flies 320 miles against it. Let $r$ be the plane's speed. The equation is $\frac{480}{r+40} = \frac{320}{r-40}$. Solving gives $160r = 32000$, so $r=200$ mph.

• A boat travels 45 miles downstream with a 4 mph current in the same time it travels 27 miles upstream. Let $b$ be the boat's speed. The equation is $\frac{45}{b+4} = \frac{27}{b-4}$. Solving gives $18b = 288$, so $b=16$ mph.

Property

For journeys with multiple parts where the total time is known, the sum of the times for each part equals the total time. Using the formula $t = \frac{D}{r}$, the equation is:

This allows you to find an unknown rate when given the distances of each part and the total duration of the trip.

Examples

• Maria trained for 4 hours. She ran 10 miles and biked 30 miles. Her biking speed is 10 mph faster than her running speed, $r$. The equation is $\frac{10}{r} + \frac{30}{r+10} = 4$. Solving gives $4r^2 = 100$, so $r=5$ mph.

• A truck drives for 6 hours, covering 120 miles on a highway and 90 miles on a side road. Its highway speed, $r$, was 25 mph faster than on the side road. The equation is $\frac{120}{r} + \frac{90}{r-25} = 6$.

Property

When one part of a journey takes a specific amount of time longer than another, set the longer time equal to the shorter time plus the difference. Using the formula $t = \frac{D}{r}$, the equation is:

Examples

• A cyclist rides 30 miles downhill and then returns uphill. His uphill speed is 10 mph slower than his downhill speed, $r$. The uphill trip took 2 hours longer. The equation is $\frac{30}{r-10} = \frac{30}{r} + 2$. Solving gives $r=15$ mph.

• A jogger runs 10 miles to a park and gets a ride back home. The ride is 15 mph faster than her jogging speed, $r$, and the jog took 1.5 hours longer than the ride. The equation is $\frac{10}{r} = \frac{10}{r+15} + 1.5$.

Property

To find the time it takes for two people or machines to complete a job together, add their individual work rates. If person A takes $t_1$ hours and person B takes $t_2$ hours, their combined time $T$ is found with:

The term $\frac{1}{t}$ represents the fraction of the job completed in one hour.

Examples

• An expert can build a deck in 8 hours, and an apprentice in 12 hours. Together, their time $T$ is found by $\frac{1}{8} + \frac{1}{12} = \frac{1}{T}$. This gives $3T + 2T = 24$, so $5T = 24$, and $T = 4.8$ hours.

• One pipe can fill a tank in 3 hours, and a second pipe can fill it in 6 hours. Working together, their time $T$ is found by $\frac{1}{3} + \frac{1}{6} = \frac{1}{T}$. This gives $2T + T = 6$, so $3T = 6$, and $T = 2$ hours.

Property

To find how long it would take one person to do a job alone, use the same work formula. If you know Person A's time ($t_1$) and the combined time ($T$), you can solve for Person B's time ($t_2$):

Here, $t_2$ is the unknown variable.

Examples

• Two chefs can prepare a meal in 3 hours. One chef can do it alone in 4 hours. To find the other chef's time, $t$, we solve $\frac{1}{4} + \frac{1}{t} = \frac{1}{3}$. The LCD is $12t$. So $3t + 12 = 4t$, which means $t=12$ hours.

• Together, two hoses fill a pool in 6 hours. One hose alone takes 15 hours. To find the other hose's time, $t$, solve $\frac{1}{15} + \frac{1}{t} = \frac{1}{6}$. The LCD is $30t$. So $2t + 30 = 5t$, which gives $3t=30$, and $t=10$ hours.