Mathematics » Rational Expressions and Equations » Solve Uniform Motion and Work Applications

# Solving Uniform Motion Applications

## Solving Uniform Motion Applications

We have solved uniform motion problems using the formula $$D=rt$$ in previous tutorials. We used a table like the one below 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 would solve the equation for t and get the formula $$t=\frac{D}{r}$$.

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.

## 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.

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 $$D=r\bullet t$$ , we solve for t and get $$\frac{D}{r}$$. 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. $$\phantom{\rule{7.2em}{0ex}}\frac{200}{r-30}=\frac{300}{r+30}$$ We multiply both sides by the LCD. $$200\left(r+30\right)=300\left(r-30\right)$$ $$\left(r+30\right)\left(r-30\right)\left(\frac{200}{r-30}\right)=\left(r+30\right)\left(r-30\right)\left(\frac{300}{r+30}\right)$$ Simplify. $$\phantom{\rule{4.45em}{0ex}}\left(r+30\right)\left(200\right)=\left(r-30\right)\left(300\right)$$ $$\phantom{\rule{4.8em}{0ex}}200r+6000=300r-9000$$ Solve. $$\phantom{\rule{7.3em}{0ex}}15000=100r$$ $$\phantom{\rule{7.8em}{0ex}}150=r$$ 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 $$\phantom{\rule{0.8em}{0ex}}150+30=180\text{mph}\phantom{\rule{1.2em}{0ex}}\frac{300}{180}=\frac{5}{3}$$ hours Headwind $$\phantom{\rule{0.2em}{0ex}}150-30=120\text{mph}\phantom{\rule{1.2em}{0ex}}\frac{200}{120}=\frac{5}{3}$$ hours The times are equal, so it checks. The plane was traveling 150 mph.

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

## 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?

### Solution

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

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 $$D=r\bullet t$$ , we solve for t and get $$t=\frac{D}{r}$$. 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. $$\phantom{\rule{4.2em}{0ex}}\frac{8}{r}+\frac{24}{r+4}\phantom{\rule{0.5em}{0ex}}=\phantom{\rule{0.5em}{0ex}}3$$ Solve. $$\phantom{\rule{0.4em}{0ex}}\begin{array}{}\\ \hfill r\left(r+4\right)\left(\frac{8}{r}+\frac{24}{r+4}\right)& =\hfill & 3\bullet r\left(r+4\right)\hfill \\ \hfill 8\left(r+4\right)+24r& =\hfill & 3r\left(r+4\right)\hfill \\ \hfill 8r+32+24r& =\hfill & 3{r}^{2}+12r\hfill \\ \hfill 32+32r& =\hfill & 3{r}^{2}+12r\hfill \\ \hfill 0& =\hfill & 3{r}^{2}-20r-32\hfill \\ \hfill 0& =\hfill & \left(3r+4\right)\left(r-8\right)\hfill \end{array}$$ $$\left(3r+4\right)=0\phantom{\rule{1em}{0ex}}\left(r-8\right)=0$$ $$r=-\frac{4}{3}\phantom{\rule{1em}{0ex}}r=8$$ Check. $$\require{cancel}\cancel{r=-\frac{4}{3}}$$$$r=8$$ A negative speed does not make sense in this problem, so $$r=8$$ is the solution. Is 8 mph a reasonable running speed? Yes. $$\begin{array}{ccccccc}\text{Run 8 mph}\hfill & & & \phantom{\rule{0.6em}{0ex}}\frac{8\phantom{\rule{0.2em}{0ex}}\text{miles}\phantom{\rule{0.2em}{0ex}}}{8\phantom{\rule{0.2em}{0ex}}\text{mph}\phantom{\rule{0.2em}{0ex}}}=1\phantom{\rule{0.2em}{0ex}}\text{hour}\phantom{\rule{0.2em}{0ex}}\hfill & & & \\ \text{Bike 12 mph}\phantom{\rule{0.2em}{0ex}}\hfill & & & \frac{24\phantom{\rule{0.2em}{0ex}}\text{miles}\phantom{\rule{0.2em}{0ex}}}{12\phantom{\rule{0.2em}{0ex}}\text{mph}\phantom{\rule{0.2em}{0ex}}}=2\phantom{\rule{0.2em}{0ex}}\text{hours}\phantom{\rule{0.2em}{0ex}}\hfill & & & \\ & & & \text{Total 3 hours}\hfill & & & \text{Jazmine’s running speed is 8 mph.}\hfill \end{array}$$

Once again, we will use the uniform motion formula solved for the variable t.

## Example

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

### Solution

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

We fill in the chart to organize the information.

 We are looking for Hamilton’s downhill speed. Let $$r=$$ Hamilton’s downhill speed. His uphill speed is 8 miles per hour slower. Enter the rates into the chart. $$h-8=$$ Hamilton’s uphill speed The distance is the same in both directions, 12 miles. Since $$D=r\bullet t$$ , we solve for t and get $$t=\frac{D}{r}$$. We divide the distance by the rate in each row, and place the expression in the time column. Write a word sentence about the time. He took 2 hours longer uphill than downhill. The uphill time is 2 more than the downhill time. Translate the sentence to get the equation. Solve. $$\begin{array}{ccc}\hfill \frac{12}{h-8}& =\hfill & \frac{12}{h}+2\hfill \\ \hfill h\left(h-8\right)\left(\frac{12}{h-8}\right)& =\hfill & h\left(h-8\right)\left(\frac{12}{h}+2\right)\hfill \\ \hfill 12h& =\hfill & 12\left(h-8\right)+2h\left(h-8\right)\hfill \\ \hfill 12h& =\hfill & 12h-96+2{h}^{2}-16h\hfill \\ \hfill 0& =\hfill & 2{h}^{2}-16h-96\hfill \\ \hfill 0& =\hfill & 2\left({h}^{2}-8h-48\right)\hfill \\ \hfill 0& =\hfill & 2\left(h-12\right)\left(h+4\right)\hfill \\ \hfill h-12& =\hfill & 0\phantom{\rule{1em}{0ex}}h+4=0\hfill \\ \hfill h& =\hfill & 12\phantom{\rule{2em}{0ex}}\require{cancel}\cancel{h=4}\hfill \end{array}$$ Check. Is 12 mph a reasonable speed for biking downhill? Yes. Downhill $$\phantom{\rule{1.5em}{0ex}}12\phantom{\rule{0.2em}{0ex}}\text{mph}\phantom{\rule{3.3em}{0ex}}\frac{12\phantom{\rule{0.2em}{0ex}}\text{miles}}{12\phantom{\rule{0.2em}{0ex}}\text{mph}}=1\phantom{\rule{0.2em}{0ex}}\text{hour}$$ Uphill $$\phantom{\rule{1.5em}{0ex}}12-8=4\phantom{\rule{0.2em}{0ex}}\text{mph}\phantom{\rule{1.5em}{0ex}}\frac{12\phantom{\rule{0.2em}{0ex}}\text{miles}}{4\phantom{\rule{0.2em}{0ex}}\text{mph}}=3\phantom{\rule{0.2em}{0ex}}\text{hours}$$ The uphill time is 2 hours more than the downhill time. Hamilton’s downhill speed is 12 mph.