## 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. |