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 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 \(D=rt\) 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.

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 \(D=r\bullet t\) , 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.

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 \(D=r\bullet t\) , 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.

The above figure show 2 diagonal, parallel lines pointing in opposite directions. The top line points to the right, and downward and has “12 miles” written beneath it. The bottom line points to the left and upward, and has, “ 8 miles per hour slower, 2 hours longer” written beneath it.

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.

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