8.8 Solve uniform motion and work applications

Solve uniform motion applications

We have solved uniform motion problems using the formula $D=rt$ in previous chapters. 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.

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.	$\frac{200}{r-30}=\frac{300}{r+30}$
We multiply both sides by the LCD. $200(r+30)=300(r-30)$	$(r+30)(r-30)\left(\frac{200}{r-30}\right)=(r+30)(r-30)\left(\frac{300}{r+30}\right)$
Simplify.	$(r+30)\left(200\right)=(r-30)\left(300\right)$
	$200r+6000=300r-9000$
Solve.	$15000=100r$ $150=r$
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 $150+30=180\text{mph}\frac{300}{180}=\frac{5}{3}$ hours
Headwind $150-30=120\text{mph}\frac{200}{120}=\frac{5}{3}$ hours
The times are equal, so it checks.	The plane was traveling 150 mph.

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

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.	$\frac{8}{r}+\frac{24}{r+4}=3$
Solve.	$\begin{array}{}\\ \hfill r(r+4)(\frac{8}{r}+\frac{24}{r+4})& =\hfill & 3\bullet r(r+4)\hfill \\ \hfill 8(r+4)+24r& =\hfill & 3r(r+4)\hfill \\ \hfill 8r+32+24r& =\hfill & 3r^2+12r\hfill \\ \hfill 32+32r& =\hfill & 3r^2+12r\hfill \\ \hfill 0& =\hfill & 3r^2-20r-32\hfill \\ \hfill 0& =\hfill & (3r+4)(r-8)\hfill \end{array}$
	$(3r+4)=0(r-8)=0$
	$r=-\frac{4}{3}r=8$
Check. $\cancel{r=-\frac{4}{3}}$ $r=8$
A negative speed does not make sense in this problem, so $r=8$ is the solution.
Is 8 mph a reasonable running speed? Yes.
$\begin{array}{ccccccc}\text{Run 8 mph}\hfill & & & \frac{8\text{miles}}{8\text{mph}}=1\text{hour}\hfill & & & \\ \text{Bike 12 mph}\hfill & & & \frac{24\text{miles}}{12\text{mph}}=2\text{hours}\hfill & & & \\ & & & \text{Total 3 hours}\hfill & & & \text{Jazmine’s running speed is 8 mph.}\hfill \end{array}$

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