Example 8.81 from Elementary Algebra, 8.8 Solve Uniform Motion and Work Applications

Solve Uniform Motion and Work Applications: example 8.81

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?

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Link can ride his bike 20 miles into a 3 mph headwind in the same amount of time he can ride 30 miles with a 3 mph tailwind. What is Link’s biking speed?

How did it go?

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 $D = r ∙ t$ , 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. $200 (r + 30) = 300 (r - 30)$	$(r + 30) (r - 30) (\frac{200}{r - 30}) = (r + 30) (r - 30) (\frac{300}{r + 30})$
Simplify.	$(r + 30) (200) = (r - 30) (300)$
	$200 r + 6000 = 300 r - 9000$
Solve.	$15000 = 100 r$ $150 = r$
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 $150 + 30 = 180 mph \frac{300}{180} = \frac{5}{3}$ hours
Headwind $150 - 30 = 120 mph \frac{200}{120} = \frac{5}{3}$ hours
The times are equal, so it checks.	The plane was traveling 150 mph.

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 $D = r ∙ t$ , 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. $200 (r + 30) = 300 (r - 30)$	$(r + 30) (r - 30) (\frac{200}{r - 30}) = (r + 30) (r - 30) (\frac{300}{r + 30})$
Simplify.	$(r + 30) (200) = (r - 30) (300)$
	$200 r + 6000 = 300 r - 9000$
Solve.	$15000 = 100 r$ $150 = r$
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 $150 + 30 = 180 mph \frac{300}{180} = \frac{5}{3}$ hours
Headwind $150 - 30 = 120 mph \frac{200}{120} = \frac{5}{3}$ hours
The times are equal, so it checks.	The plane was traveling 150 mph.

Solve Uniform Motion and Work Applications: example 8.81

Solution

Your turn

Solve Uniform Motion and Work Applications: example 8.81

Solution

Your turn