Example 8.82 from Elementary Algebra, 8.8 Solve Uniform Motion and Work Applications

Solve Uniform Motion and Work Applications: example 8.82

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?

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.

Dennis went cross-country skiing for 6 hours on Saturday. He skied 20 mile uphill and then 20 miles back downhill, returning to his starting point. His uphill speed was 5 mph slower than his downhill speed. What was Dennis’ speed going uphill and his speed going downhill?

How did it go?

Your turn

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

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

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

Solve Uniform Motion and Work Applications: example 8.82

Solution

Your turn

Solve Uniform Motion and Work Applications: example 8.82

Solution

Your turn