Example 8.83 from Elementary Algebra, 8.8 Solve Uniform Motion and Work Applications

Solve Uniform Motion and Work Applications: example 8.83

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.

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.

Kayla rode her bike 75 miles home from college one weekend and then rode the bus back to college. It took her 2 hours less to ride back to college on the bus than it took her to ride home on her bike, and the average speed of the bus was 10 miles per hour faster than Kayla’s biking speed. Find Kayla’s biking speed.

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 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 $D = r ∙ t$ , 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.	$\begin{matrix} \frac{12}{h - 8} & = & \frac{12}{h} + 2 \\ h (h - 8) (\frac{12}{h - 8}) & = & h (h - 8) (\frac{12}{h} + 2) \\ 12 h & = & 12 (h - 8) + 2 h (h - 8) \\ 12 h & = & 12 h - 96 + 2 h^{2} - 16 h \\ 0 & = & 2 h^{2} - 16 h - 96 \\ 0 & = & 2 (h^{2} - 8 h - 48) \\ 0 & = & 2 (h - 12) (h + 4) \\ h - 12 & = & 0 h + 4 = 0 \\ h & = & 12 h = 4 \end{matrix}$
Check. Is 12 mph a reasonable speed for biking downhill? Yes.
Downhill $12 mph \frac{12 miles}{12 mph} = 1 hour$
Uphill $12 - 8 = 4 mph \frac{12 miles}{4 mph} = 3 hours$
	The uphill time is 2 hours more than the downhill time. Hamilton’s downhill speed is 12 mph.

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 $D = r ∙ t$ , 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.	$\begin{matrix} \frac{12}{h - 8} & = & \frac{12}{h} + 2 \\ h (h - 8) (\frac{12}{h - 8}) & = & h (h - 8) (\frac{12}{h} + 2) \\ 12 h & = & 12 (h - 8) + 2 h (h - 8) \\ 12 h & = & 12 h - 96 + 2 h^{2} - 16 h \\ 0 & = & 2 h^{2} - 16 h - 96 \\ 0 & = & 2 (h^{2} - 8 h - 48) \\ 0 & = & 2 (h - 12) (h + 4) \\ h - 12 & = & 0 h + 4 = 0 \\ h & = & 12 h = 4 \end{matrix}$
Check. Is 12 mph a reasonable speed for biking downhill? Yes.
Downhill $12 mph \frac{12 miles}{12 mph} = 1 hour$
Uphill $12 - 8 = 4 mph \frac{12 miles}{4 mph} = 3 hours$
	The uphill time is 2 hours more than the downhill time. Hamilton’s downhill speed is 12 mph.

Solve Uniform Motion and Work Applications: example 8.83

Solution

Your turn

Solve Uniform Motion and Work Applications: example 8.83

Solution

Your turn