1Cademy - Finding Speed Using Known Distances and Time Difference

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Problem-Solving Strategy for Distance, Rate, and Time Applications

Example

Finding Speed Using Known Distances and Time Difference

Apply the distance, rate, and time problem-solving strategy to find an unknown speed when the distances for two segments are known and the difference in their travel times is given, producing a rational equation that often leads to a quadratic equation.

Problem: Hamilton rode his bike downhill $12$ miles to the ocean and then rode uphill $12$ miles to return home. His uphill speed was $8$ mph slower than his downhill speed. It took him $2$ hours longer to get home than it took him to get to the ocean. Find his downhill speed.

Read and draw: Sketch the route with an arrow downhill ( $12$ miles) and an arrow uphill ( $12$ miles). Create a rate–time–distance table.
Identify: Hamilton's downhill speed.
Name: Let $h$ = Hamilton's downhill speed in mph. His uphill speed is $h - 8$ . The distance in both directions is $12$ miles. Using $t = \frac{D}{r}$ , divide each distance by its rate to fill in the time column:

	Rate (mph)	Time (hrs)	Distance (miles)
Downhill	$h$	$\frac{12}{h}$	$12$
Uphill	$h - 8$	$\frac{12}{h - 8}$	$12$

Translate: The uphill time is $2$ hours longer than the downhill time: $\frac{12}{h - 8} = \frac{12}{h} + 2$
Solve: Multiply both sides by the LCD, $h(h - 8)$ , to clear the fractions: $h(h - 8) \cdot \frac{12}{h - 8} = h(h - 8) \left( \frac{12}{h} + 2 \right)$ $12h = 12(h - 8) + 2h(h - 8)$ $12h = 12h - 96 + 2h^2 - 16h$ $0 = 2h^2 - 16h - 96$ Divide by $2$ : $0 = h^2 - 8h - 48$ Factor: $0 = (h - 12)(h + 4)$ So, $h = 12$ or $h = -4$ . Discard the negative speed since it is not physically meaningful.
Check: Is $12$ mph a reasonable downhill biking speed? Yes. Downhill time is $\frac{12}{12} = 1$ hour. Uphill speed is $12 - 8 = 4$ mph, so uphill time is $\frac{12}{4} = 3$ hours. The uphill time ( $3$ hours) is $2$ hours more than the downhill time ( $1$ hour). $\checkmark$
Answer: Hamilton's downhill speed is $12$ mph.

This example illustrates the known-distances-with-time-difference scenario. Because individual times are unknown, each is represented as a rational expression. Equating one time to the other plus the given difference produces a rational equation, which, after clearing denominators, can be solved using standard quadratic techniques.

Updated 2026-05-01

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OpenStax Intermediate Algebra

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