1Cademy - Solving a Uniform Motion Application: Jogging Rates on Flat and Hilly Trails

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

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Solving a Uniform Motion Application: Jogging Rates on Flat and Hilly Trails

Apply the distance, rate, and time problem-solving strategy to find an unknown jogging rate when two segments of a trip have different distances and speeds, and their time difference is known.

Problem: Victoria jogs $12$ miles to the park along a flat trail and then returns by jogging on an $20$ mile hilly trail. She jogs $1$ mile per hour slower on the hilly trail than on the flat trail, and her return trip takes her two hours longer. Find her rate of jogging on the flat trail.

Read and draw: Sketch the trip with a $12$ -mile flat trail and a $20$ -mile hilly trail. Create a rate-time-distance table.
Identify: Victoria's jogging rate on the flat trail.
Name: Let $r$ = jogging rate on the flat trail. Her rate on the hilly trail is $r - 1$ . The time on the flat trail is $\frac{12}{r}$ and the time on the hilly trail is $\frac{20}{r - 1}$ .
Translate: The hilly trail took $2$ hours longer than the flat trail, so the flat trail time plus $2$ equals the hilly trail time: $\frac{12}{r} + 2 = \frac{20}{r - 1}$
Solve: Multiply both sides by the least common denominator, $r(r - 1)$ : $12(r - 1) + 2r(r - 1) = 20r$ $12r - 12 + 2r^2 - 2r = 20r$ $2r^2 + 10r - 12 = 20r$ $2r^2 - 10r - 12 = 0$ Divide by $2$ : $r^2 - 5r - 6 = 0$ $(r - 6)(r + 1) = 0$ The solutions are $r = 6$ and $r = -1$ . Since speed cannot be negative, $r = 6$ mph.
Check: Flat trail time: $\frac{12}{6} = 2$ hours. Hilly trail time: $\frac{20}{6 - 1} = \frac{20}{5} = 4$ hours. The return trip took $4 - 2 = 2$ hours longer. $\checkmark$
Answer: Victoria's rate of jogging on the flat trail was $6$ mph.

Updated 2026-05-01

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OpenStax Intermediate Algebra

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