1Cademy - Solving a Uniform Motion Application: Biking and Bus Speeds

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

Example

Solving a Uniform Motion Application: Biking and Bus Speeds

Apply the distance, rate, and time problem-solving strategy to find unknown speeds when different modes of transportation are used and a time difference is known.

Problem: 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.

Read and draw: Sketch the journey with $75$ miles by bike and $75$ miles by bus. Create a rate-time-distance table.
Identify: Kayla's biking speed.
Name: Let $r$ = biking speed. The bus speed is $r + 10$ . The biking time is $\frac{75}{r}$ and the bus time is $\frac{75}{r + 10}$ .
Translate: The bus ride took $2$ hours less than the bike ride, so the biking time minus $2$ equals the bus time: $\frac{75}{r} - 2 = \frac{75}{r + 10}$
Solve: Multiply both sides by the least common denominator, $r(r + 10)$ : $75(r + 10) - 2r(r + 10) = 75r$ $75r + 750 - 2r^2 - 20r = 75r$ $-2r^2 - 20r + 750 = 0$ Divide by $-2$ : $r^2 + 10r - 375 = 0$ $(r + 25)(r - 15) = 0$ The solutions are $r = -25$ and $r = 15$ . Since speed cannot be negative, $r = 15$ mph.
Check: Biking time: $\frac{75}{15} = 5$ hours. Bus time: $\frac{75}{15 + 10} = \frac{75}{25} = 3$ hours. The bus ride was $5 - 3 = 2$ hours shorter. $\checkmark$
Answer: Kayla's biking speed was $15$ mph.

Updated 2026-05-01

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OpenStax Intermediate Algebra

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