Example

Try It 7.97: Solving a Uniform Motion Application for 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 7575 miles home from college one weekend and then rode the bus back to college. It took her 22 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 1010 miles per hour faster than Kayla’s biking speed. Find Kayla’s biking speed.

  1. Read and draw: Sketch the journey with 7575 miles by bike and 7575 miles by bus. Create a rate-time-distance table.
  2. Identify: Kayla's biking speed.
  3. Name: Let rr = biking speed in mph. The bus speed is r+10r + 10. The distance for both trips is 7575 miles. Because t=Drt = \frac{D}{r}, the biking time is 75r\frac{75}{r} and the bus time is 75r+10\frac{75}{r + 10}.
Rate (mph)Time (hrs)Distance (miles)
Bikerr75r\frac{75}{r}7575
Busr+10r + 1075r+10\frac{75}{r + 10}7575
  1. Translate: The bus ride took 22 hours less than the bike ride, so the biking time minus 22 equals the bus time: 75r2=75r+10\frac{75}{r} - 2 = \frac{75}{r + 10}
  2. Solve: Multiply both sides by the least common denominator, r(r+10)r(r + 10): r(r+10)(75r2)=r(r+10)(75r+10)r(r + 10) \left(\frac{75}{r} - 2\right) = r(r + 10) \left(\frac{75}{r + 10}\right) 75(r+10)2r(r+10)=75r75(r + 10) - 2r(r + 10) = 75r 75r+7502r220r=75r75r + 750 - 2r^2 - 20r = 75r 2r220r+750=0-2r^2 - 20r + 750 = 0 Divide by 2-2: r2+10r375=0r^2 + 10r - 375 = 0 Factor the trinomial: (r+25)(r15)=0(r + 25)(r - 15) = 0 The solutions are r=25r = -25 and r=15r = 15. Since speed cannot be negative, discard r=25r = -25. Therefore, r=15r = 15 mph.
  3. Check: Biking time: 7515=5\frac{75}{15} = 5 hours. Bus time: 7515+10=7525=3\frac{75}{15 + 10} = \frac{75}{25} = 3 hours. The bus ride was 53=25 - 3 = 2 hours shorter. \checkmark
  4. Answer: Kayla's biking speed was 1515 mph.

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Updated 2026-05-25

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