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Cyclist 1 travels 95 m in 27 s. Cyclist 2 travels 87 m in 22 s. Cyclist 3 travels 106 m in 26 s. Cyclist 4 travels 108 m in 24 s. Which cyclist has the greatest average speed?
ACyclist 1
BCyclist 2
CCyclist 3
DCyclist 4
Answer & Solution
Correct answer: D. Cyclist 4
1. Divide distance by time for each cyclist to compare average speeds.
2. Cyclist 1: 95/27 = 3.52 m/s. Cyclist 2: 87/22 = 3.95 m/s.
3. Cyclist 3: 106/26 = 4.08 m/s. Cyclist 4: 108/24 = 4.50 m/s.
4. Cyclist 4 has the largest value of the four, so Cyclist 4 is fastest on average.
_Source: OpenStax Physics (CC BY 4.0), Ch 2 "Motion in One Dimension", section 2.2 Speed and Velocity_
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