During a 60-minute walk, a hiker moves 2 km north, then 2 km east, and finally 2 km south. Evaluating the hiker’s average velocity compared to average speed for the entire walk, which observation is most accurate?
The average velocity is less than the average speed due to the difference between displacement and total distance.
Average speed = total distance ÷ total time, while average velocity = displacement ÷ total time, where displacement is the straight-line distance from the starting point to the final point.
Total distance = 2 + 2 + 2 = 6 km. Going 2 km north and then 2 km south cancels out, so the hiker ends 2 km east of the start; displacement = 2 km east. In 60 minutes (1 h), average speed = 6 km/h and average velocity = 2 km/h towards the east.
So the average velocity (2 km/h) is less than the average speed (6 km/h), because the displacement (2 km) is smaller than the total distance (6 km). The path is three sides of a square, not a closed triangle, so the velocity is not zero.
Check: the hiker’s final position is 2 km due east of the start, so displacement ÷ distance = 2 ÷ 6 = 1/3, and indeed 2 km/h ÷ 6 km/h = 1/3.
Hence, The average velocity is less than the average speed due to the difference between displacement and total distance.
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