A student records the total distance covered as 200 meters and the displacement as 120 meters in 50 seconds during a straight-line walk. Which statement best characterizes the relationship between average speed and average velocity in this situation?
Average speed is greater than average velocity because total distance exceeds displacement
Average speed = total distance ÷ total time, and the magnitude of average velocity = displacement ÷ total time. Both use the same time, so the larger of distance and displacement gives the larger quantity.
Average speed = 200 ÷ 50 = 4 m/s. Average velocity = 120 ÷ 50 = 2.4 m/s. So average speed (4 m/s) is greater than average velocity (2.4 m/s).
This happens because the total distance of 200 m is more than the displacement of 120 m; the walker must have turned back along the line for part of the walk.
Check: the ratio of average speed to average velocity is 4 ÷ 2.4 = 200 ÷ 120 = 5/3, which is exactly the ratio of distance to displacement, confirming that the speed is larger because the distance is larger.
Hence, Average speed is greater than average velocity because total distance exceeds displacement.
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