Is average speed the same as the magnitude of average velocity?
Short answer: Average velocity = displacement/time. A full lap ends where it began, so displacement is 0.
The trap
The wrong idea
“Average speed is always equal to the magnitude of average velocity.”
It sounds reasonable, and a lot of JEE/NEET aspirants carry it into the exam. Here's a one-question check, then the explanation that makes it click.
Quick check
Try it before you peek
You run one full lap of a 400 m track in 80 s. Your average velocity is:
- 5 m/s
- 0 m/s
- 2.5 m/s
Show the answer
B. 0 m/s
Average velocity = displacement/time. A full lap ends where it began, so displacement is 0.
The click
Average speed = total distance travelled ÷ total time. Average velocity = displacement ÷ total time, where displacement is the straight-line change in position from start to finish, with a direction. Distance counts every metre you run; displacement only cares where you ended up compared with where you started.
On one full lap of a 400 m track, the distance is 400 m, so the average speed is 400/80 = 5 m/s. But you finish exactly at the starting line, so the displacement is 0 and the average velocity is 0 m/s. Both numbers are correct; they answer different questions.
The straight line between two points is the shortest path, so the magnitude of displacement can never exceed the distance. That means average speed ≥ |average velocity|, always. They are equal only when the motion is along a straight line without turning back.
Careful: this is about averages. At any single instant, speed is exactly the magnitude of the instantaneous velocity. The gap opens only when you average over a path that curves or reverses. In exam questions, words like "returns", "round trip", "lap" or "semicircle" signal that distance and displacement differ.
Got this one? The Knot Scan finds the misconception you don't know you have, in 8 questions.