Why Isn't Average Speed Equal to Average Velocity?
1D Kinematics: Average Velocity vs. Average Speed
Average Velocity depends only on the start and end positions — it is displacement ÷ time. The slope of the secant line on the x(t) graph gives exactly this value.
Average Speed uses the total path length — distance ÷ time. Try moving back and forth: displacement shrinks while distance keeps growing. Average speed ≥ |Average velocity|, always.
Turning points are where velocity changes sign — marked with a violet dot and drop-line on the graph. At that instant the object momentarily stops before reversing direction.
v̄ = Δx / Δt | s̄ = d / Δt | slope of secant = avg velocity
© The Science CubeWhy Does Average Speed Come Out Bigger Than Average Velocity?
In short: Average velocity is displacement divided by time, Δx/Δt, so it only counts the start and end positions. Average speed is distance divided by time, d/Δt, and counts every bit of the path. Distance never decreases, so average speed is never smaller than the size of the average velocity. They match only when the particle never turns back.
Common doubts
Can average velocity be zero while average speed is not? Yes. Drag the particle out and back to where it started: the displacement is zero, but the distance is not, so average speed is positive.
Can average speed be negative? No. Distance is never negative, so average speed is never negative. Average velocity can be negative when the net displacement points in the negative direction.
What does the secant line show? On an x(t) graph, the slope of the line joining two points is the average velocity between those two times.
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