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Is speed the magnitude of the velocity vector or the travel distance divided by the travel time?

A common example shows the problem: circular motion. "When something moves in a circular path (at a constant speed …) and returns to its starting point, its average velocity is zero but its average speed is found by dividing the circumference of the circle by the time taken to move around the circle. This is because the average velocity is calculated by only considering the displacement between the starting and the end points while the average speed considers only the total distance traveled." (https://en.wikipedia.org/wiki/Speed)

In this case the magnitude of the velocity vector is zero but the speed is positive.

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We must distinguish between instantaneous and average quantities.

Average velocity is the total displacement divided by time, while average speed is the total distance divided by time,

$$\vec{v}_{av}=\frac{\Delta\vec{x}}{\Delta t},\quad\quad v_{av}=\frac{\Delta s}{\Delta t}$$

Instantaneous speed is just the magnitude of instantaneous velocity,

$$v(t)=|\vec{v}(t)|$$ However, $\Delta s\neq|\Delta\vec{x}|$. So, the average speed is not the magnitude of the average velocity.

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  • $\begingroup$ Yes, but then the definition in physics books that speed is the magnitude of velocity is misleading at best. $\endgroup$
    – RG1
    Oct 13, 2017 at 21:16
  • $\begingroup$ They just talk about instantaneous quantities. $\endgroup$ Oct 13, 2017 at 22:53

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