Consider a particle that moves around the coordinate grid. After $t$ seconds, it has the position $$ S(t)=(\cos t, \sin t) \quad 0 \leq t \leq \pi/2 \, . $$ The particle traces a quarter arc of length $\pi/2$ around the unit circle. This means that the average speed of the particle is $$ \frac{\text{distance travelled along the arc of the circle}}{\text{time}}=\frac{\pi/2}{\pi/2} = 1 \, . $$ However, since the motion of the particle is circular, the distance travelled is not the same as the displacement. The displacement of the particle would be $\sqrt{2}$, and so the average velocity would be $$ \frac{\text{straight line distance from initial position}}{\text{time}} = \frac{\sqrt{2}}{\pi/2} = \frac{2\sqrt{2}}{\pi} \text{ at angle of $\frac{3}{4}\pi$ with the positive $x$-axis} \, . $$ Here is the part I don't quite understand: over an interval, the average speed of the particle is different from the magnitude of its velocity. In the above example, the former is $1$, whereas the latter is $\frac{2\sqrt{2}}{\pi}$. However, the magnitude of the instantaneous velocity of the particle is the same as the instantaneous speed: here, they are both equal to $1$. We can mathematically prove this by considering the following limit $$ |S'(t)| = \lim_{h \to 0}\frac{|S(t+h)-S(t)|}{|h|}=\lim_{h \to 0}\frac{\sqrt{\left(\sin(t+h)-\sin t \right)^2+\left( \cos(t+h)-\cos t\right)^2}}{|h|} \, , $$ which turns out to be equal to $1$. Hence, the magnitude of the instantaneous velocity is $1$. And clearly, the instantaneous speed of the particle is $$ \lim_{h \to 0}\frac{h}{h} = 1 \, , $$ since the distance travelled along the arc between $S(t+h)$ and $S(t)$ is simply $h$ units. However, will this always be the case? Is the magnitude of the instantaneous velocity of a particle always equal to its instantaneous speed?

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    $\begingroup$ To my knowledge the definition of speed is that it is the magnitude of the velocity, $v := |\vec v|$. $\endgroup$
    – Semoi
    Commented Feb 8, 2021 at 18:16
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    $\begingroup$ What you are discovering here has nothing to do with physics but how any smooth and continuous function can appear linear when seen on a small enough scale. $\endgroup$
    – Triatticus
    Commented Feb 8, 2021 at 18:37
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    $\begingroup$ @Triatticus Thanks, that makes sense. Is there a precise way of formulating this mathematically? And if so, is there a way of proving this statement as a theorem? $\endgroup$
    – Joe
    Commented Feb 8, 2021 at 21:52

1 Answer 1


By definition, $$\left|\text{instantaneous velocity}\right| = \text{instantaneous speed}.$$

However, \begin{aligned} \left|\text{average velocity}\right| &= \left|\frac{\text{displacement (i.e., change in position)}}{\text{time elapsed}}\right|\\ &= \frac{\left|\text{displacement (i.e., change in position)}\right|}{\text{time elapsed}}\\ &\leq \frac{\text{distance travelled}}{\text{time elapsed}}\\ &= \text{average speed}. \end{aligned}

  • $\begingroup$ For example, for constant-speed motion along one complete circle, the average velocity is zero whereas the average speed is nonzero. $\endgroup$
    – ryang
    Commented Jan 17, 2023 at 18:52

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