# Is there a difference between instantaneous speed and the magnitude of instantaneous velocity?

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?

• To my knowledge the definition of speed is that it is the magnitude of the velocity, $v := |\vec v|$. Feb 8, 2021 at 18:16
• 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. Feb 8, 2021 at 18:37
• @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?
– Joe
Feb 8, 2021 at 21:52

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}