Conceptual Reason
- Average speed uses total distance travelled, while average velocity uses displacement (straight-line change in position) over a finite time interval. Because the path can curve or reverse, distance ≥ |displacement|, so average speed ≥ |average velocity|.
- For instantaneous quantities, we look at motion over an extremely small time interval \(\Delta t \to 0\). In such an infinitesimally short interval, the particle has no time to bend its path appreciably or change direction significantly; its motion is effectively along a straight line.
- Therefore, over this tiny interval:
- Distance travelled ≈ magnitude of displacement.
- So, instantaneous speed (distance per unit time) equals the magnitude of instantaneous velocity (displacement per unit time).
Mathematical Argument
- Let \(s(t)\) be the distance travelled along the path and \(\vec{r}(t)\) the position vector.
- Instantaneous velocity: \[ \vec{v} = \frac{d\vec{r}}{dt} \]
- Instantaneous speed: \[ v_{\text{speed}} = \frac{ds}{dt} \]
- For motion along a curve, \(\dfrac{ds}{dt} = \left|\dfrac{d\vec{r}}{dt}\right| = |\vec{v}|\).
Hence, at any instant, the instantaneous speed is exactly the magnitude of instantaneous velocity, even though their average values over a finite time interval may differ.