Related to Distinguish between instantaneous speed and instantaneous velocity

I understand that the average velocity is given by the displacement divided by the change in time, and it is a vector quantity. Similarly, the average speed is given by the distance traveled divided by the change in time, and it is a scalar quantity.

I am also aware that the instantaneous speed is the magnitude of the instantaneous velocity.

However, I am unsure how to prove this relationship using the average formulas mentioned above. How can we show that the infinitesimal displacement has the same magnitude as the infinitesimal distance traveled?

I mean, prove that $|\lim_{\Delta t\to 0} \frac{displacement}{\Delta t}|=\lim_{\Delta t\to 0} \frac{distance\,traveled}{\Delta t}$.

This is analogue to discuss why, with a decreasing time interval, we obtain, in the limit, the instantaneous scalar acceleration, which is precisely the magnitude of the instantaneous vector acceleration.

Thank you.

  • $\begingroup$ What is there to prove? These are definitions. $\endgroup$
    – d_b
    Commented Feb 26 at 17:55
  • $\begingroup$ @d_b I mean, prove that $|\lim_{\Delta t\to 0} \frac{displacement}{\Delta t}|=\lim_{\Delta t\to 0} \frac{distance\,traveled}{\Delta t}$ $\endgroup$ Commented Feb 26 at 17:57
  • $\begingroup$ Are you asking about the Pythagorean theorem? $\endgroup$
    – John Doty
    Commented Feb 26 at 17:58
  • $\begingroup$ @JohnDoty sorry, I could not understand your question, could you explain please? $\endgroup$ Commented Feb 26 at 17:59

2 Answers 2


If A and B are two points with a finite separation on the path of a moving particle, the segment of path between these points need not be straight, so, for this segment:

$$\text{distance gone}\geq \text {|displacement|}.$$

Dividing by the time taken to go from A to B we have,

$$\text{speed}\geq \text {|velocity|}.$$

However as we take B closer and closer to A the segment of path approaches a straight line, so the distance is the same as the magnitude of the displacement, and the $\geq$ becomes simply = in both relationships.


The direction of the tangent to the path of the particle in space at any point is the direction in which the particle covers the distance in an infinitesimal amount of time. To obtain the direction of displacement vector at the point you can choose any point in space as the origin and take axes arbitrarily. In the vicinity of the point on the path choose any other point on the path. Subtract those position vectors(which is the same as joining them head to head) and you will obtain that the displacement vector is in the direction of the tangent of the curve as the variable point approaches the fixed point.


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