# Instantaneous speed x 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.

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

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