# How does instantaneous velocity or acceleration have any other numerical value than 0? [duplicate]

Instantaneous velocity is defined as the limit of average velocity as the time interval ∆t becomes infinitesimally small. Average velocity is defined as the change in position divided by the time interval during which the displacement occurs.

When the time interval is infinitesimally small, there shouldn't be any considerable change in position. Thus the instantaneous velocity should be 0.

• Have you ever taken calculus?
– user93237
Sep 9, 2019 at 3:24
• @McFluff I think Zeno of Elea might agree with you... Sep 9, 2019 at 7:02
• This belongs in mathematics stackexchange. Sep 9, 2019 at 7:50
• Yes, I have taken calculus. But just the basics. Sep 9, 2019 at 11:55
• @McFluff Zeno was an ancient Greek philosopher that had exactly the same idea. I was hoping you might Google him... lmgtfy.com/?q=zeno+of+elea Sep 9, 2019 at 12:06

Suppose you are travelling at a uniform velocity and you cover 1 meter in 1 second. Your average velocity is

$$\frac{1\ {\rm m}}{1\ {\rm s}} = 1 \frac{\rm m}{\rm s}.$$

If you consider a 1 millisecond interval within that 1 second, you cover 1 millimeter. Your average velocity in that milisecond is

$$\frac{1\ {\rm mm}}{1\ {\rm ms}} = 1 \frac{\rm m}{\rm s}.$$

If you consider a 1 microsecond interval within that 1 second, you cover 1 micron. Your average velocity in that microsecond is

$$\frac{1\ {\rm \mu m}}{1\ {\rm \mu s}} = 1 \frac{\rm m}{\rm s}.$$

If you consider a 1 nanosecond interval within that 1 second, you cover 1 nanometer. Your average velocity in that nanosecond is

$$\frac{1\ {\rm nm}}{1\ {\rm ns}} = 1 \frac{\rm m}{\rm s}.$$

No matter how small a time interval you consider, the distance traveled is proportionally reduced, and the average velocity covered in that time remains 1 m/s, rather than falling to 0.

Therefore we say that the limit of the average velocity, as the time interval approaches zero, is a non-zero value (1 m/s in my example).

$$v_\text{average}=\frac{\Delta s}{\Delta t}$$ $$v_\text{instantaneous}=\lim_{\Delta t\to0}\frac{\Delta s}{\Delta t}$$

• If the time interval gets infinitesimally small $$\Delta t\to 0$$, then you are dividing with something very, very tiny - so the number should become very big: $$\frac{\cdots}{\Delta t}\to \infty \quad\text{ when } \quad\Delta t\to0$$
• If the change in position gets infinitesimally small $$\Delta s\to 0$$, then you are multiplying with something very, very tiny - so the number should become very small: $$\frac{\Delta s}{\cdots}\to 0 \quad\text{ when } \quad\Delta s\to0$$

Now, what if both happen at the same time, $$\frac{\Delta s}{\Delta t}$$? What if, as in your case, the $$\Delta s$$ is tied to $$\Delta t$$ so that when one becomes very small, the other one does as well? Then how do you know, which of them that affects the number the most? The denominator or the numerator? Does the number become very large or very tiny?

$$\frac{\Delta s}{\Delta t}\to\text{ ?}\quad\text{ when } \quad\Delta t\to0$$

You seem to be assuming that the tiny change in position $$\Delta s$$ is the one that dominates, so the result should go towards $$0$$ - but why wouldn't you assume the tiny time interval $$\Delta t$$ to dominate instead, so the result goes towards infinity $$\infty$$?

The answer is that anything can happen, depending on the values. It depends on the exact relationship between them. If the result goes towards an infinitely large number, we say that it is diverging. If it stabilises at some number, we say that it is converging. In work with physics you will often see it converging, since you will often deal with values that are interdependent and that "balance off" at some resulting number. In the case of velocity, the result does indeed converge towards some value, which we then choose to call the instantaneous velocity.

This is what calculus is all about: the mathematical discipline of going towards - converging towards - a limit and then figuring out what that limit is.