5
$\begingroup$

This question already has an answer here:

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.

$\endgroup$

marked as duplicate by my2cts, WillO, John Rennie, Aaron Stevens, AccidentalFourierTransform Sep 9 at 23:53

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 42
    $\begingroup$ Have you ever taken calculus? $\endgroup$ – Samuel Weir Sep 9 at 3:24
  • 14
    $\begingroup$ @McFluff I think Zeno of Elea might agree with you... $\endgroup$ – Oscar Bravo Sep 9 at 7:02
  • 1
    $\begingroup$ This belongs in mathematics stackexchange. $\endgroup$ – my2cts Sep 9 at 7:50
  • 2
    $\begingroup$ Yes, I have taken calculus. But just the basics. $\endgroup$ – McFluff Sep 9 at 11:55
  • 12
    $\begingroup$ @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 $\endgroup$ – Oscar Bravo Sep 9 at 12:06
39
$\begingroup$

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

  • 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.

$\endgroup$
40
$\begingroup$

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).

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.