I want to know why, in kinematics, is velocity described as $v = \frac{\Delta x}{\Delta t}$, and why it is not described as any other expression (like a multiplication), why does a division is the one that fits for velocity.

  • $\begingroup$ How would you define velocity? $\endgroup$ – Ofek Gillon Nov 2 '18 at 19:21
  • $\begingroup$ Once you have developed the means to measure distances and the means to measure time, you're bound to notice that for a person walking, the distance travelled is roughly proportional to the time spent walking. And then you notice that it's the same for a person riding a camel, or a horse, etc. but the constant of proportionality is different. Pretty soon, you'll want a name for that constant of proportionality. In the English language, we call it "speed" or "velocity". $\endgroup$ – Solomon Slow Nov 2 '18 at 20:58
  • $\begingroup$ Hint: You're asking why we describe speed with a division, but a related and relevant question could be, "Why did we invent division?" $\endgroup$ – Solomon Slow Nov 2 '18 at 21:01

First, we need to define what speed is. Let's say, we define speed as a number of meters an object moves per one second.

Then we try to figure out how to calculate the speed given some distance in meters and some time in seconds it takes for an object to move by that distance. Say, it is $\Delta x$ meters and $\Delta t$ seconds.

So, we are thinking: if it takes an object $\Delta t$ seconds to move by $\Delta x$ meters, this distance must be $\Delta t$ times greater than the number of meters the object moves per one second, which is our definition of speed. Also, by definition of multiplication, if $C$ is $B$ times greater than $A$, $C$ could be expressed as a product of $A$ and $B$: $C=A\times B$.

Based on that, $\Delta x$ must be a product of speed, $v$, and $\Delta t$:

$\Delta x=v \times \Delta t$

From here: $v=\frac {\Delta x}{\Delta t}$.

  • $\begingroup$ But, when you define speed as no. of meters an object moves per one second, you are actually stating the last division formula. This looks like a circular argument. As I see it, the key is: why would we want to define speed like that? $\endgroup$ – FGSUZ Nov 3 '18 at 14:56
  • $\begingroup$ @FGSUZ I think it is a matter of interpretation of the question. Since it is so basic, it is hard to tell what is assumed to be known and what needs to be determined. My interpretation was that we already know what the velocity (or speed) is, i.e., we know what we need to calculate, but we don't know how. $\endgroup$ – V.F. Nov 3 '18 at 16:46

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