I have a rather basic, maybe even dumb, question. I was wondering why speed is defined as it is:

$s = d/t$

Of course, what the equation means is not too difficult to understand. However, there are many ways that d and t could be related, for instance:

$s = d + t$

I am not sure who the first person to define speed was, but I was wondering how they made the decision to define speed as distance divided by time.

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    $\begingroup$ Suppose I go one meter in one second, call that speed $v$. Now suppose I go one meter in two seconds. Doesn't that sound like the speed should be half, i.e. $v/2$? $\endgroup$
    – DanielSank
    Jul 30, 2017 at 18:10
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    $\begingroup$ @dts I get it: you want to add distance with time, ie [L] with [T]. I don't think that's quite supported. At least all books that I've read till the university level say that only similar quantities can be added. Maybe you've found a new theory. $\endgroup$ Jul 30, 2017 at 18:29
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    $\begingroup$ @dts speed is speed. You cannit ask why it's that. Feynman had said that Physics doesn't find answers to why always. I could ask why quarks have flavours, or why electron is fundamental. But these are stupid questions. $\endgroup$ Jul 30, 2017 at 18:33
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    $\begingroup$ It is a definition. There's no why to a definition. If I define "wibble" as "foo" divided by "bar", that's just a definition. Speed just happens to be a useful definition, which wibble is not. Adding quantities of with different units makes no sense. $\endgroup$ Jul 30, 2017 at 19:17
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    $\begingroup$ Also, I wonder why the word "garage" is defined as a structure where cars are parked. Of course, that definition is not too difficult to understand. But the word "garage" could have had many other meanings. It could have meant "three quarters of a pizza", for example. I'm not sure who the first person to define "garage" was, but I was wondering how they made the decision to define it as they did, instead of differently. $\endgroup$
    – WillO
    Jul 31, 2017 at 0:48

6 Answers 6


The definition of speed (please, let me call it velocity hereinafter) is not random at all.

It seems you understand that it must depend on the distance $d$ and the time $t$, so I'll skip to the next stage.

Evidently (for a constant $t$) velocity increases if $d$ does; and (for a constant space) $v$ decreases if $t$ rises. That constrains the ways we can define it. For example, your example of $d+t$ is authomatically discarded. You could say $d-t$, that satisfies the growing conditions.

Then we apply the reasoning in the limit case. For a 0 distance, velocity must be 0 independently of time (unless time is 0 too), that discards any sums. If the time to reach the space is infinite, the velocity must be 0. That's forcing $t$ to be a denominator.

So we deduce it's a fraction, but how can we sure there are not powers of those quantities? We impose the linearity of space. It doesn't make sense that the velocity is different if you pass from 50 to 60, or from 70 to 80 in the same time. If all points in space are equivalent, there cannot be distinctions like these, so using the numerator $\Delta d$ guarantees that all points in space are equivalent. If it were $\Delta d^2$ the result would be different from 70 to 80 and from 50 to 60, for example. That's againts the obvious principle that we can set the origin where we want (we must be able to measure from the point we choose, as we do everyday with a simple ruler, placing it where we want). The same reasoning applies to time.

So they must be a fraction, and there cannot be other powers than 1. The only possible difference is a constant factor

$s=k \frac{\Delta d}{\Delta t}$

And this is what speed (or velocity) is, after all. The constant is actually the unit factor. It depends on what units you are using. I hope this is useful to you.

  • $\begingroup$ This is exactly what I was looking for! Thank you so much! $\endgroup$
    – dts
    Jul 30, 2017 at 20:48
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    $\begingroup$ This seems to pre-assume what velocity/speed is though. You say "Evidently (for a constant t) velocity increases if d does; and (for a constant space) v decreases if t rises. That constrains the ways we can define it. " But that already comes from the definition that speed is distance traveled during a set amount of time. $\endgroup$
    – JMac
    Jul 30, 2017 at 20:48
  • $\begingroup$ I'm so glad this was useful, as I don't use to know enough to help. @JMac That's a nice note. I guess you are right, it's true, I pre-assumed what $v$ is. After all, I think that the question didn't mean why we define a physical quantity like that, but "how and why our everyday experience yiedls that definition" . This is probably more philosophy but... I'm from the ones who think that space and time are innate ideas, and so its relation is acquired by experience. I think I only did a Socrates act: I only made explicity what was probably already inside our minds. Thanks again for your note $\endgroup$
    – FGSUZ
    Jul 30, 2017 at 20:53
  • $\begingroup$ @FGSUZ I just find this addresses a misconception. The fact is, the only "experience" that has to do with it is that we choose to say "velocity is a measure of distance per time" the same way we choose to define everything else. There's no everyday experience that makes us decide "yes, this we shall call speed!", it could have been called anything. When talking about speed you know more than just that we're talking about distance and time, we know that by definition we are talking about $v \equiv \frac d t$ it is equation we ourselves define. It's good it helped OP I guess though. $\endgroup$
    – JMac
    Jul 30, 2017 at 21:32
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    $\begingroup$ I was taught that "speed" was a scalar, and "velocity" a vector. So if you're talking about the scalar "distance" as the "d" in the equation, then you'd better be talking about "speed" rather than "velocity", or you're doing it wrong. $\endgroup$ Jul 31, 2017 at 0:11

The measure of distance over time is useful in physics.

Like many useful measures, it was given a name; in this case speed.

  • $\begingroup$ But why did we name this quantity "speed" rather than some different quantity? Humans have had a notion of speed for much longer than we've been dividing distances by times. $\endgroup$ Jul 31, 2017 at 11:19
  • $\begingroup$ @TannerSwett Why does it matter what we named it? We've known that spatial change relative to elapsed time is an important quantity so we gave it a name. The question asked why it's called speed, not why speed is an important quantity. Although we didn't always explicitly divide distance by time, that's exactly what our minds processed movement as, so naturally we made some definition for different aspects of it. $\endgroup$
    – JMac
    Jul 31, 2017 at 11:23
  • $\begingroup$ @TannerSwett Also, the human notion of speed is exactly space covered over time. $\endgroup$
    – gented
    Jul 31, 2017 at 11:43
  • $\begingroup$ My point is, I feel like this answer misses the point of the question. @JMac, it doesn't matter what we named it, and I didn't ask why we named it that. I asked why we chose this quantity, rather than some other quantity, as being the correct quantity corresponding to the pre-existing word "speed". $\endgroup$ Jul 31, 2017 at 11:58
  • $\begingroup$ In other words, there are two different concepts of "speed". One is the intuitive "swiftness" that we automatically get an impression of by looking at a moving object; call that speed-1. The other is distance divided by time; call that speed-2. The two concepts are equivalent, of course, but the OP is asking how do we know that they're equivalent, and you're not answering that. $\endgroup$ Jul 31, 2017 at 12:01

It is nothing but a name given to rate of change of distance with time. If you know the speed and any other quantity (distance or time), then you can find the third one.

P.S. You can add only dimensionally same quantities. So $s = d + t$ is wrong.

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    $\begingroup$ Although the accepted answer is fine, I think the postscript here deserves some attention. $\endgroup$
    – T. C.
    Jul 31, 2017 at 10:07

Imagine you have a car. I travel a mile in the car. But in what amount of time? If I travel a mile in an hour, that's a very slow car. But if I travel a mile in a minute, that's a decent car.

Let's say we have a decent car, and it traveled a mile in a minute. How far could we go over an hour? Well, there are 60 minutes in an hour, so we go 60 times the distance we went in the first minute - 60 miles in an hour.

What we basically just did is set up a proportion - 1 mile corresponded with 1 minute, so what distance corresponds with 60 minutes? We write this out mathematically as $$\frac{1\text{ mile}}{1\text{ minute}} = \frac{x\text{ miles}}{60\text{ minutes}}$$

(You solve this by "cross-multiplying" - 60 minutes * 1 mile = x miles * 1 minute, and then we'd divide both sides by a minute, so here, basically the units just cancel, and we get 60 * 1 miles = 60 miles.)

Now, imagine we said we wanted to measure how 'fast' the car is going, and we'll call that speed. It's obviously a relation between distance and time ($d$ and $t$). We've already seen above that distance is proportionate to time, that is, it's represented by division.

Let's look at this a different way. If we travel a larger distance in a smaller time, the speed is higher. If we travel a shorter distance in a longer time, the speed is lower.

When we think about a number divided by another number, when the number on top (the numerator) is bigger than the number on the bottom (the denominator) the result of the division (the quotient) comes out bigger, like in 8 / 2 = 4 vs. 6 / 2 = 3. When the denominator is bigger, the result comes out smaller, like in 6 / 2 = 3 vs. 6 / 3 = 2.

In other words, division satisfies the properties the representation of speed needs to have - when $d > t$, $d/t$ (the speed) is large. When $d < t$, the speed is smaller.

A final way to think about it. We talk about a car's speed in miles per hour, or kilometers per hour. Miles/kilometers are units of distance. Hours are units of time. So we have $d/t$ again.


In short, speed is the rate of change of distance over time, and the equation is derived from calculus.

Strictly speaking, s=d/t is not true in general. Speed is the absolute value of the velocity, which is defined as the rate of change of the displacement with respect to time. For the 1 dimensional case velocity is given by:


Taking things a step further, acceleration is the rate of change of velocity:


Now, if you have no acceleration, the velocity can be calculated by solving the integral:


Here, $C_{1}=v$, keeping things simple. The displacement is then:


Now, if d=0 at t=0, $C_{2}$ must also equal zero, so:


Or, equivalently:


Speed is the absolute value of this, i.e: $s=|d/t|$

If acceleration isn't zero, speed is $s=|at+v_{0}|$ where $v_{0}$ is the initial velocity. In this case it becomes awkward to define it in terms of the distance traveled. Acceleration can change over time as well, leading to a more complex relationship.

  • $\begingroup$ Thank you for the answer! I have been thinking about this definition too. I have seen many textbooks simply say that v=d/t, and it seems like they have some intuition that I do not. So would this be the "formal" proof that v=d/t (for constant acceleration)? $\endgroup$
    – dts
    Jul 31, 2017 at 3:22
  • $\begingroup$ I suppose it is the formal proof. I think textbooks like to avoid calculus to keep things simple, but I believe they are wrong to do it. Showing velocity and acceleration as rates with respect to time is more intuitive, IMHO. $\endgroup$ Jul 31, 2017 at 3:26
  • $\begingroup$ I know many people write $\frac{dx}{dt}$ instead of the IMO better $\frac{\mathrm{d}x}{\mathrm{d}t}$, but in case of $\frac{dd}{dt}$, those italic ds are really confusing. Mind if I edit them to roman style? $\endgroup$ Jul 31, 2017 at 11:03
  • $\begingroup$ Go ahead. I wasn't sure how to do it in Mathjax. $\endgroup$ Aug 2, 2017 at 2:59

When you're developing a physical theory, you're free to define your quantities as you like. You won't get away with $s = d + t$ since dimensions of addends don't match, but you can still come up with a whole bunch of equations, e.g. $s = d × t$.

In the end, physical theories are useful insofar they can describe the real world and predict what happens. Speed (or velocity) defined as $s = d / t$ is very useful for this: objects having the same velocity share a lot of interesting properties, like having a constant distance between them, or going from start to finish in an equal amount of time. Speed defined as $s = d × t$ just doesn't predict anything useful (or very little), that's why nobody defines it like this.


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