I'm a philosophy student (I, regrettably, don't know calculus or much physics). Last year I spent some time learning how work, power, speed, velocity, energy, force, and acceleration relate. But I was never able to fit my understanding of acceleration into my understanding of the world. I think my biggest challenge was understanding how you can have (m/s)/s. I.e. the 'per-second per-second' part doesn't make sense to me — and how that represents an increase in speed. Does it express, every second a thing travels X meters/second faster? That's my best guess, but I see a lot of problems with that guess, so I presume that it's incorrect.

In words, why is acceleration expressed as (m/s)/s? How does that expression relate (if at all) to the everyday notion of acceleration?

  • 17
    $\begingroup$ Your guess is basically correct. Which are the problems that this guess generate to you? $\endgroup$ Jan 30 '14 at 14:42
  • 2
    $\begingroup$ @IgnacioVergaraKausel this guess is only correct if the acceleration would be constant and in the same direction as the velocity (special relativity aside). $\endgroup$
    – fibonatic
    Jan 30 '14 at 19:54
  • 1
    $\begingroup$ Acceleration is simply expressed as velocity per seconds. And Velocity as meter per seconds. So the SI notation is m/s/s. But the '/s/s' part doesn't have sens by itself, don't be confusing about SI unit notation. $\endgroup$
    – FabiF
    Jan 31 '14 at 16:28
  • 2
    $\begingroup$ Perhaps it will help to realize that slowing down is also acceleration. When you put your foot on the accelerator in a car, your speed goes up. When you put your foot on the break, your speed goes down. Both are accelerations; the magnitude of the acceleration tells you how quickly you are gaining or losing speed. $\endgroup$ Jan 31 '14 at 19:04
  • 1
    $\begingroup$ Now take it a step further. You step on the accelerator; does the acceleration go from zero acceleration to some non-zero acceleration instantly? Or does the acceleration itself change over time as your foot goes down? Obviously the latter. The same with braking. So just as acceleration tells you how velocity is changing with time, jerk tells you how acceleration is changing with time. When you feel like someone is driving jerkily, your body is measuring the jerk, not the acceleration - it is measuring how fast the acceleration is changing. $\endgroup$ Jan 31 '14 at 19:08

10 Answers 10


Maybe it will be even clearer to you if one explained it in a more fundamental way, but for this, we need a little bit of senior grade mathematics. I am assuming you have heard of derivatives; if my assumption is false, I am sorry for that, but in this case this answer might not be helpful to you.

Let's get clear about something important (but rather philosophical) first. This speed and acceleration stuff isn't real. It is some sort of thought experiment that is quite useful in that it helps describing our world.

Let's take some object - without restriction and for the sake of simplicity, let's assume it's an apple - and push it around (in your head). What is happening? The position of the object changes over time, so here we've got a connection of two fundamental physical units, distance and time. You can speak of distance as a function over time (that means, you can plot it with the x axis being the time axis and the distance at a given time are the y values).

Now, let's have a look at the speed (and now, again for sake of simplicity, assume the object is travelling in a straigt line, otherwise you'll get some more general vector spaces that might be nasty to imagine). How do you calculate average speed? So, if the apple would have been travelling at the same speed all the time, how big would this speed need to be?

Basically, the formula is $v_{average} = \frac{\text{distance}}{\text{time}}$ (quite intuitive, I think). But again, this is already of theoretical nature. It is not some sort of "inherent property", but physicists have "invented" it to describe processes.

If you don't want to calculate the average speed for the whole distance, but only for a certain period of time, the formula is still $v_{average} = \frac{\text{distance}}{\text{time}}$, but of course, you have to change the values for time and distance accordingly.

Here's a picture:

Average speed

$\Delta$ is the Greek letter Delta and means "difference" - difference between start and end distance and start and end time. The straight line in the picture is called a secant and its slope is equal to the average distance. (Just believe me on this one - I don't know how to make it appear more plausible at the moment.)

Now you can ask the question of speed at a certain moment and you have to realize that the equation above won't work any more.Looking at only one point, the difference between start and end time and start and end position is zero. Now, you are not allowed to divide by zero, and that's a problem.

Imagine this geometrically: you are moving one of the two points along the curve until the two points are identical. The secant from above has always been dependent on two points. Nor, there's only one, so theoretically, there's an infinite amount of lines that go through this one point. However, only one line (well, supposing all this is differentiable - ignore that) does actually give us what we want. It should be the tangent to the curve. Now, that's what we call speed. All the slopes of the tangents in a point of the curve form a new graph which gives you speed over time, which is the derivative of position with respect to time.

Completely analoguously, if something is travelling, you might want to know how the speed changed. For example, imagine an inclined plane with our apple on it. Depending on the material of the plane and the slope of it, the apple might become faster (friction not so big), remain constant (friction equal to gravitation that pushes the apple "downwards") or may become slower (lots of friction).

enter image description here

This is described with acceleration. If the speed is constant, acceleration is zero, because nothing happens. If the object becomes faster, the acceleration is positive, because acceleration is the rate of change of speed. Similarly, if the apple becomes slower, there is negative acceleration. Now, to measure the average acceleration, we do the same thing as above: $\text{time} * \text{acceleration} = \text{velocity} \implies \text{acceleration} = \frac{\text{velocity}}{\text{time}}$. Now just look at the units: On the right side, you already have meters per second for speed, and now you are looking at the change of this speed over time. this gives you (meters per second) per second.

By the way, you can apply exactly the same ideas I mentioned before (secant, tangent, derivative) to the velocity graph and you will see that acceleration is the derivative of speed.

By the way, I would really encourage you to keep reading and thinking about physics, mathematics and the other sciences. It is always good to work interdisciplinary and I think it is crucial for a philosopher to know what those science people seem to "know" about everything out there. I have seen too many philosophers building theories that just - well - didn't match reality.

I think, this youtube series on the topic is quite well done and you might enjoy watching it.

  • 2
    $\begingroup$ This speed and acceleration stuff isn't real what do you mean it's not real?! Momentum (mass times velocity) is a very real quantity. A bullet in front of your face standing still is very different to a bullet in front of your face approaching you at hundreds of meters per second. Momentum is as real as the physical position! $\endgroup$
    – cheshire
    Jan 31 '14 at 14:58
  • 1
    $\begingroup$ Agree with cheshire. distance as in distance traveled is less real than speed. Distance is just the spatial component of the difference between two positions in space-time. $\endgroup$
    – MSalters
    Jan 31 '14 at 15:59
  • 1
    $\begingroup$ @cheshire and MSalters: I'm pretty sure he means "real" as in tangible or pseudo-tangible. Distance can be seen and measured with something solid, but speed/velocity and acceleration has to be measured over time and calculated. $\endgroup$
    – Izkata
    Jan 31 '14 at 16:45
  • 1
    $\begingroup$ I agree that this bit is a little dodgy. I am drawing analogies here to what Dedekind said: "God made the natural numbers; all else is the work of man." n apples (with n being a natural number) really denotes a set of apples while half an apple does not, but has to be defined to apply to the real world. Now, position is somewhat similar to natural numbers, although even here you need to define a metric (and finding the right one is not exactly trivial). Still, derivatives involve limits and, therefore, the infinite (or infinitesimal). (And I did not say position and distance are real.) $\endgroup$
    – chaosflaws
    Jan 31 '14 at 18:23
  • $\begingroup$ I think he means the descriptive concepts aren't real. Arguably, real things exist independently of the things that conceive of them; whereas concepts do not. In any case, that sentence was the one that corrected my thinking, so thank you for it. $\endgroup$
    – Hal
    Feb 1 '14 at 4:42

Your interpretation is correct if acceleration is constant, and the motion is in a straight line. The object will change its velocity by that much every second.

A quick example: If you drop an object, its acceleration will be about $9.8~\text{(m/s)/s}$. This means after one second it's traveling at $9.8~\text{m/s}$, after two seconds it's traveling at $19.6~\text{m/s}$, and so on.

As a side note, most often people "do math" on the units so that (m/s)/s is written m/s$^2$. This hides the interpretation of acceleration, though. Your way of writing it is more clear and is just as correct.

(The interpretation gets a little trickier if acceleration is not constant or not in a straight line. In the latter case, one could have a constant speed but a changing velocity due to direction change. But the interpretation still has to do with a change in velocity per unit of time.)

  • $\begingroup$ Thank you. The problem I saw with my understanding: I presume, that (m/s)/s or m/s^2 are complete definitions of acceleration. However, I do not see anything in those expressions that defines acceleration as the accumulation of speed. I.e. in English we might express (m/s)/s as the number of meters a person travels in a (or in x) second(s), in a (or in x) second(s). To me it seems like the notion would be better expressed (and I don't know better, which is why I presume I'm misunderstanding) as the difference between the speed of a thing at one instant and its speed at another instant. $\endgroup$
    – Hal
    Jan 30 '14 at 17:06
  • 10
    $\begingroup$ Acceleration is not an accumulation of speed (much like the upward speed is not accumulation of the elevation). An accumulation of speed over an interval (i.e. the integral of acceleration) is the speed difference between the final and initial moment and it has units of velocity, m/s, not the units of acceleration. Acceleration is not accumulation; acceleration is the rate by which the speed is increasing at a given moment. The whole point of the acceleration is that it is the derivative of the velocity, and the derivative is sort of the opposite thing than accumulation (integral)! $\endgroup$ Jan 30 '14 at 17:36
  • 4
    $\begingroup$ @Hal I think a better english statement would be "the number of (meters per second) that an object accumulates in a second". The parentheses you used in your question are very helpful in guiding intuition. $\endgroup$
    – Kyle Oman
    Jan 30 '14 at 19:24
  • $\begingroup$ @LubošMotl thank you. That actually helps quite a bit. But, one core part of my question remains: acceleration is the rate by which the speed is increasing at a given moment. How does the expression (m/s)/s convey anything about an increase in anything? $\endgroup$
    – Hal
    Jan 30 '14 at 22:50
  • 2
    $\begingroup$ Maybe you are asking why "per" is represented by the "slash", "/". It's because this sign means mathematically "over", the word indicating dividing (division) and division is the opposite of multiplication. For example, if someone accelerates at the acceleration 10 m/s/s for 7 seconds, then he changes his speed by 10 m/s/s times 7 s = 70 m/s (that was multiplication). Note that one of the "s" canceled because s/s=1. This relationship may be reverted. If the speed change is 70 m/s and it takes 7 seconds to achieve it, the (average/constant) acceleration is 70 m/s / (7 s) = 10 m/s/s. $\endgroup$ Jan 31 '14 at 6:58

The problem I saw with my understanding: I presume, that (m/s)/s or m/s^2 are complete definitions of acceleration. However, I do not see anything in those expressions that defines acceleration as the accumulation of speed. I.e. in English we might express (m/s)/s as the number of meters a person travels in a (or in x) second(s), in a (or in x) second(s). To me it seems like the notion would be better expressed (and I don't know better, which is why I presume I'm misunderstanding) as the difference between the speed of a thing at one instant and its speed at another instant. –

But, one core part of my question remains: acceleration is the rate by which the speed is increasing at a given moment. How does the expression (m/s)/s convey anything about an increase in anything?

You can’t really talk about speed without specifying a time. You can’t say an object is traveling 10 meters. You have to specify how long it takes that object to travel that 10 meters. It could be moving 10 meters per second, or per minute, or per year. Speed has to have a time unit.

We can say that an object is moving 10 m/s at time A, and 20 m/s at time B. The speed increased by 10 m/s. But this tells us nothing about acceleration. Acceleration is about how long it takes an object to change speed. If our object only took one second to change speed from 10 m/s to 20 m/s, it accelerated very quickly. If it took 10 minutes, it accelerated much more slowly. The time it took for the change to happen is what the 2nd “per second” is talking about. Acceleration (m/s/s) is about how much the speed (meters per second) changes per second.

Here’s an example to go along with the explanation. The acceleration due to gravity is 9.8 meters per second per second. I’m going to round that up to 10 for simplicity. Suppose you go to the top of a tall building and drop a bowling ball off. Its initial speed is 0 m/s. Gravity pulls it downward. After 1 second its speed is 10 m/s. After 2 seconds, its speed is 20 m/s. After 3 seconds, it’s moving 30 m/s, etc. The bowling ball’s speed is changing by 10 meters per second for (or per) each second it falls: 10 meters per second per second.

I hope this helps. Physics can be hard to wrap your head around, especially if your training is not in the hard sciences.


To add to BMS' answer, the words "speed" and "velocity" have different meanings in science. An object's speed is a number telling how fast it's going. The velocity indicates both speed and the direction of motion, which makes it a "vector quantity." For example, a rocket's speed could be 100 mph and its velocity could be {70.7 mph in the x-direction and 70.7 mph in the y-direction} . These two perpendicular velocity components combine per the formula $V_{tot} = \sqrt{V_x^2 + V_y^2} $

  • 3
    $\begingroup$ And that's why police enforces speed limits, but not velocity limits - you are still free to choose which way to go (within the boundary conditions of the road) :) $\endgroup$ Jan 31 '14 at 1:45
  • 2
    $\begingroup$ @WojciechMorawiec makes you wonder if someone actually tried that defense... $\endgroup$
    – user
    Jan 31 '14 at 12:32

We have a terrible "everyday understanding of acceleration":

I think your real problem is that we don't have a very good everyday understanding of acceleration. We spend most of our time going about the same speed. The one place we do commonly think about acceleration is cars. High-performance cars will often brag about their acceleration like "goes from 0 to 60 mph in 6.2 seconds". They mix time units, using both hours and seconds, "miles per hour" is distance / time, and "in 6.2 seconds" is 1 / time. We can put this in Google to translate it to 5.58 m/s/s, but that's a much harder-to-interpret number (especially for an American!).

But in "0 to 60 mph in 6.2 seconds", hopefully, the units do make sense. From a stop, it takes 6.2 seconds to get up to freeway speed. Stopping a car is another place where we think a lot about changes in velocity, but even there we don't tend to focus on (de)acceleration, rather we focus on the distance covered while stopping---which would require a couple integrals to calculate based on acceleration data!

A little extra info: the rate of change of acceleration is jerk, which always makes me think of being on a roller coaster. Often, at the end of a roller coaster, as you pull up to the loading platform the cars are lightly braked, so your decelerating just a little bit. Then they put on a hard brake and stop you, which very quickly gives you a big negative acceleration, then no acceleration (big jerk). And usually the jerk bumps your head against the headrest.

Explanation with units:

Let's define

Rate of Change: amount of change divided by length of time for change.

Whatever units you use to measure a quantity, those are the units used for measuring differences in those quantities.

For velocity, which is the rate of change of position, we use meters to measure position. If you go 20 meters in 5 seconds, then your (average) velocity is 20 m / 5 s = 4 m/s.

Pretty much whenever time is in the bottom of a fraction, you've got a rate of change for whatever else there is. This way, m/s is a rate of change for meters (position).

Velocity, as we said above, is measured in m/s. If you're going 4 m/s, and then 10 seconds later you're going 9 m/s, your velocity has changed so clearly you've accelerated! Subtraction tells us the change in velocity, 9 m/s - 4 m/s = 5 m/s, but to get the rate of change we need to divide by the time it took for the change to happen: 10 seconds. 5 m/s / 10 s = 0.5 m/s/s.


Does it express, every second a thing travels X meters/second faster?

Essentially, yes. It's telling you that the speed is increasing at a rate of X meters/second per second.
Of course, the rate can increase or decrease, even after a mere 0.4 seconds. So X is the rate right now.


Newton stated that an object at rest will stay at rest, and one in motion will stay in motion (the concept of inertia). To change this requires force. Applying force will result in a change of velocity (speed with direction) over the given time interval that the force is applied. This is acceleration.

One can accelerate in any direction, although from our personal experiences driving, we usually think of accelerating in the direction our car is pointing. However, changing direction is an acceleration. So, is changing speed without a change in direction.

You can sense these forces when you apply the brakes in your car, or when you turn your car onto a side street without stopping.

It helps to consider the force that is generating the change in velocity when thinking of acceleration. It makes sense that acceleration starts when the force is applied and ends when the force is withdrawn. That becomes the time interval during which velocity changes.


The units of velocity is $\mathrm{m/s}$. The rate of change of velocity (velocity/time) is $\mathrm{(m/s)/s = m/(s \cdot s)}$, just what we call acceleration. So if I increase my velocity from $50$ to $60\: \mathrm{m/s}$ over 10 seconds my acceleration has been $1\: \mathrm{m/(s \cdot s)}$.


ensions: we have a unit for length L, one for time T and some few others. Now velocity has "dimensions" L/T, you could for instance choose to measure velocity in units of miles per minute since mile has dimension of a length L and minute has dimensions of time T.

Now we ask how does velocity change in time. That is, per unit of time T, in whatever units you choose for time. And the word "per T" is synonymous with the mathematical operation "1/T". Hence the words "velocity per time" should have the translation "velocity 1/T". The object I just wrote is very close to the definition of acceleration. What is the dimension of this object? Well, as we agreed upon above velocity has dimensions of L/T hence we expect the object to have dimension $L/T\cdot 1/T$ ie, $$\frac{L}{T^2}. $$ The last formula us obtained via just simple arithmetic.

Now let's choose some units for the dimensions: for instance we can choose to measure length in meters m and time in seconds s hence velocity would have the units of m/s and acceleration units of m/s/s.


I believe that the question derives from the "every day" understanding of "acceleration." To a person unfamiliar with physics, acceleration implies an increase in speed, and deceleration, a decrease in speed. In addition, the dependency of the change of speed on time is a blur, ignored, or unimportant.

In physics, we learn that acceleration (positive or negative) is the change in speed in a given amount/interval of time. So, if we concentrate on the units used to measure these components, we have: speed, in meters per second; and time, in seconds. Therefore, the units of acceleration are: (meters per second), (per second), which mathematically is written as (m/s)(/s).

To reiterate, if the speed of an object changes (up or down), this change obviously happens in an interval of time, and this is defined as acceleration. The units of measurement are a consequence of this definition!


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