Distance is paired with Displacement and it seems to be a bigger idea than just the magnitude of Displacement. Speed is paired with Velocity. I have always thought that there is not such pairing with Acceleration. I would teach: Acceleration is a vector, and we can talk about the magnitude of Acceleration, but we cannot talk about a scalar that describes the change in Speed over time.

Is there a scalar counterpart to Acceleration?

P.S. First question on physics.stackexchange. Please excuse me if the question is inappropriate.

Edit to Question: I think I need to explain the question better. If I walk to the store and back home, my displacement is zero and my velocity would also be zero (regardless of how long it took me). Since the store was some distance away, my total distance traveled for this situation would be positive and so my speed would be positive. I would then conclude that speed is not the magnitude of velocity, but something different (just like distance is a different concept than displacement). In this way I can say distance & speed are scalar (not magnitudes of vectors). Now that I define my question a bit better, I think I see the problem. It seems distance is a summary (i.e. sum of) of the magnitudes of a group of vectors (displacements). Speed is the summary of the magnitudes of the first derivative (with respect to time) of the group of vectors. I doubt there is (or is there a need for) a similar summary of the magnitudes of the second derivative of the group of displacements. I welcome your thoughts.

  • $\begingroup$ Centripetal acceleration has a magnitude but no fixed direction. $\endgroup$ May 2 '19 at 18:59
  • $\begingroup$ Centripetal acceleration has a well-defined direction at each instant. It just happens to change with time. $\endgroup$
    – G. Smith
    May 2 '19 at 20:03
  • $\begingroup$ The OP seems to he asking about whether there is a word that means the magnitude of the acceleration. The answer is no. $\endgroup$
    – G. Smith
    May 2 '19 at 20:05
  • $\begingroup$ "Since the store was some distance away, my total distance traveled for this situation would be positive and so my speed would be positive" No, this last part is not true. Because, speed is an instantaneous value. You've got to pick some moment during the trip, and check what the speed is at that moment. If you pick the end of the trip, where you are back home sitting still in your armchair, then your speed is zero. $\endgroup$
    – Steeven
    May 9 '19 at 8:31
  • $\begingroup$ True, I should have said 'non-negative', but I don't think that changes the question (hopefully that doesn't). $\endgroup$ May 10 '19 at 15:27

The way I read this question it's asking for an unambiguous word to use to describe only the scalar part of acceleration. To the best of my knowledge this word doesn't exist in the English language.

The thing is, Acceleration is a vector but the scalar part is also acceleration. The word is overloaded. Why? Well vector came along later.

vector (n.)
"quantity having magnitude and direction," 1846
vector | etymonline.com

acceleration (n.) "act or condition of going faster," 1530s
acceleration | etymonline.com

As you can see, in the English language at least, acceleration as a scalar without direction has a few years on the word vector. The problem is that we never introduced a word as the pair of acceleration to give an unambiguous way to distinguish between the two ideas. So when we started using vectors we just overloaded the word acceleration. You can use it to mean either one. Which means readers have to figure out the meaning from context.

If I walk to the store and back home, my displacement is zero

Well it is now.

and my velocity would also be zero (regardless of how long it took me). So long as right now you're holding still and not heading for the backdoor on your way to the pool.

Since the store was some distance away, my total distance traveled for this situation would be positive


and so my speed would be positive.

Er huh? You mean your average speed? Your speed now? These are not the same thing.

Distance is paired with Displacement and it seems to be a bigger idea than just the magnitude of Displacement. Speed is paired with Velocity.

I think this might be the root of the problem. This is not the same kind of pairing.

Rate of distance is speed. Rate of displacement isn't velocity. Velocity doesn't know or care where you started. Velocity doesn't know or care how fast you were going when you started. Velocity is about how fast you're going now and which way. No, rate of displacement is average velocity.

The difference between distance and displacement is that displacement is "as the crow flies". Displacement can be measured between two position points. Distance is measured over an infinite continuum of position points that represent everywhere you've been.

enter image description here

If I was going to organize these concepts they'd look like this:

| Measured at one point in time | Measured at two points in time         |
| Scalar        | Vector        | Scalar            | Vector             |
| Distance      | Position      | Displacement      | Displacement       |
| Speed         | Velocity      | Average Speed     | Average Velocity   |
| Acceleration  | Acceleration  | Ave. Acceleration | Ave. Acceleration  |

The difference between Speed and Average Speed:

It's possible to move 1 mile in an hour and at the moment that hour ends be going 60 miles per hour. This difference has nothing to do with dimensions. It has to do with driving like my Grandma.

For more about this look up the fundamental theorem of calculus

  • $\begingroup$ It sounds like your saying speed can have direction? I'm also confused about your table. wouldn't instantaneous velocity be the change in position (i.e. displacement) by dt? Please clarify. $\endgroup$ May 10 '19 at 15:34
  • $\begingroup$ @clausvalca226 if it's instantaneous there is no change in position. There is only one position. That doesn't mean you aren't moving in that instance. I'll update the table. $\endgroup$ May 10 '19 at 16:11
  • $\begingroup$ @clausvalca226 better? $\endgroup$ May 10 '19 at 16:20
  • $\begingroup$ None of the quantities in your table is dimensionless. You probably mean "scalar" and "vector" quantities. Dimension has nothing to do with direction. Or lack of it. You also send to have personal definitions for velocity. Or maybe for "rate". Velocity is the rate of displacement (derivative in respect to time). $\endgroup$
    – nasu
    May 25 '19 at 13:14
  • 1
    $\begingroup$ No, distance has nothing to do with the definition of velocity. Distance is not related to motion. Motion means change in position and change in position is measured by displacement. $\endgroup$
    – nasu
    May 26 '19 at 2:40

The magnitude of acceleration is scalar, same as the magnitude of velocity (speed) which is scalar. It's just that the magnitude of acceleration doesn't seem to be that useful a concept, so we don't have a word for it.

That's because speed tells you a lot about how quickly you overcome some distance in normal space and that idea is natural for us. Acceleration tells you the same thing, but in velocity-space instead of normal space and since we don't live in velocity space it is unnatural for us to think about it. But its the same concept.


g-force. Example: pilots worry about g-force, but not the direction.

Furthermore: in space-craft engineering $g$ is a commonly used unit. For instance, the Mars Exploration Rovers (the ones that were dropped on the surface in airbags) were designed to tolerate 40 g on the 1st bounce, and this (plus margin) was the mark to which internal components were tested. Likewise, sustained hypersonic entry and transient parachute-deploy decelerations were discussed in g.

Launch and thruster induced vibrations were quantified via acceleration spectral density in "g-squared per Hertz". I believe this also standard in earthquake-related structural engineering.

  • 2
    $\begingroup$ I don't think that's true; the direction of g-forces is critically important to a pilot. You can endure much more laterally than vertically and much more toward the feet than the head. $\endgroup$
    – Xerxes
    May 2 '19 at 20:19
  • 1
    $\begingroup$ @Xerxes But we still often just say "He experienced 2G acceleration", without mentioning the direction. And they train for high-G flight in centrifuges, where the direction is constantly changing. $\endgroup$
    – Barmar
    May 2 '19 at 20:27
  • 1
    $\begingroup$ Up, down, north, south, east, or west don't matter at all. However, pilot body orientation maters. Red out happens at much lower G then black out. It's about having either to much or to little blood in your brain, respectively. Has nothing to do with acceleration working differently based on direction. It has to do with having your feet some distance away from your head. $\endgroup$ May 2 '19 at 20:56
  • 1
    $\begingroup$ @Xerxes which is why they call that "negative" g's to distinguish it from the more tolerable positive g's. $\endgroup$
    – JEB
    May 2 '19 at 22:50
  • 1
    $\begingroup$ @candied_orange When the Mars Exploration Rovers were being designed, we all expected 40 g on the 1st bounce. No direction was mentioned. Also: power spectral densities for vibration caused by thrusters was discussed in g-squared per Hertz. No direction, no meters...so I think it is entirely reasonably to say "g" satisfies the OP's request. $\endgroup$
    – JEB
    May 3 '19 at 16:45

Not a facetious answer, and not entirely general, but how about Gs (as in multiples of the Earth's surface gravitational acceleration)?


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