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I understand what acceleration is, and I know the formula, and I understand it's a vector. I just don't understand how the equation works exactly. I'm kind of picky, I know, but bear with me.

Velocity is the amount of distance traveled during the amount of time, $\frac{s}{t}$. That makes sense. But how on earth is acceleration related to the distance divided by time squared? Where does the squared come in?

I mean, yes, I can prove it mathematically. $\frac{\text{distance/time}}{\text {time}}$ is $\frac{\text{distance}}{\text{time}^2}$. But why? How is it possible to square time? Can I even assume to understand the individual components separately? Or do I have to assume I'm dividing a vector by time and just view it that way? Is it the amount of velocity absorbed during a different amount of time, or something?

I can't just plug in numbers and say I understand physics, even when I understand the end result. It's like A->B->C. I understand A and C, but where did the B come in? How was this proven? Maybe there's a proof online or something? All I could find was the proof for centripetal acceleration...

Basically what I'm asking is how each of the variables relate to one another separately, and how that all works out.

I really need to understand physics, or at least to the point where I can manipulate the equations in my mind to spatially transpose graphs of the accurate calculations on reality. I'm never going to make it over the hurdle in gym class if I can't even understand how fast I'm accelerating.

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Thanks for the fancy symbols. :) –  Genevieve Ccio Jan 5 '13 at 9:01
    
"how on earth is acceleration the velocity divided by time squared?" you want to re-write this, I can't edit it cause it's too small. –  kηives Jan 5 '13 at 17:08
    
Does it help to think about it in terms of $\frac{\rm mph}{\rm s}$ or $\frac{\rm km/h}{\rm s}$ ? I think the confusion stems from the fact the differentiation is not division although it looks like that in terms of units. –  ja72 Mar 5 '13 at 17:00
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8 Answers

up vote 9 down vote accepted

It's simpler than you (probably) think.

In your example of defining speed: this is a change of position $s$ in a time $t$. The units of distance are metres and the units of time are seconds, so the units of velocity are metres per second. So far so good.

Now consider acceleration: this is a change of velocity $v$ in a time $t$, so the units of acceleration are $v/t$. The units of velocity are metres/sec and the units of time are seconds, so the units of acceleration are (metres/sec)/sec or metres/sec$^2$.

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The most intuitive way of understanding acceleration is to understand it in terms of Taylor series expansions

$$\sum_{n=0}^{\infty} \dfrac{f^{(n)}(u)}{n!} (t-u)^n$$

A good entry level discussion on how one applies the Taylor series expansion to the question of position-velocity-acceleration can be found in this short paper from this website attributed to S.A. Fulling.

If we review Taylor's Theorem we start out with evaluating the series with $u=0$, Wolfram shows the expansion as:

$$f(t) = f(0) + tf'(0) + \dfrac{t^2}{2!}f''(0) + \dots + \dfrac{t^{(n-1)}}{(n-1)!}f^{(n-1)}(0) + \int_0^t \dfrac{(t-u)^{(n-1)}}{(n-1)!}f^{(n-1)}(u) du$$

If you look at the first three terms, you should see the similarity to:

$$x(t) = f(t) =x_0 + v_ot + \dfrac{1}{2}a_0t^2$$

Now first consider the expansion of the $\frac{1}{n!}$ inverse factorial terms.

$$\dfrac{1}{0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \dfrac{1}{4!} + \dfrac{1}{5!} + \dots = \dfrac{1}{1} + \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} + \dots$$

then consider the sum of only the terms after the $\dfrac{1}{2}$ term:

$$\sum_{n=3}^{\infty} \dfrac{1}{n!} = \lim_{n \to \infty} \dfrac{1}{n!} = e - 2.5 = 0.218282\dots$$

The summation of all the inverse factorial terms after $\frac{1}{2!}$ is only $0.218282\dots$ which is clearly less than $\frac{1}{2}$, so unless the higher order derivatives of $f^{n}(u)$ when $u=0$ are substantial, then the overall effect of the higher order derivatives will never be greater than the first non-linear factor $t^2$.

Returning to the Taylor expansion, the goal of course is determining the values of $x_0$,$v_0$,$a_0$. These are frequently given, or can be determinable by observation, but in any case can be understood as constants obtained from integration. For instance, as explained in the above exercise, one would start by integrating some arbitrary constant for the third derivative of some function assuming the real integral was less than the integral of an arbitrary constant:

$$\int_0^t f'''(u) du \le \int_0^t M du$$ such that $$f''(t) + a_0 \le Mt$$

performing these integrals for successive derivatives we can eventually find the form of the equation,

$$|f(t)-x_0 + v_0t + \dfrac{1}{2}a_0t^2| \le \dfrac{1}{6}M|t|^3$$

Per the exercise, this shows that the the graph of $f(t)-x_0 + v_0t + \frac{1}{2}a_0t^2$ will be between two curves $\pm \frac{1}{6}M|t|^3$ which should be very close together at $t=0$.

Since $M = f'''(0)$, if $$f'''(0) = \epsilon$$ with $$\epsilon \approx \frac{1}{\infty}$$ then we could say $$f(t)\approx x_0 + v_0t + \frac{1}{2}a_0t^2$$

In addition, if the derivative expansion is shown to have higher order terms that cancel, or are sufficiently suppressed, then the higher order terms can be ignored.

However, the specific values of the $x_0$,$v_0$,$a_0$ will not be known unless domains of integration and boundary conditions are specified.

In any case, one can continue to expand the position variable with respect to time and find that "jerk" is the third power or time and "Jounce" (or "Snap") is the fourth power in the expansion. These higher order contributions, and all other higher order contributions, have a decreasing contribution to the overall equation attributable to the $n!$ factorial term in the denominator of the summation. This is a good analogy to understanding the concept of the coupling constant and how the coupling can decrease as one expands perturbatively about a solution to an equation (e.g. a function).

So acceleration is the first non-linear term in the expansion of the function relating position to time. Since higher order terms do not contribute more than the first non-linear term (acceleration) for certain $f(t)$, their effects can be ignored in most circumstances (although in some situations, such as elevator design, jerk is a consideration, and when one gets into design space craft, the higher order derivatives must be considered as well).

It should be noted that $f(t)$ has units of position (e.g. m = meters), since the Taylor expansion is for the $t$ variable, and the $t$ variable is raised to a higher order in each successive term, the constant must carry units which cancel out the units of time for each term. Since $a_0$ is associated with $t^2$, it must carry units of $\frac{m}{s^2}$ in order to ensure that $f(t)$ is in units of $m$.

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The current answers have already covered every angle rather well, but perhaps the following wording and example can still contribute slightly.

Velocity is how much the position of an object changes per unit of time (this unit is free for you to choose). Acceleration then stands to velocity as velocity stands to position: it is how much the velocity of an object changes per unit of time (again free to choose, but be consistent).

For example, consider an object that travels from point A to point B, both being 5 meters apart (as the eagle flies). If the object takes 5 seconds to travel from A to B, then its average velocity has been 5 meters per 5 seconds (and this average velocity points in the direction of $\mathbf{AB}$. Or if you want to use 1 second as a unit of time: 1 meter per 1 second ($1\,\mathrm{m/s}$). This is a commonly used unit of time. But if you want, you can use 1 day as well. Then in this example the average velocity becomes $86400\,\mathrm{m/day}$. You could even use something odd like 5 minutes and 23 seconds as a unit of time and call it a $\mathrm{Mia}$. Then the example gives $323\,\mathrm{m/Mia}$.

Anyway, if we now know that the object did not always travel at a speed of $323\,\mathrm{m/Mia}$ (which was $1\,\mathrm{m/s}$ remember), we might want to know what the velocity profile looked like over time. How did the velocity change per unit of time? (both in direction as in magnitude) Let's say we only know the velocity at the beginning (in A, e.g. $0\,\mathrm{m/s}$) and at the end (in B, e.g. $2\,\mathrm{m/s}$) and that the direction has not changed (for simplicity). Then we know that the (average) change in velocity is $2\,\mathrm{m/s}$. We also already know it took 5 seconds for the object to get from A to B, so the average acceleration is [2 meters per second] per 5 seconds. Or mathematically:

$$\frac{2\,\mathrm{m/s}}{5\,\mathrm{s}} = \frac{2\,\mathrm{m}}{5\,\mathrm{s}^2} = 0.4\,\mathrm{m/s^2}$$

You can look at it like this as well: define a new unit of velocity (like the $\mathrm{Mia}$ for time) and call it the $\mathrm{Wouter}$, equal to $1\,\mathrm{m/s}$. Then the average change in velocity is $2\,\mathrm{Wouter}$ and the average acceleration is $2\,\mathrm{Wouter}$ per $5\,\mathrm{seconds}$ or, when we use $1\,\mathrm{s}$ as a unit of time: $0.4\,\mathrm{Wouter}$ per $1\,\mathrm{second}$. We could now calculate how many $\mathrm{Wouter}$ per $\mathrm{Mia}$ this corresponds to, but let's not.

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Squaring time makes no sense, true. There is no magic behind it.

If the body is moving under constant speed then the distance traveled is $S=v t$ , for different trajectories we will have different $v$'s so after some time of investigation we find out that to obtain speed (as a PARAMETER) of motion we need to divide space over time.

Then our friend comes and tells us that he observed motions where $S=a t^2$ , so the space traveled is proportional to time squared (as a parameter), then we sat down thought thought and decided to find out this new parameter of motion $a$ same way as before just by dividing space over time, but in this case over time squared.

Imagine that you don't know dimensions of velocity and acceleration but you just observed 2 different type of motion - linear and parabolic. Space has nothing to do with time so to make dimensional sense out of $S=a t^2$ you will need to invent something.

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You asked for the derivation or proof for the formula of acceleration. One thing you have to understand about observing the physical reality out there, is that you can't start from scratch. It simply isn't possible. Phenomenon in nature has to be converted into a plausible, understandable and very much tangible concept. Therefore, basic concepts in physics like motion: velocity, acceleration, speed are all DEFINED. The second thing you have to understand is how these concepts are inter-related. If you observe the motion of a particle, you will notice that it covers a certain distance in a certain interval of time. Therefore you understand that the amount of distance covered is proportional to the time during which the motion takes place. From this you can infer that a certain constant or quantity, not directly observable in nature can be defined, in order to better understand the type of motion being described. From this was born the concept of velocity, that defined itself as being the ratio of the change in position of an object to the change in time taken for the change in position to be made. But as you further observe the motion of objects, you come to realize that not even velocity remains constant. If you take a look at your speedometer when you're driving a vehicle you'll doubtless notice that the speed that you're traveling at doesn't remain constant. It fluctuates depending on how much you step on it. So physicists introduced another concept that sought to define change in velocity itself. Observing changing velocity made physicists realize that objects that change velocities can cover different distances in same intervals of time. So the concept of acceleration was born, that DEFINED the rate of change of velocity itself. If you want to better understand these concepts it's always advisable to buy some books on classical mechanics, and to especially study graphs of motion. Compare the graphs of an objects motion with respect to time, the same object's velocity with respect to time and the same object's acceleration to time. Relax, let the concepts sink in, think about it, and physics becomes the easiest thing in the world. It should be, after all, it describes a world that you yourself have lived in, experienced and understood for years.

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Part of the problem is the word time and how it is misused. Time can mean, among other things, the reading on a clock at a particular moment or the difference in two such readings. The former should be more correctly called a clock reading while the latter should be more correctly called a duration.

Another part of the problem is that authors use the same symbol, usually just $t$, to represent both of these quantities! While $t$ is fine for a clock reading, a better symbol for duration is $\Delta t$.

I fault the authors of physics textbooks, especially introductory textbooks, for perpetuating this obfuscation and its resulting confusion to students.

Anyway, there is hardly ever (I can't think of one as I write this) a case in physics when we need to substitute a clock reading into a kinematic or dynamic equation. In every case I can think of, what we really deal with is duration $\Delta t$, and there are two reasons for this. The first is that the numbers on a clock, although astronomical in origin, are otherwise completely arbitrary and thus have no inherent physical significance. The second is that the difference between two clock readings does indeed have great physical significance. It is the latter, duration, that appears in the definitions of velocity, acceleration, and any other physical quantity that involves "time" (note the quotes).

Now, to specifically address your question about the meaning of acceleration, consider that the correct articulation of the definition of average acceleration is the change in an object's velocity compared to the duration through which that change in velocity happens. That is far more physically accurate than the sloppy and overly simple distance over time. I always make sure my students learn the more correct definition; it prevents so many errors.

Next, look at the unit for acceleration. It must be the unit of velocity ($\mathrm{m}/\mathrm{s}$) compared to the unit of duration ($\mathrm{s}$). This can correctly be written as the obtuse (and nearly devoid of physical meaning) combination $\mathrm{m}/\mathrm{s}^2$ but this hides the meaning so clearly articulated by the better definition given above. So, write the unit of acceleration as $\frac{\mathrm{m}/\mathrm{s}}{\mathrm{s}}$ and read it from the bottom up. An acceleration of, say, $3\;\frac{\mathrm{m}/\mathrm{s}}{\mathrm{s}}$ would be read and interpreted as

"For every second of duration, the object's velocity changes by $3\;\mathrm{m}/\mathrm{s}$."

Remember that acceleration is a vector quantity so there could, in general, be both a change in magnitude and a change in direction. Any such arbitrary change can be decomposed into a change that is parallel (or antiparallel) to the original acceleration and a change that is perpendicular to the original acceleration. My point is we should also quote a direction, but I didn't do that here.

Your point about the meaning of time squared is a great question and one that should be brought up by more students. Time squared means nothing. Duration squared, on the other hand, is physically meaningful and is the basis of our physical understanding of motion.

You bring up another issue, and that is the term distance. It too has been somewhat abused, but I'll save that for another time if anyone asks.

I really wish the community would hold textbook authors accountable for this serious abuse of terminology and notation.

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Full support for distinguishing "clock reading $t$" from "duration". However, since "clock readings" are not necessarily "good" (scaled-isometric wrt. the corresponding durations) the symbols $\tau$ or $s$ seem more suitable to denote durations, than $\Delta t$. Likewise regarding "distance" vs. "positions" or somesuch. More nitpicks: decompose change of velocity (not of acceleration); "duration squared" is hardly meaningful itself (while "frequency drift rate" is, for instance), but trying to express this distinction through units and writing style sucks ... –  user12262 Mar 7 '13 at 21:42
    
I never said change in acceleration. I was speaking generically about any change to any vector. Regarding the rest, I've no idea what you're talking about. Duration squared is physically meaningful. –  user11266 Mar 10 '13 at 17:08
    
"I never said change in acceleration. I was speaking generically [...]" That's hard to appreciate given a sentence starting with "Remember that acceleration is a vector quantity so [...]". "Duration squared is physically meaningful." In contrast, "Distance squared", for instance, even has an own name: "Area (of a rectangle)", of some rectangle itself, or commensurate and equivalent for other shapes. "Regarding the rest [...]" Again: I fully agree with the first paragraph of your answer. But $\Delta t$ is proportional to duration only for good clocks, not for clocks in general. –  user12262 Mar 13 '13 at 18:43
    
Actually, I already disagree with the end of the final sentence of the first paragraph of your answer. No: the difference in readings should not be called "duration", but rather "coordinate difference". You noted (second sentenc) that the readings are attached to what you call "moments" (of a given clock). (Einstein referred instead to "positions of the indicator", for instance). The term "duration" should be reserved to quantify the relation between pairs of such "moments" (of a given participant); independent of any particular readings (or coordinates) attached to them. –  user12262 Mar 15 '13 at 6:54
    
Where were you when I was in highschool. –  Shivan Dragon Mar 28 at 13:21
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it is nothing but the rate of change of velocity with time.u dont have to square the time. suppose a body is going with constant velocity,it means it is covering equal distance in a particular time always.then velocity is constant,so acceleration will be zero.

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Please use proper capitalisation and avoid unnecessary abbreviations in future contributions. This helps improve readability. –  user9886 Jan 7 '13 at 11:16
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Looking at this mathematically is one way to do it, but you could also look at it in a more intuitive manner.

As you correctly pointed out, the velocity of something ($v$) is the distance travelled ($s$) divided by the time taken ($t$).

Acceleration is defined to be how fast your velocity is changing. So in terms of units, you would ask how many metres per second ($m$/$s$) do you gain every second ($s$), or $m/s/s$ (metres per second, per second), which is mathematically equivalent to $m/s^2$.

You don't actually have to "square time" to find the acceleration, you just have to figure out how your velocity is changing with time.

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In fact, many times you will hear (in speech) accelerations given as "9.8 meters per second per second." But in writing, that's often rather verbose, and writing units as $m/s/s$ is also a waste of space. Particularly if you have character/page limits. –  tpg2114 Jan 5 '13 at 13:28
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