# Talk:Acceleration/Archive 1

## Change in acceleration

"When either speed or direction are changed, there is a change in acceleration. |class=start

This is not always true. Change in acceleration is a different unit as acceleration itself (see Jerk) is different from velocity which is different for merely time or space by itself. Change in direction or velocity/speed requires the presence of acceleration, but not the change of it. I could be traveling at 10m/s[E] have an acceleration of 1m/s^2[W] for 9 seconds and then up after those 9 seconds with a velocity of 1m/s[E] and then my change of acceleration could occur to become 1m/s^2[E] for 9 seconds and after the course of those 9 seconds I would return to the original velocity of 10m/s[E]. I think I know what was meant to be said but the final phrasing is poor. Only acceleration not the change of acceleration is required to change velocity. Stoutpuppy 22:19, 4 March 2006 (UTC)

I'm having trouble with thing article -- At the start you define acceleration as a change in velocity over time (meaning that it cannot be instantaneous) and then later describe it as the derivitive of velocity with respect to time (the instantaneous acceleration). I'm gonna try to make that clearer.--Adam (http://www.ifobos.com) 00:57, 15 December 2005 (UTC)

Good idea. It's wrong anyway, to accelerate is to change the velocity over time, not the acceleration. That would be s. -- Tarquin

I'll merge later today unless someone beats me to it. -- Tarquin 04:03 Aug 31, 2002 (PDT)

Higher derivative nomenclature ref: http://math.ucr.edu/home/baez/physics/General/jerk.html

While acceleration and deceleration are similar as far as being the inverse of each other.. when it gets to the Force involved hitting an object, it's a whole different can of worms, since the deceleration is dependent upon the Young's Modulus of the object (compliance) which raises the deceleration forces to extremely high values as the time value drops to microseconds. There should be a more complete explanation of deceleration/target interactions to help clarify this for folks..

In the first formula (a=dv/dt) I don't see a definition for d. Perhaps this is obvious to those more math literate than me. Perhaps some examples in various units on a linked page would help all of these acceleration and gravity articles.

Maybe you like a = deltav/delta/t better. And that's correct, except that the time interval is reduced to zero.70.244.238.120 (talk) 19:47, 2 May 2010 (UTC)

On an instantaneous basis, acceleration is the instantaneous rate of change in the velocity as the time interval approaches zero. Over a time period the change in velocity change is proportional to an integral (summary) of the force and resulting acceleration value over the accumulative time period.WFPM (talk) 00:17, 30 April 2010 (UTC)

## imprecision

The article erroneously states that accelerated frames are equivalent to non-accelerated frames. Instead, accelerated frames are equivalent to frames in a gravitational field, although nowadays such gravitaitional fields that don't originate with masses are called "pseudo-gravitational fields"[1], and physical acceleration is now regarded as "absolute". In fact the last remark that "space-curvature" must be taken into account for accelerated frames (contrary to non-accelerated frames outside of gravitational fields) already contradicts the statement that accelerated frames are equivalent to non-accelerated frames. Harald88 23:18, 25 November 2006 (UTC

idk... —Preceding unsigned comment added by 96.225.193.42 (talk) 04:51, 16 October 2007 (UTC)

## Do the Maths

The first calculus formula in this article states that: acceleration equals the first and second derivative of a velocity-time curve. Isn't that impossible? Or is it just me?--BusinessMan11 19:44, 11 April 2007 (UTC)

Only if the velocity is zero or ex :)--Cronholm144 00:57, 26 June 2007 (UTC)
Read again. Acceleration is the first derivative of velocity w.r.t. time, and the second derivative of displacement w.r.t. time. As velocity ${\displaystyle v={\frac {dx}{dt}}}$, then ${\displaystyle a={\frac {dv}{dt}}={\frac {d}{dt}}{\frac {dx}{dt}}={\frac {d^{2}x}{dt^{2}}}}$. Stannered 20:28, 11 April 2007 (UTC)
The article did indeed say what BusinessMan purported it to. I fixed it. It was a mistake in presentation and I don't quite know how people missed it.--Cronholm144 00:57, 26 June 2007 (UTC)

## Instantaneous velocity?

I'm no physicist, but since when does "acceleration" refer to instantaneous velocity (as stated in the intro)?? Having taken college level physics classes, I'm fairly confident that this is not correct. Perhaps they meant to say "instantaneous acceleration" or "the change in instantaneous velocity." Cmw4117 19:02, 7 June 2007 (UTC)

## Poor English in the decriptor below the Time-Velocity graph in the intro.

The blurb under the graph at the top-right of the page includes tautologies. The text "time rate of change" could simply state "rate of change" (since rate already includes time). The text "velocity and/or direction" seems to miss the basic premise that the word velocity includes a directional component.

Can't agree that the word velocity includes a directional component because it's a nondirectional word.70.244.238.120 (talk) 19:53, 2 May 2010 (UTC)

In addition to the tautologies, the blurb would be more simply understood if the second concept, about the slope-velocity correlation in the graph, was a second sentence. —Preceding unsigned comment added by 121.127.195.146 (talk) 13:25, August 30, 2007 (UTC)

## math teaches us something new?

So acceleration is velocity over time, which simplifies to distance over time (squared). Can this give us a new concept of acceleration - in terms of distance and time squared? what does time squared mean? it is a hard concept to grasp, for me.

let's say you had a distance, and a time, and were supposed to find the acceleration - lets say a car went 100 miles in 1 hour - then it's average acceleration would be 100miles/hour/hour. LoL - what does this mean? acceleration is the rate of change of an object's velocity, with respect to time, so - 100 miles per hour per hour would boil down to the object's velocity changed 100 miles per hour IN one hour?

LoL, is this correct? BriEnBest (talk) 06:14, 15 December 2007 (UTC)

That is not correct. The article doesn't say that acceleration is velocity over time. It states that it is the rate of change of velocity. Average acceleration is the change in velocity over the change in time. In your example, you don't state if the velocity is changing or not. All you can get from your example is the average velocity (which is distance traveled over change in time - giving an average velocity of 100 mph). PhySusie (talk) 13:04, 15 December 2007 (UTC)

i'm sorry it should be delta velocity over time. and that is implied when it says that acceleration is (equivalent to) distance over time squared, because that is where that equation comes from - (change in velocity) over time = (change in distance over time) over time. BriEnBest (talk) 22:29, 15 December 2007 (UTC)
and it does say that acceleration is velocity over time (just not in math) when it says that it is the rate of change of velocity (because the rate of change is with respect to time). that means change in velocity over time. BriEnBest (talk) 22:33, 15 December 2007 (UTC)

Anyways, what i'm trying to get at is this: the equation, change in distance over time squared was not originally conceived - it was simplified from the equation change in velocity over time. It is easy to understand acceleration from the perspective of the latter equation (velocity over time). However, what I'm getting at is understanding acceleration from the perspective of the former equation - which only uses distance and time squared. It is hard for me to relate (in my mind) acceleration to distance over time squared, in terms of distance and time squared. I should try thinking that distance over time IS velocity. And that velocity over time (distance over time over time) is acceleration. The distance an object travels in a certain amount of time give an average velocity. If it's starting velocity was zero and it's ending velocity was x then the change in velocity would be x. if it's time was t then it's average acceleration would be x/t, which means total distance over t squared. so, we can therfor say that if it's starting velocity was zero, then it's acceleration was total distance over time (sqaured), but what IS time squared? BriEnBest (talk) 22:43, 15 December 2007 (UTC) Insert non-formatted text here hehe —Preceding unsigned comment added by 115.147.2.116 (talk) 10:22, 10 September 2008 (UTC)

To keep this concept straight, you need to work with the displacement formula, which is:

      S sub T = Linear (1 Dimensional) Displacement amount
S sub T = S sub zero + V1 times T  + 1/2 times a times T squared (for a constant a value)
And then V sub T = V1 + a times T   (= Initial plus accumulated velocity)
And the value of a is, of course, a constant


So that's the way it works in 1 dimension. But in his "Principia Mathematica" Isaac Newton figured out and reported that if you organize displacement values into 2 orthogonal (right angular) directions, you can do the displacement calculations in each direction independently of what is going on in the other orthogonal direction. And the net acceleration, velocity, and displacement values are then the vector sum of the independent calculations.WFPM (talk) 15:30, 9 May 2010 (UTC)

## Deceleration

I can see my physics professor rolling his eyes and harumphing whenever I hear the word "deceleration". He was very adamant about the vernacular of physics; that there was no room for ambiguities in a precise science (He also would state about using "amperage, voltage, and wattage" that it was acceptable for electricians, but in physics one had to learn "flow, force, and power"). Acceleration is the rate of change in velocity. It may be negative or positive. Since acceleration is derived from latin, and "deceleration" originated in 1900, I would guess the original science never assumed it would be necessary to dumb it down any.

I guess this has something to do with cars in the US. Everybody says "accelerate" for speed up, although they most commonly say 'slow down' or 'stop' for the negative. Even in car tests they list 'acceleration' and 'braking'. Msjayhawk (talk) 05:35, 13 February 2009 (UTC)

acceleration is a motion of changing direction. —Preceding unsigned comment added by 122.55.184.82 (talk) 06:46, 4 May 2009 (UTC)

Acceleration is Dv/Dt and has nothing to do with direction.

## In Relativity

The article states that the theory of general relativity sees all accelerated frames to be equal, like frames going at different velocities. I know this to be false; the only 'force' that general relativity sees as not affecting a reference frame is that of gravity, precisely because it is not really a force; it is the curvature of spacetime, and all objects following a path through curved spacetime are indeed moving through a geodesic, which would be analogous to a straight line, as in special relativity. This, however, is not true for the other forces. I would be happy to make the necessary edits in the section, but will allow for somebody to contradict me before I do; I may, after all, be a little rusty on relativity. --Slartibartfast1992 22:14, 12 June 2009 (UTC)

There, it's done. If anybody feels that the section I just wrote is wrong, please discuss. --Slartibartfast1992 02:06, 22 June 2009 (UTC)

I have some ideas about your edits, but no time to write about them right now. I'll let you know later. MarcusMaximus (talk) 01:02, 23 June 2009 (UTC)
I have some objections to your conception of objects "feeling" forces, and "belonging to" or "being in" reference frames. I also think the word "indistinguishable" should be left in, because that is how I've always seen relativity described.

MarcusMaximus (talk) 06:18, 23 June 2009 (UTC)

So, what sentences do you suggest changing and what do you suggest changing them to? Have you made the changes already? May I add also that I get my view of reference frames and 'feeling' forces from Taylor and Wheeler's Exploring Black Holes, which we might want to use as a source. In addition, I'd like to state that the paragraph as previously written was dead wrong in assuming that objects accelerated by forces (not including gravity) can be included in reference frames indistinguishable from those without any acceleration by a force (again, not including gravity). I say not including gravity because the object itself doesn't feel accelerated, meaning it is not being physically accelerated, but geometrically accelerated according to a reference frame. In that way, gravity is the only 'force' that can be said to be compatible with inertial reference frames. --Slartibartfast1992 03:09, 25 June 2009 (UTC)
I haven't made any changes. My post above is actually two posts, one saying that I'll post later and then my later post. Forgive me for being only negative in my comments, I don't mean to just tear you down, but I haven't given this topic much thought about how to construct a good pedagogical explanation--still working on it. I appreciate what you are trying to get out of the article, so let's see if we can hammer something out together.
I haven't read that book you refer to, but here are the questions that come to mind when I read the section that I believe need to be resolved in order to have an effective article:
• I think it is generally not a good idea to talk about an object "feeling" a force. We are talking about dynamics, not sensory perception--what does it mean to "feel" a force versus not to feel one, particularly for inanimate objects?
• I also think it is problematic to say that "the force of gravity is not felt by an object in free-fall" immediately after saying that "gravity is not actually a force".
• What is a "force of acceleration", and is it distinct from another type of force?
• What does it mean to "belong to" a reference frame, or to be "accelerated into another reference frame"? I have always been under the impression that any object's kinematical state can be represented in any reference frame using a set of simple relations (although there might be relativistic versions of those equations). In fact, as I understand relativity, it is a theory specifically intended to explain how the motions of objects can be expressed/measured in multiple different reference frames.
• Why do you say that inertial reference frames "are dependent on the fact that all objects in that frame travel at the same, constant speed"?
PS, the link in your final sentence to "free-float" is to a business-related article that doesn't pertain to this topic. MarcusMaximus (talk) 03:56, 25 June 2009 (UTC)
Alright, to tackle them one by one, numerically,
1) Einstein talked about the fact that all forces accelerating an object make that object feel acceleration, i.e., someone accelerating in a car will feel himself being pulled back. This is to differentiate between physical acceleration (one that is felt, and that is the same regardless the coordinate frame you view it from) and geometric acceleration, which depends on the particular reference frame and varies or may even not exist in some. The latter may also be called relative acceleration.
2)Good point, honest blunder on my part. I'll just fix it by not mentioning it as a force, but rather as just 'gravity', since the acceleration from a force must be felt in order for it to actually be a force. Gravity provides relative acceleration; if analyzing a particle under the effects of gravity from a free-fall frame, it will be apparent that it's moving in a linear fashion and is not really being accelerated. See the article on proper acceleration; it has an animation precisely on this at some point in the article.
3)Hmmm, I don't know what I meant right there. I'll fix that now. Essentially, there are only three forces in nature; electromagnetic, weak and strong. Gravity, as was already covered, isn't really a force.
4)To belong to an inertial reference frame is to be travelling at the speed at which the frame is going or, in other words, for your speed in that frame to be 0. Of course it can be represented in any frame, but what I mean is for the object to actually set the arbitrary 0 for an inertial reference frame. You may also say that I'm dealing with the observer of that frame.
5) In this, I'm talking about special or general-relativistic reference frames, in which the velocity is perfectly constant because it is not altered by any real forces (gravity, recall, is not a force; hence the inclusion of general-relativistic frames of reference).
I see where I may have made some mistakes or gotten my concepts mix up; don't fret, I'll fix them right now. --Slartibartfast1992 05:21, 25 June 2009 (UTC)
The problem I have with the discussion of "feeling" acceleration versus not feeling it, and with the concept of "observers", is that it makes the whole thing subjective. It doesn't describe what is actually going on; it describes what a person would feel or see if they were present, making certain assumptions about that person's knowledge or lack thereof about dynamics and relativity.
What is the difference between "feeling" an acceleration and not feeling one? It seems to me that the only reason you "feel" a force is that it causes a compression sensation on the side of your body that is touching your chair (in the case of a rocket). If you were being pulled by your hands, you'd feel a tension sensation in your wrists and shoulders. The rest of your body feels it because the force is physically transmitted between the chemical bonds, causing stress and strain in your tissues. If you could find a way to apply a force equally and proportionally to all the particles in your body, you wouldn't "feel" it at all, because there wouldn't be any internal stress or strain--that's all your nervous system is equipped to measure.
Does this make sense? How do we describe the difference between acceleration caused by forces and acceleration caused by gravity without talking about "feeling"? MarcusMaximus (talk) 16:03, 25 June 2009 (UTC)
You can't talk about it for a layman (which is what we're trying to do here) without using the word 'feeling'. The correct term might be 'tidal force', or more correctly 'tidal acceleration', meaning that the velocity on the forward part of the driver's body is less than the velocity on his back, meaning that he feels a compression. To explain it in terms of tidal force for the average Wiki reader would be a challenge, which you are welcome to embark on with the help of your sandbox. Feelings do describe what is actually going on because they relate directly to tidal acceleration; indeed, it's the only type of acceleration that all frames will agree on, since the object can't feel two different accelerations at once. This makes it precisely 'what's going on'.
As for observers, they're an essential part of relativity. You really can't make it any simpler than that, since it is the section on relativity after all. What I'm talking about there is the observer on the reference frame, or, as it is sometimes referred to, the 'laboratory'.
Your description of why you don't feel a force is exactly the argument used by Newtonian Mechanics to support its belief that gravity really is a force; because it acts the same on all mass, a body doesn't feel tension. General Relativity has proven this view wrong, along with the rest of Newtonian Mechanics. Right now, you're probably sitting tight in your chair, you feel yourself compressed towards the ground; you feel the same acceleration as if you were being accelerated at 9.8 m/s^2 on a spaceship in flat spacetime. Why do you think you feel this acceleration, when you seem not to be moving on the surface of Earth? Indeed, Newtonian Mechanics will argue that gravity and the force of the ground are equal and opposite, and that is why you don't move; they neutralize. In reality, you are feeling an acceleration; a compression on the part of your body closer to the chair or ground. This is not because gravity is accelerating you downwards; gravity isn't a force, it can't do that. It's because the chair is accelerating you upwards, throught he electromagnetic force in its chemical bonds. That is why 'feeling' is essential to the relativistic interpretation of gravity, in that it's not a force but the curvature of spacetime.
When thinking it about it, you'll conclude that 'feeling' (or if you prefer, the sensation of tension or compression) is not subjective at all. As for observers, they're also not subjective; relativity is all about observers and their reference frames! --Slartibartfast1992 06:15, 26 June 2009 (UTC)
In free fall, without physical contact with a surface, you can't feel gravity, though I don't recall an instance in which this experimental setting, without contact, is replicable for other forces. They all seem to rely on physical contact to transmit their force, and this contact seems to be the basis of the feeling. Maybe an example of the other forces could be the force of Earth's rotation. We don't feel it, similarly to how we don't feel gravity in free fall, though it is a force. So I think that Einstein's statement about gravity being different in feeling is incorrect. JonatasM (talk) 06:29, 16 October 2010 (UTC)

I deleted two paragraphs in the section on the relation to relativity, which were not needed, and contained errors. I recommend adding formula for acceleration in polar coordinates in the article. Michael9422 (talk) 16:00, 5 September 2009 (UTC)

It's true that such a large description was unnecessary, but if you could point out the errors, which I'm interested in, I would appreciate it. --Slartibartfast1992 03:16, 18 September 2009 (UTC)
Hi Slartibartfast1992. There were two minor grammatical errors in those paragraphs - twice using "it" instead of "she". And also, there was the incorrect assertion that "Special relativity applies only to nonaccelerating objects". I will quote from Wolfgang Rindler's book "Essential Relativity" (1977 second edition), section 2.14 page 47, "In fact, special relativity applies to all physics in inertial frames, and general relativity simply reduces to special relativity in such frames." Michael9422 (talk) 04:06, 19 September 2009 (UTC)

I'm not sure what was intended, but "forces felt by objects ... are actually feeling themselves being accelerated" is not worded clearly. The forces don't feel themselves. Jablomih (talk) 12:26, 29 April 2010 (UTC)

I don't think the present paragraph is clear. What exactly is the relation to relativity that you are trying to make? Newton's laws of motion also results in no force being 'felt' in free-fall for the simple reason that each particle of an object accelerates equally -- regardless of their masses. I think what may be a better point to make in this section is Einstein's equivalence principle which states that an accelerating frame with no gravitational field is physically indistinguishable from a frame 'at rest' in relation to a gravitational field? Stating that gravity is not a force is somewhat confusing without describing more about general relativity, nor do I believe is it necessary. For the purposes of this article it's probably sufficient to assume that gravity is a force. Benlansdell (talk) 08:15, 1 August 2010 (UTC)

## Classical Mechanics box

I do not think the 'Classical Mechanics' box belongs in the article. Perhaps there should be a 'Classical Mechanics' link in the 'See Also' section. A non-technical reader is going to be overwhelmed with details. Am I alone? Michael9422 (talk) 18:14, 29 March 2010 (UTC)

Acceleration is one of the fundamental concepts of classical mechanics, so it should definitely include the classical mechanics box. There really isn't any other kind of acceleration I can think of except the classical mechanics kind. MarcusMaximus (talk) 22:47, 8 May 2010 (UTC)
My concern is that some readers, especially young readers, will be intimidated by the mathematics. For comparison, I read the 'acceleration' articles of several older printed encyclopedias, and none of them contained calculus.Michael9422 (talk) 16:13, 7 July 2010 (UTC)
Out of curiosity, what encyclopedias were those? They sound incomplete to me. And are you arguing that this article should not contain any calculus, or that it should also contain a description of acceleration that doesn't use calculus? MarcusMaximus (talk) 22:54, 11 July 2010 (UTC)
The printed encyclopedias that I mentioned are the World Book (~1970) and the Encyclopedia Americana (2004). My objection is that the 'classical mechanics' box does not belong because it is beyond the scope of this article, and because it has its own Wikipedia entry.Michael9422 (talk) 18:40, 22 July 2010 (UTC)

You don't have to worry about acceleration as long as it's constant acceleration. where it gets complicated is where you have a variable force and thus a variable acceleration value. And that's when you get into differential equations, which are complicated. Of course you know that, but some people don't.WFPM (talk) 15:44, 9 May 2010 (UTC)

I'm not sure what you mean by "You don't have to worry about acceleration as long as it's constant acceleration." Constant acceleration is an ultra-qualified special case, and almost never occurs in nature. MarcusMaximus (talk) 01:27, 10 May 2010 (UTC)

I mean that the "acceleration" property of motion just naturally falls out of the displacement equations as you keep differentiating it with respect to the time interval. It's just the second differential rate of change of the displacement value with respect to the time interval value. And when it gets complicated is when you want about how fast {with respect to time) can you change that value, which is the third derivative. And we don't have a name for the time rate of change of the acceleration value, which would be Da/Dt, because the equation doesn't contain a third order function of distance with respect to time.WFPM (talk) 13:16, 10 May 2010 (UTC)

See the article on jerk. I think you just might not have run across this stuff in your experience. Equations of motion almost always start with acceleration as a function of force, but jerk is often calculated because it's a quantity of interest in mechanisms, electromagnetics, and inertial measurement. MarcusMaximus (talk) 15:16, 10 May 2010 (UTC)

Cute and interesting. I remember doing environmental tests on battery powered electronic devices at Redstone in New Jersey, which involved dropping from a 4 feet onto flat 2 x 4's on the concrete floor. And one on each face and edge and corner for 26 total drops. We called that drop testing.WFPM (talk) 22:21, 10 May 2010 (UTC) And I think that the impact quantity property associated with these tests was valued by the number of "g" acceleration values that were created by the drop impact, which would be an amount of acceleration value, I think. Or would it?WFPM (talk) 22:27, 10 May 2010 (UTC) And what about impact? Aren't we going to explain it to death like the others?WFPM (talk) 22:42, 10 May 2010 (UTC) And in order to get a jerk quantity value you would have to create a continuous Da/Dt functional line and not a Dv/Dt line, like in your graph, so that you could take the first derivative of it.WFPM (talk) 22:52, 10 May 2010 (UTC)

Yes, the g-level would be the "amount" of acceleration, or what is commonly called magnitude. By the way, you don't need a continuous da/dt function nor a graph to estimate a derivative. MarcusMaximus (talk) 04:15, 11 May 2010 (UTC)

My concept of a derivative is based on the idea that the derivative value was the value of the function dx/dt + delta x - dx/dt as the incremental delta t value is reduced to zero. And if I don't have a formula for the dx/dt value, I wouldn't be able to figure it out. But I know that Newton did so your probably right.WFPM (talk) 13:24, 11 May 2010 (UTC) But since the first derivative of a curved velocity versus time line gives you the acceleration value, I figured you might need to create something like a tangent to a curve in an acceleration versus time line in order to arrive at a rate of change of acceleration (jerk) value.WFPM (talk) 13:43, 11 May 2010 (UTC)

## Intro for laymen

This article needs an intro for laymen. It's not a bad intro for a laymen's explanation except that it immediately uses the word 'vector' and 'kinematics' in its first few sentences. This may be 100% correct in these distinction, but don't you think that maybe these articles should "share space" with those trying to learn? Vectors are an abstract mathematical tool, and kinematics would take hours trying to explain to a seventh grader. Yet a grade-schooler can understand the basic meaning of acceleration. Why exclude all those masses from understanding these articles. and display it as an archive for the elite? Please dumb down the first several paragraphs before you get to the heady stuff later on. I know it's difficult because some people will generally attack an article for minor inaccuracies, and 'dumbing down' introduces approximations and ballpark explanations. But if the 'entry-level' approach is obvious as to its intent, even the nitpicky attackers will understand and will back-off. —Preceding unsigned comment added by 99.147.240.11 (talk) 17:09, 5 September 2010 (UTC)

I gave it a try. Have a look and tell us what you think. MarcusMaximus (talk) 21:48, 5 September 2010 (UTC)
Wow, quick response. Thanks for the change. Not so intimidating anymore. By the way, the best example of an 'intro for the rest of us' that I can recall is the article on the 'mean value theorem'. Whoever wrote that did an excellent job. 99.147.240.11 (talk) 15:24, 6 September 2010 (UTC) —Preceding unsigned comment added by 99.147.240.11 (talk) 15:06, 6 September 2010 (UTC)

${\displaystyle a={\frac {v^{2}}{2S}}}$. For example, let's say that a=10 (m/s)/s, then after first second stone speed will be 10 m/s and average speed 10/2=5 m/s, so stone will fall 5 m. After next second stone speed will be 20 m/s and average speed will be (10+20)/2=15 m/s and stone will fall 15 m in second second. In third second stone will fall with average speed (20+30)/2=25 m/s and will fall 25 m and after third second stone speed will be 30 m/s. And so on. After ten seconds stone speed will be 100 m/s and stone average speed in second number ten will be (100+90)/2=95 m/s, so stone will fly S=5+15+25+35+45+55+65+75+85+95=500 m after ten seconds. So now we know time t=10 s and speed v=100 m/s and falling distance S=50 m. So from formula ${\displaystyle a={\frac {v^{2}}{2S}}={\frac {100^{2}}{2\cdot 500}}={10000 \over 1000}=10\;(m/s^{2})}$. Also you can check this way ${\displaystyle S={\frac {v^{2}}{2a}}={100^{2} \over 2\cdot 10}={10000 \over 20}=500\;(m)}$. Also ${\displaystyle S=t\cdot {v \over 2}=10\cdot {100 \over 2}=500\;(m)}$. Also ${\displaystyle a={\frac {v^{2}}{2S}}={v^{2} \over 2t\cdot {v \over 2}}={v^{2} \over tv}={v \over t}={100 \over 10}=10\;(m/s^{2})}$. In this case v=t*10=t*a. Another example, a=10 m/s/s, t=100 s, v=t*a=100*10=1000 m/s. S=t*v/2=100*1000/2=100*500=50000 m. Also ${\displaystyle S={\frac {v^{2}}{2a}}={1000^{2} \over 2\cdot 10}={1000000 \over 20}=50000\;(m)}$. Also a=v/t=1000/100=10 m/s/s. —Preceding unsigned comment added by 84.240.9.58 (talk) 17:59, 10 September 2010 (UTC)

## Merge proposal: Uniform acceleration into Acceleration

I've added the merge tags in support of discussion that started in 2008 and had support today on Talk:Uniform acceleration, which is an essentially unsourced article on a narrow special case. I don't see a reason to keep it separate, given how little content there is. Dicklyon (talk) 05:50, 4 December 2010 (UTC)

• Support. That article is indeed just a special case. DVdm (talk) 11:18, 4 December 2010 (UTC)