# Talk:Acceleration

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## Uniform acceleration

I don't understand the equation in the article, is it possibly wrong?

${\displaystyle \mathbf {s} =\mathbf {u} t+{{1} \over {2}}\mathbf {a} t^{2}=\mathbf {u} t+{{1} \over {2}}{{v} \over {t}}\mathbf {t} ^{2}={(\mathbf {u} +{{1} \over {2}}\mathbf {v} )t}}$
You have substituted a = v/t, but v is not a t. The final speed v is given by v = u + a t. Hope this helps. - DVdm (talk) 17:22, 12 September 2012 (UTC)

I recommend addition of a second plot: Displacement vs. Time – a Parabola. It will reduce confusion. Also, because free-fall acceleration is described, it will make the point clear that the Linear v vs. t corresponds with a parabolic Displacement vs. Time. — Preceding unsigned comment added by SirHolo (talkcontribs) 18:48, 15 January 2016 (UTC)

## "Planar decomposition" of acceleration into tangential and normal component

The section of the article on tangential and centripetal acceleration contains the statement that "the acceleration of a particle moving on a curved path on a planar surface can be written using the chain rule of differentiation... etc" (immediately before the reference number 4).

This claim that such decomposition of acceleration is valid only for planar curves is further supported by the statement "Extension of this approach to three-dimensional space curves that cannot be contained on a planar surface leads to the Frenet–Serret formulas", which appears at the end of the section.

Would the authors care to explain why they believe that decomposition into tangential and normal acceleration cannot be done in the same way for "space curves"? I see nothing in the derivation presented that would support such a claim.--Ilevanat (talk) 23:33, 17 October 2012 (UTC)

## Decomposition of acceleration into tangential and normal component is the same for space curves as well

Having established that the author of the article section under disscusion will not answer my question (see User talk:Brews ohare), and due to my determination not to edit work of others without consent, I can only here inform the interested readers that the tangential and normal acceleration formulas in the article are equally valid for planar and spatial curves. See, for example, the following link: http://www.math.oregonstate.edu/home/programs/undergrad/CalculusQuestStudyGuides/vcalc/velacc/velacc.html

--Ilevanat (talk) 22:49, 25 October 2012 (UTC)

## Correction of some minor errors and of a common misconception related to planar-spatial curves

Allowing for the possibility that the previous author is forbidden even to discuss the physics-related articles, I have finally decided to make corrections described above.--Ilevanat (talk) 01:16, 26 November 2012 (UTC)

## Name Change

I suggest that the name of this page be changed to Coordinate acceleration so as to contrast it with Proper acceleration. KingSupernova (talk) 15:29, 9 January 2013 (UTC)

I think the current article (Acceleration) can safely remain as it is now, in a more general, non-relativistic context, in which Coordinate acceleration already is redirected to Proper acceleration, as it should. - DVdm (talk) 16:47, 9 January 2013 (UTC)
That I guess makes sense for the article title, but I think coordinate acceleration should definitely redirect to acceleration and not proper acceleration. KingSupernova (talk) 13:56, 10 January 2013 (UTC)
I'm not so sure about that, as coordinate acceleration is only used in the context of non-Galilean relativity, where it is opposed to proper acceleration. Strictly speaking, perhaps we should have an article relativistic acceleration to which both coordinate and proper would redirect, but I don't think we'll find consensus for that.

What say others? - DVdm (talk) 14:15, 10 January 2013 (UTC)

Since most of the page on acceleration is really talking about proper acceleration and neither page really make the differece clear, I think the best thing would be to merge both pages into one, under acceleration, with different sections for the different kinds. However this would probably cause the page to be too long and complicated, so what might be simpler would be to clarify acceleration, making clear the differences between the two, and remnoving a lot of the information about proper acceleration and giving links to proper acceleration instead. Some other people's thoughts would be helpful here. KingSupernova (talk) 15:28, 11 January 2013 (UTC)
No way. The current name is what most people looking for an article on acceleration would expect to see. Dger (talk) 19:00, 11 January 2013 (UTC)
Well it doesn't really matter as the page they searched for would redirect to the right one. KingSupernova (talk) 03:19, 14 January 2013 (UTC)

## What is acceleration?

In any differentiable manifold ${\displaystyle Q}$ (even in the Newtonian spacetime), in order to define acceleration, a connection ${\displaystyle \Gamma }$ (i.e., a covariant derivative ${\displaystyle \nabla }$) must be given. Newton uses inertial frames because in his mathematical apparatus there is no concept of affine space, connection, parallelization (cf. teleparallelism ). Lagrange accelerations[1] and Lagrange forces are not vectors but their difference yes. Hence, Euler-Lagrange equations are tensorial. In Newton gravity theory, gravitational forces are vectors, in Albert Einstein GR gravity theory gravitational-inertial forces are Lagrange forces. In other words, the concept of acceleration is a covariant derivative concept, an additional structure on ${\displaystyle Q}$.Mgvongoeden (talk) 12:41, 22 April 2015 (UTC)

Notes
1. ^ G. Giachetta, "Jet Methods in Nonholonomic Mechanics", Journal of Mathematical Physics, 33, pag. 1652, (1992).
Do you have some kind of proposal to improve the article? I mean, see wp:talk page guidelines. - DVdm (talk) 18:11, 20 April 2015 (UTC)

## Where to place this?

Have been reverted 2 times. Where to place it? ${\displaystyle \mathbf {a} =\lim _{{\Delta t}\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {d\mathbf {v} }{dt}}=v{\frac {d\mathbf {v} }{dx}}}$
Moreover, I feel that using "t" is time interval instead of just time would be better. If we are here to improve the pedia, why not be clear that nobody in audience has any difficulty?
117.248.121.6 (talk) 11:36, 16 May 2015 (UTC)

The v dv/dx expression, referring to a derivative w.r.t. position, just does not seem to fit in the text where the topic is derivative w.r.t. time, as here and here. It is of course true, but is it useful? Do you have a source where it is actually used? If so, we could perhaps include it—and the source—in a separate subsection. - DVdm (talk) 11:50, 16 May 2015 (UTC)
Re your remark about time interval: usually the variable t is called time, whereas an explicit time interval is represented by Δt. - DVdm (talk) 12:04, 16 May 2015 (UTC)
@DVdm: t is called time, but in the formula, we put the time interval as the value. And, it always remains to the reader that hoe s/he remembers it. I feel the explicit correct/ wrong can't be decided. And that formula: My sir had taught me. I too thought it had no use. But a month later (3rd of May, 2015) in AIPMT, the question was find acceleration. v(x) = k (x^[-2n]) and there I found the use of this formula's use for the very first time. So, shall I include it?
117.248.121.6 (talk) 12:55, 16 May 2015 (UTC)
No, without a source that establishes its notability, let's not include it. - DVdm (talk) 12:58, 16 May 2015 (UTC)
By the way, this was a good find. However, I relinked time to the relevant article Time in physics. - DVdm (talk) 13:26, 16 May 2015 (UTC)

## Identities?

Would it be helpful or just complicate things to have a table of identities for uniform acceleration? I know I keep finding myself on this page for them:

${\displaystyle v=v_{0}+at}$
${\displaystyle s=v_{0}t+{\frac {1}{2}}at^{2}={\frac {v_{0}+v}{2}}t}$
${\displaystyle |v|^{2}=|v_{0}|^{2}+2\,a\cdot s}$

so we have time as a function of v and a, time as a function of distance traveled and velocities:

${\displaystyle t={\frac {v-v_{0}}{a}}={\frac {2s}{v_{0}+v}}}$

and we have the acceleration required to change speed in a given time and the acceleration required to go a distance from a given velocity in a given time:

${\displaystyle a={\frac {v-v_{0}}{t}}=2{\frac {s-v_{0}t}{t^{2}}}}$

also, distance traveled given velocities and a:

${\displaystyle s={\frac {v^{2}-v_{0}^{2}}{2a}}}$

Others? At least they are now here for my future reference. :-) —Ben FrantzDale (talk) 12:03, 16 October 2015 (UTC)

Could be helpful, provided (1) all the variables are carefully described, and provided (2) we work in one dimension (negative/positive "magnitudes" only, and vectors a, v and v0 parallel), in which case the third equation becomes
${\displaystyle v^{2}=v_{0}^{2}+2as}$
Otherwise it would get too complicated. And of course, as we don't want the readers to have to go verify for themselves, provided (3) we include a source, which should be easy to find. - DVdm (talk) 13:07, 16 October 2015 (UTC)

## Wrong equation in Uniform acceleration?

The equation:

${\displaystyle \mathbf {s} (t)=\mathbf {s} _{0}+\mathbf {v} _{0}t+{\tfrac {1}{2}}\mathbf {a} t^{2}={\frac {\mathbf {v} _{0}+\mathbf {v} (t)}{2}}t}$

seems wrong, since the last part does not contain ${\displaystyle \mathbf {s} _{0}}$. The equation seems to mixing up displacement and position. See http://www.open.edu/openlearn/science-maths-technology/science/physics-and-astronomy/describing-motion-along-line/content-section-1.6.1#ueqn001-050 — Preceding unsigned comment added by MLópez-Ibáñez (talkcontribs) 18:33, 27 February 2016 (UTC)

You are right and I've fixed it, -- 20:44, 27 February 2016 (UTC)
I think it is still wrong to talk about initial displacement. Displacement at time t is the difference between position at time t and initial position. Thus initial displacement has to be zero by definition. MLópez-Ibáñez (talk) 00:18, 28 February 2016 (UTC)
I've added words to clarify that, in this case, s is displacement from the origin, not displacement from the initial position. -- 00:36, 28 February 2016 (UTC)