Talk:Acceleration

Uniform acceleration

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

$\mathbf {s} =\mathbf {u} t+{{1} \over {2}}\mathbf {a} t^{2}=\mathbf {u} t+{{1} \over {2}}{{v} \over {t}}\mathbf {t} ^{2}={(\mathbf {u} +{{1} \over {2}}\mathbf {v} )t}$

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 (talk • contribs) 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 $Q$ (even in the Newtonian spacetime), in order to define acceleration, a connection $\Gamma$ (i.e., a covariant derivative $\nabla$) 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 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 $Q$.Mgvongoeden (talk) 12:41, 22 April 2015 (UTC)

Notes

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)

---

I'm experiencing the same dichotomy in the definition of acceleration, but in a different sense... a person (I won't give his real name unless mods request it... I bet you didn't know 'Gandalf61' denies the fundamental physical laws, eh? That brings into question everything he's edited here...) who's been attempting to claim 2LoT can be violated at the quantum scale (for the record, 2LoT is even more rigorously observed at the quantum scale than macroscopically) has been backed into a deeper and deeper corner, to such an extent that we're now arguing over whether acceleration is a rank-1 tensor or a rank-2 tensor. He claims all derivatives of position (velocity, acceleration, jerk, snap, crackle, pop, lock, drop) are rank-1 tensors because apparently a temporal derivative is a 'special' kind of derivative that doesn't have to comport with differential calculus. He's edited this page to remove all references to the tensor rank of acceleration. I've reverted his changes.

As evidence that a derivative increases resultant tensor rank:

---

http://openaccess.thecvf.com/content_cvpr_2018/papers/Kim_High-Order_Tensor_Regularization_CVPR_2018_paper.pdf "In general, taking a derivative of a tensor increases its order by one: The derivative of function f is a vector, a first-order tensor. Similarly, the derivative of a second order tensor h is a third order tensor ∇gh."

---

https://en.wikipedia.org/wiki/Tensor_derivative_(continuum_mechanics)#Gradient_of_a_tensor_field

"Derivatives of scalar valued functions of vectors Let f(v) be a real valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the vector defined through its dot product..."

"Derivatives of vector valued functions of vectors Let f(v) be a vector valued function of the vector v. Then the derivative of f(v) with respect to v (or at v) is the second order tensor defined through its dot product..."

"Derivatives of tensor valued functions of second-order tensors Let F(S) be a second order tensor valued function of the second order tensor S. Then the derivative of F(S) with respect to S (or at S) in the direction T is the fourth order tensor..."

"The gradient of a tensor field of order n is a tensor field of order n+1."

---

In other words, the derivative of a scalar is a vector, the derivative of a vector is a rank 2 tensor, and the 2nd derivative of a rank 2 tensor is a rank 4 tensor, because the gradient of a tensor field of order n is a tensor field of order n+1.

Of course, the gradient is dual to the derivative, one cannot have a derivative without a gradient, nor can one have a gradient without a derivative.

---

Gandalf61 stated in his change of the web page that "acceleration is derivative of velocity wrt time, *not* a gradient of a vector field". That is wrong. http://clas.sa.ucsb.edu/staff/alex/VCFAQ/vectorFields/vectorFields.htm The technical definition of a vector field is a map from R^3 to R^3 What this means is we can assign a 3 dimensional vector to every point in R^3. We can think of the vector field as an ordered set of 3 functions: F = (f1(x,y,z), f2(x,y,z),f3(x,y,z)). Here the functions f1, f2 and f3 are ordinary scalar functions of x, y, and z.

Some common vector fields in physics: Electric Fields

Magnetic Fields

Gravitational Fields

Wind Velocity <---

Fluid Velocity <---

https://en.wikiversity.org/wiki/Acceleration_field "Acceleration field is a two-component vector field..."

If Gandalf61 wishes to sabotage Wikipedia pages in support of his denial of the fundamental physical laws and his kooky take on reality, then his entire history of edits should be reviewed to ensure they actually comport with reality.

To RealOldOne2... the person doing the editing of this page to remove any reference to tensor rank is none other than our CFACT friend, Dave Burton, with whom you've had several encounters.

Ah, I see now... the new user 'RealOldOne2' has hijacked a well-known 'nym... it's none other than 'Gandalf61' aka Dave Burton, attempting yet again to sabotage a Wikipedia page to comport with his kooky fundamental-physical-law denying take on reality. Can an administrator take this nutter 'Gandalf61' aka 'RealOldOne2' down, please?

---

Where to place this?

Have been reverted 2 times. Where to place it? $\mathbf {a} =\lim _{{\Delta t}\to 0}{\frac {\Delta \mathbf {v} }{\Delta t}}={\frac {d\mathbf {v} }{dt}}=v{\frac {d\mathbf {v} }{dx}}$
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:

$v=v_{0}+at$

$s=v_{0}t+{\frac {1}{2}}at^{2}={\frac {v_{0}+v}{2}}t$

$|v|^{2}=|v_{0}|^{2}+2\,a\cdot s$

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

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

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:

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

also, distance traveled given velocities and a:

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

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

$v^{2}=v_{0}^{2}+2as$

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:

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

seems wrong, since the last part does not contain $\mathbf {s} _{0}$. 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 (talk • contribs) 18:33, 27 February 2016 (UTC)

You are right and I've fixed it, -- Dr Greg  talk  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. -- Dr Greg  talk  00:36, 28 February 2016 (UTC)

External links modified

Hello fellow Wikipedians,

I have just modified one external link on Acceleration. Please take a moment to review my edit. If you have any questions, or need the bot to ignore the links, or the page altogether, please visit this simple FaQ for additional information. I made the following changes:

• Added archive https://web.archive.org/web/20101108131449/http://www.measurespeed.com/acceleration-calculator.php to http://measurespeed.com/acceleration-calculator.php

When you have finished reviewing my changes, you may follow the instructions on the template below to fix any issues with the URLs.

This message was posted before February 2018. After February 2018, "External links modified" talk page sections are no longer generated or monitored by InternetArchiveBot. No special action is required regarding these talk page notices, other than regular verification using the archive tool instructions below. Editors have permission to delete these "External links modified" talk page sections if they want to de-clutter talk pages, but see the RfC before doing mass systematic removals. This message is updated dynamically through the template {{source check}} (last update: 5 June 2024).

• If you have discovered URLs which were erroneously considered dead by the bot, you can report them with this tool.
• If you found an error with any archives or the URLs themselves, you can fix them with this tool.

Cheers.—InternetArchiveBot (Report bug) 16:45, 1 December 2017 (UTC)

category=category

Can the category of this vital article be category?--Dthomsen8 (talk) 19:39, 28 September 2019 (UTC)

Acceleration is a vector

I have reverted repeated revisions by anonymous contributor 71.135.47.16 (talk · contribs) which claim that acceleration is not a vector/rank-1 tensor but is instead a rank-2 tensor. This seems to be based on the mistaken assumption that acceleration is the gradient of velocity and so should have a tensor rank that is one greater than that of velocity. Instead, of course, acceleration is the derivative of velocity with respect to time and so has the same rank as velocity, which in turn has the same rank as position displacement i.e. all three quantities are vectors/rank 1 tensors. Gandalf61 (talk) 08:56, 17 March 2020 (UTC)

Position isn't really a vector! (Displacement in Euclidean space maybe, but not position).TR 09:25, 17 March 2020 (UTC)

Agreed. My bad. Fixed above. Gandalf61 (talk) 11:23, 17 March 2020 (UTC)