![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
"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)
Why don't you merge this article with acceleration? Rcouto
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)
does anyone here know about acceleration residuals? i was reading a scientific paper about doppler shifts and it talks about residuals, however, it assumes that the reader knows what they are. after a google search, all of the results make the same assumption, so i can't find an actual explanation anywhere. i'm not sure if this article is where it should go, but i'd really appreciate someone defining residuals.
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)
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)
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)
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)
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)
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)
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.
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. -- Slarti bartfast 1992 22:14, 12 June 2009 (UTC)
There, it's done. If anybody feels that the section I just wrote is wrong, please discuss. -- Slarti bartfast 1992 02:06, 22 June 2009 (UTC)
MarcusMaximus ( talk) 06:18, 23 June 2009 (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)
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)
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)
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 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)
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)
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)
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)
. 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 . Also you can check this way . Also . Also . 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 . 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)
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)
![]() | This page is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
![]() | This is an archive of past discussions. Do not edit the contents of this page. If you wish to start a new discussion or revive an old one, please do so on the current talk page. |
Archive 1 |
"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)
Why don't you merge this article with acceleration? Rcouto
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)
does anyone here know about acceleration residuals? i was reading a scientific paper about doppler shifts and it talks about residuals, however, it assumes that the reader knows what they are. after a google search, all of the results make the same assumption, so i can't find an actual explanation anywhere. i'm not sure if this article is where it should go, but i'd really appreciate someone defining residuals.
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)
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)
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)
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)
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)
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)
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.
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. -- Slarti bartfast 1992 22:14, 12 June 2009 (UTC)
There, it's done. If anybody feels that the section I just wrote is wrong, please discuss. -- Slarti bartfast 1992 02:06, 22 June 2009 (UTC)
MarcusMaximus ( talk) 06:18, 23 June 2009 (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)
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)
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)
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 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)
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)
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)
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)
. 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 . Also you can check this way . Also . Also . 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 . 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)
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)
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