Talk:Acceleration
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[edit] Change in acceleration
"When either speed or direction are changed, there is a change in acceleration." 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.
[edit] acceleration residuals
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.
[edit] 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)
