Is a(t) = 0 a constant acceleration?

[byosys]

This has been bugging me for a little while now, but I just got around to posting it. My physics teacher said that an acceleration equal to 0 was NOT a constant acceleration as the was no acceleration to speak of. I contended that given a(t) = 0 for all t that a(x) - a(y) = 0 for all x and y (that exist of course) and therefore an acceleration of 0 was indeed constant. My teacher tried to explain it to me several times (which I kept trying to refute) and in the end basically said "deal with it".


So, do you consider an acceleration (or velocity or anything else you can think of) of zero to be constant.

 

[tex1138]

You've got my vote. In any calculus I have taken, a constant is any function whose derivative is zero--definitely true in this case. On top of that, all of the constant acceleration equations in my physics textbook hold for a = 0.

You didn't indicate whether this was high school, or college, or what. I tend to assume high school due to you saying "teacher" instead of "professor". My sister had a math teacher (who was also the physics teacher, although she didn't have him for that class) who would routinely mess up example problems and then, when he was called on it by students, invent some b.s. conclusion to the problem to try and imply that there was no mistake. Thank $deity I got through before he was hired. But anyway, the moral of the story is that it would be a lot more productive use of your time to seek out more advanced sources of information and study up on your own than to argue with someone whose mind is already made up.

 

[BrownTown]

Yeah, technically I think your right, but its not really worth it to argue about such symantics in my view, teachers (and professors with PHDs for that matter) make mistakes like the rest of us, I kinda get annoyed when people try to look smart correcting professors when they make a mistake. Now, I deffinitely symathise with you if you have a teacher who will not admit they are wrong, but its often OK to just let it go if you know you are right and not annoy the teacher seeing as whether or not they are wrong they still determien your grade .

 

[DrPizza]

<-- physics teacher

I agree that a(t)=0 for all t is a constant acceleration. tex1138 summed up the #1 reason why I'd say it's a little more than simple semantics.
And, I seem to recall a high stakes exam that had "constant acceleration" for the correct answer when acceleration was 0. (I think the choice was C) constant acceleration of 0 m/s^2 )

 

[BrownTown]

I say semantics here becasue often the world of words and numbers gets blurred in mathmatical fields. There is a scientific deffinition for acceleration, and there is a deffinition used in real life. In the scientific deffinition then clearly an acceleration of 0 is a cosntant acceleration. In the real world deffinition something is accelerating if its speed is changing, if its not then there is no acceleration. The argument that you are actually acclerating with a constant zero function just seems like an overcomplication of the obvious fact that you aren't acceleration to most people. So, if you are in school, or taking a test you should say you have an acceleration of zero, and the teacher should have told you that you are correct. In the real world youd probably be better jsut saying your not accelerating. Although to be honest I realise I'm just digging myself in a hole now trying to justify my previous statement. So read what DrPizza said and disregard my sleep deprived rants.

 

[silverpig]

It's the semantics of arguing a constant acceleration of 0 vs no acceleration.

I think the teacher's argument was that if there is no acceleration, it can't be constant, increasing, etc. Sort of like holding up an empty hand and saying that it contains a massless apple...

I would say you were right though, but I'd also say it's not really worth arguing over.

 

[ruffilb]

Originally posted by: DrPizza
<-- physics teacher

I agree that a(t)=0 for all t is a constant acceleration. tex1138 summed up the #1 reason why I'd say it's a little more than simple semantics.
And, I seem to recall a high stakes exam that had "constant acceleration" for the correct answer when acceleration was 0. (I think the choice was C) constant acceleration of 0 m/s^2 )

 

[byosys]

This was in high school AP Physics. Her argument was based around the fact that if a(t) = 0, then there is no acceleration "to be" constant which didn't make sense to me, hence the argument (I'm not one to let little details like that go easily). Anyway, I'm glad to see that I'm not the only one that thinks like I do. Thanks.

 

[BassBomb]

if acceleration is zero, then the velocity is constant

 

[BrownTown]

byosys, you are right, be happy in your glorious triumph over the education system!

 

[Born2bwire]

Originally posted by: silverpig
It's the semantics of arguing a constant acceleration of 0 vs no acceleration.

I think the teacher's argument was that if there is no acceleration, it can't be constant, increasing, etc. Sort of like holding up an empty hand and saying that it contains a massless apple...

I would say you were right though, but I'd also say it's not really worth arguing over.

Yeah, while 0 m/s/s is technically a constant acceleration, it also means that there is no acceleration. So I could see either one being valid but in the context of high school courses, I would probably go with the teacher on this one. Over technicality when trying to teach the basics only leads to confusion. Besides, as silverpig said, this isn't something that is worth arguing over.

 

[byosys]

Originally posted by: BrownTown
byosys, you are right, be happy in your glorious triumph over the education system!

Damn straight I will be!

Originally posted by: BassBomb
if acceleration is zero, then the velocity is constant

And?

 

[msparish]

a(t)=0
a'(t)=0 as well

byosys 1
teacher 0

 

[DrPizza]

Originally posted by: BassBomb
if acceleration is zero, then the velocity is constant

Accept, according to the teacher, that's false if the constant velocity = 0...

And, a logical extension of this is if velocity = 0, then the position function is constant.

Ask your teacher what it means if the position remains at zero - does it mean that there's "no position"?

 

[eLiu]

Zero is a constant. Fine, there's "no acceleration to speak of"...and there is constantly no acceleration to speak of, lol. Constant... as in non-changing.

He's trying to draw some English into a mathematical concept--bad.

 

[stupidkid]

Uh isn't acceleration a vector quantity? So it has a direction and a magnitude and measures a change in velocity. Therefore, if velocity changes direction but stays constant in terms of magnitude wouldn't a(t) = 0 but yet the object would still technically be accelerating? I'm not totally sure but that's just my 2¢.

 

[DrPizza]

Originally posted by: stupidkid
Uh isn't acceleration a vector quantity? So it has a direction and a magnitude and measures a change in velocity. Therefore, if velocity changes direction but stays constant in terms of magnitude wouldn't a(t) = 0 but yet the object would still technically be accelerating? I'm not totally sure but that's just my 2¢.

If something travels in a circle, it can maintain the same *speed*, but as the direction is constantly changing, there must be an acceleration. The acceleration (centripetal) = the speed squared divided by the radius of the turn, and certainly doesn't = 0.

 

[BrownTown]

no, if the acceleration is 0 the objects velocity is not changing.

EDIT: aww man, way to come in and post before me so now i look like im a parrot...

FURTHER EDITING:

So, the velocity vector could be <Cos(t),Sin(t)> which has a constant magnitude of 1, but not a constant direction. The acceleration would then be <-Sin(t),Cos(t)>, which is also has a cosntant (non-zero) magnitude of 1. This is like what DrPizza described, the orbit of a body around another due to a constant force attracting them.

 

[imported_inspire]

Your physics teacher was wrong. Given a function, f(t)=c, for C some real number, the function f(t) is said to be constant.

That's the mathematical definition (Being a Mathematician, I know handy things like this.) Last time I checked, zero was a real number.

 

[Sk8orDie]

My AP Physics teacher in high school used the explination "because god said so" and "because god wanted it that way" on more than one occasion. No offense to any teachers or professor who may be reading this, but they are not always right.

 

[BrownTown]

Heh, I had a professor who would pretty much tell students they were wrong everytime they tried to correct him even when he knew they were right. They are currently the best professor I have had while at college, but maybe that's just because I can't stand those people who sit in the front row and get off on correcting the professor, or asking annoying questions that slow down the class. I think we all know that professors make mistakes, and maybe you just have a senile professor or something, but I'd say more likely when you correct a professor and they say something like "because god said so" what they are really saying is "stop being such a 5m@r+@$$".

 

[BucsMAN3K]

There is a point where common sense needs to be used.

If your acceleration is 0, and stays 0, then yes, its a constant acceleration, and you have to understand this also means constant velocity. But Mr. Common Sense says, hey, it's just easier to label this as NO acceleration and constant velocity.

I'm sure the professor understands its a constant acceleration, but he want's to make sure people understand that no acceleration is occuring as well.

 

[Agman]

I seem to recall that if a derivative is 0 the one above is constant. For example if velocity is constant, lets say 60 and you take the derivative then a(t) = 0. So there is no acceleration it just means you have constant velocity not constant acceleration.

 

[itachi]

kinda taking on what pizza said..

given two functions, f and g.. with g equal to the derivative of f..
if f is constant, g is 0..
if f can not equal 0, then the constant factor in the integral of g can not equal 0..
if the constant factor can not equal 0.. how the hell do you show that x**2 is a solution to the integral of 2x, using the integral formula?

 

[BrownTown]

see, this is the problem with "Highly Technical", the issue is resolved in the first five posts, and then a bunch of other people come in and say the same thing.