Is a(t) = 0 a constant acceleration?

[byosys]

Originally posted by: 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?

(Bolding added to the quote)

I'm not sure what "integral forumla" you mean, but the 2nd if - then statement is false. Counter example:
f(x) = x [f(x) does not equal 0]
f'(x) = g(x) = 1
int(g(x) dx) = f(x) = x + c, where c MUST be 0

 

[byosys]

Originally posted by: 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.

Yep, and threads spin into compleatly different topics than the orginal post (after the orginal question is answered of course). I don't see how that is really a problem - if people are interested (and they must be interested if they bother to post), why not let the thread go where it may (with in reason of course).

 

[itachi]

Originally posted by: byosys

Originally posted by: 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?

(Bolding added to the quote)

I'm not sure what "integral forumla" you mean, but the 2nd if - then statement is false. Counter example:
f(x) = x [f(x) does not equal 0]
f'(x) = g(x) = 1
int(g(x) dx) = f(x) = x + c, where c MUST be 0

what u showed there in the last line is what i meant by the integral formula..

i guess i wasn't really clear in those statements..
if f is constant, g is 0..
assuming f is constant - if f can't equal 0, the constant in the integral of g can't equal 0..

 

[byosys]

I'm not sure exactly what you know, so forgive me if this seems a little basic.

f(x) = x^2
g(x) = f'(x) = 2x + 0

Given that, neither f(x) nor g(x) are constant, so the conditions you listed above do not apply and c can equal zero. From there, its basic integration:

int[g(x) dx] = x^2 + c = x^2 = f(x)

Note that we can find a value for c only because we are given the initial equation. Things are a little different when constants are added/subtracted to/from the x^2

f(x) = x^2 + k where k is a constant real number
g(x) = f'(x) = 2x
int[ g(x) dx] = x^2 + c

In this case, c must equal k. If we were not given f(x) to begin or initial conditoins, we would have no way of knowing what c is.

 

[itachi]

Originally posted by: byosys
I'm not sure exactly what you know, so forgive me if this seems a little basic.

f(x) = x^2
g(x) = f'(x) = 2x + 0

Given that, neither f(x) nor g(x) are constant, so the conditions you listed above do not apply and c can equal zero. From there, its basic integration:

int[g(x) dx] = x^2 + c = x^2 = f(x)

Note that we can find a value for c only because we are given the initial equation. Things are a little different when constants are added/subtracted to/from the x^2

f(x) = x^2 + k where k is a constant real number
g(x) = f'(x) = 2x
int[ g(x) dx] = x^2 + c

In this case, c must equal k. If we were not given f(x) to begin or initial conditoins, we would have no way of knowing what c is.

err.. you're missing my point.

everything that u showed is why the statements i listed aren't valid. if an acceleration of 0 can't be considered constant, it wouldn't be a solution to the integral of 0..

 

[byosys]

Originally posted by: itachi
err.. you're missing my point.

everything that u showed is why the statements i listed aren't valid. if an acceleration of 0 can't be considered constant, it wouldn't be a solution to the integral of 0..

Oh boy...I get it now. I thought you were asking about the math, not relating it to my orginal question. You make an interesting point. Sorry for the misunderstanding.

 

[wwswimming]

Originally posted by: 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.

i agree with you.

maybe you can trick him using a limit appoach ...
lim a(t) = .0000000000000000000000000001
that's constant, right ?

keep going till you have 100 zeros.

mathematically, 10^100 is infinity, and it's inverse is zero.

what's he going to say,
a(t)= [99 zeroes]1
is constant
and
a(t)= [100 zeroes]1
is NOT constant ?

 

[BrownTown]

Originally posted by: wwswimming

Originally posted by: 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.

i agree with you.

maybe you can trick him using a limit appoach ...
lim a(t) = .0000000000000000000000000001
that's constant, right ?

keep going till you have 100 zeros.

mathematically, 10^100 is infinity, and it's inverse is zero.

what's he going to say,
a(t)= [99 zeroes]1
is constant
and
a(t)= [100 zeroes]1
is NOT constant ?

do you even understand math? Some of the posts here make me wonder...

but really, we all agree he is right here, you don't need to argue that point further

 

[byosys]

Originally posted by: wwswimming

Originally posted by: 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.

i agree with you.

maybe you can trick him using a limit appoach ...
lim a(t) = .0000000000000000000000000001
that's constant, right ?

keep going till you have 100 zeros.

mathematically, 10^100 is infinity, and it's inverse is zero.

what's he going to say,
a(t)= [99 zeroes]1
is constant
and
a(t)= [100 zeroes]1
is NOT constant ?

I'm going to echo BrownTown here, but point out where this falls appart. The bolded part is flat out wrong. 10^100 is certainly NOT infinity. Nor is 10^1,000,000 nor 10 to anyother power other than infinity. Also, 1/infinity is NOT zero either. It is indeterminate. Alot of people hear (and beleive) that 1/infinity = 0, when the proper statement is lim (1/x, x--> infinity) = 0. Hope that clears it up.

 

[Eeezee]

Originally posted by: 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.

Your intuition was correct, it is a constant acceleration. There's really no way to refute that.

I think the teacher was trying to help you for tests, homework, etc. If a textbook says "constant acceleration" then it's usually an assumption that the acceleration is non-zero. Just remember that for any tests or homework problems involving a "constant acceleration" which is more specifically probably a non-zero acceleration.

When someone says "constant acceleration" they USUALLY mean non-zero, but it is correct to mean zero as well (and is referred to as "constant velocity" to avoid confusion). I've never actually heard someone say "constant acceleration" when referring to a zero acceleration.

 

[Eeezee]

Originally posted by: 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+@$$".

Hahaha I love those professors. Usually the students are wrong and are just wasting time with questions that should be asked during office hours.