Constant Acceleration Equation - physicist please

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SUMMARY

The discussion centers on the correct equation for calculating constant acceleration from rest, with participants debating between A = V*T and A = V*T². Both equations were deemed incorrect, as acceleration is defined as a = dv/dt, leading to the correct formulation v = u + at. The conversation highlights the importance of dimensional analysis in physics, emphasizing that mixing units can lead to nonsensical results. Ultimately, the participants concluded that both initial claims were wrong and that a proper understanding of physics requires adherence to dimensional consistency.

PREREQUISITES
  • Understanding of basic physics concepts, particularly acceleration and velocity.
  • Familiarity with dimensional analysis and its application in physics.
  • Knowledge of kinematic equations, specifically v = u + at.
  • Ability to differentiate between mathematical manipulation and physical interpretation of equations.
NEXT STEPS
  • Study the derivation and application of kinematic equations in physics.
  • Learn about dimensional analysis and its significance in ensuring physical equations are valid.
  • Explore the implications of redefining base units on physical equations and measurements.
  • Investigate common misconceptions in physics related to acceleration and motion.
USEFUL FOR

Students of physics, educators teaching kinematics, and anyone interested in understanding the principles of motion and acceleration in a rigorous manner.

  • #31
Gregor said:
I guess I have to spell it out for you

a = 9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

Those are not equal at all, even the numbers are different, the units are not the same. On the far right your units are correct adn your numbers are as well, but in the middle it's just completely wron you cannot just change between multiplication and division and expect the results to be equal, Multiplication is not a commutative operation. I don't see what you're trying to even show because neither your math nor your physics makes nay sense.
 
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  • #32
Yes, that is correct.
 
  • #33
Gregor said:
1 g = 2.4516625 m / 0.5 s / 0.5 s

yes or no ?

Yes it does, but LOOK AT THE UNITS. You have also said
a = 9.80665 m/s x 0.001 s x 0.001 s
and that is certainly not a true statement because you end up with meters*seconds which is not and never will be a measure of acceleration. Pay attention to your units and don't carelessy move them around.
 
  • #34
Gregor said:
I guess I have to spell it out for you

a = 9.80665 m x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

are you saying 1 g does not = 9.80665 µm / ms / ms ?

because if you are - I'm not the one who needs help here
Gregor, if you don't have a / in the expression, you don't get a / in the answer. You are multiplying the numbers and dividing the units while simultaneously changing another number without changing the units you used to make the change.

If you want acceleration, you need to divide by s^2, not multiply by s^2 and then pretend you divided by putting in units and numbers that don't match what you actually did!

Gregor, the things you are doing here are so obviously wrong that we're having a hard time believing you aren't doing it on purpose. And if you are doing it on purpose, that's a no-no here.
 
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  • #35
a = 9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

now do you get it?

Those are not equal at all, even the numbers are different,

the first "m/s" above was a typo
look at the operations below...
9.80665 x 1 x 1 = 9.80665

1 g = 9.80665 m / s / s
9.80665 x 0.001 x 0.001 = 0.00000980665

1 g = 0.00000980665 m / 0.001 s / 0.001 s
9.80665 x 0.5 x 0.5 = 2.4516625

1 g = 2.4516625 m / 0.5 s / 0.5 s
these results are correct, so the operation works no matter what

throw any squared time interval at me and I'll give you the correct value for 1 g

you guys are getting hysterical over your dimensional analysis equations
and ignoring the fact that my end results are 100% correct
 
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  • #36
This,

9.80665 x 0.001 x 0.001 = 0.00000980665

is meaningless algebra when you take away the units.

This,

1 g = 0.00000980665 m / 0.001 s / 0.001 s

has meaning.

Your result might be correct, but it is trivial. All you have done is use a different unit base, but so what?
 
  • #37
Gregor said:
the first "m/s" above was a typo (forget it, and forget the units for now)

9.80665 x 0.001 x 0.001 = 0.00000980665

1 g = 0.00000980665 m / 0.001 s / 0.001 s

this result is correct, so the operation works no matter what you say

Who are you kidding? yes what you have now is correct, but what you said is

9.80665 m/s x 0.001 s x 0.001 s = 0.00000980665 m / 0.001 s/ 0.001 s

Your units are the same on both sides so I'm going to do what you've been doing and ignore them for a second. The number on the left hand side is 0.00000980665 but on the right hand side you had 9.80665 so there is no way those expressions could be equivalent, even ignoring the units as you have done. And just because the result of your operation is correct doesn't prove your point at all.
 
  • #38
the first "m/s" above was a typo
look at the operations below...

it pays to read all the posts

and who are you kidding?

the values are correct, the units give the values meaning
but the calculations get you there in the first place
 
  • #39
Gregor said:
the first "m/s" above was a typo
look at the operations below...


[1] 9.80665 x 1 x 1 = 9.80665

[2]1 g = 9.80665 m / s / s

[3]9.80665 x 0.001 x 0.001 = 0.00000980665

[4]1 g = 0.00000980665 m / 0.001 s / 0.001 s

[5]9.80665 x 0.5 x 0.5 = 2.4516625

[6]1 g = 2.4516625 m / 0.5 s / 0.5 s

these results are correct, so the operation works no matter what [expression numbers added for clarity]
Yes, that set works (it isn't done quite right, but it does work if we translate it). But so what? What have you proven?
 
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  • #40
you guys are getting hysterical over your dimensional analysis equations
and ignoring the fact that my end results are 100% correct

Your results might be 100% mathematically correct, but when they are 100% wrong dimensionally, then they bear no physical meaning. If they have no physical meaning, then the whole statement is trash, no matter how fancy or correct the mathematics. I hope you see why!
 
  • #41
Your units are the same on both sides so I'm going to do what you've been doing and ignore them for a second. The number on the left hand side is 0.00000980665 but on the right hand side you had 9.80665 so there is no way those expressions could be equivalent,

what on Earth are you talking about?

0.00000980665 is the result of multipying 9.80665 x 0.001 x 0.001

hello, is this thing on? :rolleyes:
 
  • #42
Gregor - SO WHAT? What does this prove? It doesn't, for example, make the things you were saying this morning any less wrong.

All you did here is multiply both sides of an equation by .000001 and discover that it is still equal to what you had before. You discovered algebra!
 
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  • #43
Your results might be 100% mathematically correct, but when they are 100% wrong dimensionally, then they bear no physical meaning. If they have no physical meaning, then the whole statement is trash, no matter how fancy or correct the mathematics. I hope you see why!

yes, absolutely

I know exactly what you're saying, and I agree with you

what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

it's a mathematical experiment, not a physics experiment
who ever said it was a physics experiment ??
 
  • #44
Gregor said:
what on Earth are you talking about?

0.00000980665 is the result of multipying 9.80665 x 0.001 x 0.001

hello, is this thing on? :rolleyes:

Don't say I'm not reading all of the posts if you're not reading mine either.
 
  • #45
Gregor said:
yes, absolutely

I know exactly what you're saying, and I agree with you

what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

it's a mathematical experiment, not a physics experiment
who ever said it was a physics experiment ??

I don't see how you've found a shortcut, you've done nothing other than screw around with units that makes a lot of your expressions physically meaningless. Why is this a physics experiment, because g has little use in mathematics but is extremeley necessary in physics.
 
  • #46
Gregor - SO WHAT? What does this prove? It doesn't, for example, make the things you were saying this morning any less wrong.

which things specifically are you referring to?
 
  • #47
what I am saying is that I have found a shortcut to recalculating the value of 1 g from any squared time interval and any length unit.

I did not see your work for this, but let us assume that you did. What does it matter? We are only interested in two units, SI, or English. Time is the same for both systems, but distance is not. So, while you may have a way to find g, it really has no importance as we do not use any system other than these two. I hope you realize all you have done is use a different base system. I could have written your equation as:

g = \frac{9.80665 \mu M}{ 1 ms * 1 ms}
 
  • #48
I don't see how you've found a shortcut, you've done nothing other than screw around with units that makes a lot of your expressions physically meaningless. Why is this a physics experiment, because g has little use in mathematics but is extremeley necessary in physics.

let's step back to the intro...

my friend and I decided to make a mathematical experiment

what would happen to the value of 1 g if the second were redefined as 1 ms

my friend insisted that the new value of 1 g would be 9.80665 mm/s/s
(9.80665 mm/ms/ms in other words)

and I said it was 9.80665 µm/s/s
(9.80665 µm/ms/ms in other words)

since s = 0.001 in this experiment

and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct :rolleyes:
 
  • #49
Because you are BOTH wrong. This is a case where you paid no attention to the units.
 
  • #50
Gregor said:
which things specifically are you referring to?
Post 17.

Regardless, I'm still wondering what you think you've discovered here. All you did was apply some algebra.
 
  • #51
Gregor said:
and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct :rolleyes:
No, Gregor - that is not what you said in your opening post. You said:
A = V*T²
And it is still wrong.

If you are wondering about the little algebra game that you played, yes, you correctly multiplied by 1 and your friend did not. Everyone saw the error you made with the equations and no one checked to see if you multiplied by 1 correctly in the second part of the post. But even still - I don't think you understand that that's all you did.
 
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  • #52
Gregor said:
let's step back to the intro...

my friend and I decided to make a mathematical experiment

what would happen to the value of 1 g if the second were redefined as 1 ms

my friend insisted that the new value of 1 g would be 9.80665 mm/s/s
(9.80665 mm/ms/ms in other words)

and I said it was 9.80665 µm/s/s
(9.80665 µm/ms/ms in other words)

since s = 0.001 in this experiment

and the algorithm I used to derive the result was A*T²

my friend used A*T

my results were right and his were wrong

and everyone in physics forums blew a fuse
because I used a non-standard formula

even though my result was correct :rolleyes:


You cannot just redefine the way we measure acceleration, I understand it if you use different units that emasure the same thing, but you are not allowed to just change it from m/s2 to ms they are not the same thing.

You are both wrong, why won't you just accept that? You didn't use a "nonstandard" formula you used one which has no physical significance as a measure of acceleration. You come and then ask this on a PHYSICS forum what did you think would happen, that we would all magically agree with you and through away hundreds of yeasr of work by people a lot smarter than you are because of this one post?
 
  • #53
cyrusabdollahi said:
I did not see your work for this, but let us assume that you did. What does it matter? We are only interested in two units, SI, or English. Time is the same for both systems, but distance is not. So, while you may have a way to find g, it really has no importance as we do not use any system other than these two. I hope you realize all you have done is use a different base system. I could have written your equation as:

g = \frac{9.80665 \mu M}{ 1 ms * 1 ms}
Apparently, that's all he was trying to do: g = 9.8 \frac{m}{s*s} * \frac{.000001}{.000001}= \frac{9.8 \mu m}{ 1 ms * 1 ms}

All that playing with the equations he did just confused him and everyone else...
 
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  • #54
Is any of this getting through, Gregor? This is quite a generous amount of ink being spilled here.
 
  • #55
yes hypermonkey, you and russ and everyone have made yourselves very clear from the start, and I appreciate your efforts.

I think we started off on a misunderstanding in any case

because when I wrote A = V*T² (an expression I wrote myself)
I used it to mean g = A*T²

the V was never meant to represent velocity alone
but rather an increase in velocity through time

I should have posted g = A*T² or even A = g*T²

I was trying to avoid A = A*T² since this would be even more ambiguousthe operation I used to derive the new value for 1 g is A*T²

9.80665 x 0.001 x 0.001

since this was a mathematical experiment, I was only interested in the numeric values, and not the units or dimensions.

since I acheived my end goal (the correct numerical value for 1 g in modified units), I have a tool to check the accuracy of operations using proper motion physics equations.

since my equation (as unorthodox as it is) yealds 100% correct values 100% of the time.

I hope I've made my point now, so there's no more misunderstandings

thanks again to Integral, Zapperz, vanesch, hypermonkey2, SaMx, Doc Al, russ_watters, cyrusabdollahi, and d_leet.

your help is very much appreciated :cool:
 
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  • #56
Gregor said:
yes hypermonkey, you and russ and everyone have made yourselves very clear from the start, and I appreciate your efforts.

I think we started off on a misunderstanding in any case

because when I wrote A = V*T² (an expression I wrote myself)
I used it to mean g = A*T²

the V was never meant to represent velocity alone
but rather an increase in velocity through time

I should have posted g = A*T² or even A = g*T²

I was trying to avoid A = A*T² since this would be even more ambiguous


the operation I used to derive the new value for 1 g is A*T²

9.80665 x 0.001 x 0.001

Oh good grief.

You see nothing wrong with saying g = at^2, even when g is defined as a gravitational acceleration and having the same dimension as an acceleration? Have you ever considered the resulting dimension of such a thing (you can no longer call it g since it is not the same creature anymore)? at^2 has the dimension of length!

So, since this is physics, what does this "length" represent?

since this was a mathematical experiment, I was only interested in the numeric values, and not the units or dimensions.

And that is why I have suggested you do your numerology in mathematics, but you insisted in being in here.

This has got to be one of the strangest thread I've ever seen on PF. I'm almost tempted to label it as crackpottery.

Zz.
 
  • #57
You see nothing wrong with saying g = at^2

In physics terms, Yes

In mathematical terms, No

the resulting values are always correct.

in any case g = at^2 doesn't need to make sense in mathematical terms
I just used it to get from A to B (fast)

I'm sure you're familiar with the expression: "the end justifies the means" ;-)

anyway, now that I have the proper equations (thanks to you and Integral)
I can do it the right way - so what's the problem?
 
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  • #58
as soon as someone can prove that the values below are not correct
then we can say that A = g*T² is mathematically false

however the values below Are correct, and therefore A = g*T² is mathematically correct (although useless for physics)

(note* the numerical value of g = 9.80665)1 g =

0.00000980665 m / 0.001 s / 0.001 s

0.000980665 m / 0.01 s / 0.01 s

0.0980665 m / 0.1 s / 0.1 s

0.392266 m / 0.2 s / 0.2 s

0.8825985 m / 0.3 s / 0.3 s

1.569064 m / 0.4 s / 0.4 s

2.4516625 m / 0.5 s / 0.5 s

3.530394 m / 0.6 s / 0.6 s

4.8052585 m / 0.7 s / 0.7 s

6.276256 m / 0.8 s / 0.8 s

7.9433865 m / 0.9 s / 0.9 s

9.80665 m / s / sI rest my case :smile:
 
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  • #59
Let me get this straight, your saying the acceleration of a particle is equal to g mulitplied by the time?
 
  • #60
no one is listening to what I'm saying (!) :confused:

the equation has no meaning in physics
all it does is provide a correct numerical value for 1 g
for any given length or time unit

this is blind maths, and the physical world has nothing to do with it

this is only useful for quick (correct) number value solutions
when playing with new length and time units

it's absolutely useless for dimensional analysis

for dimensional analysis I have the proper equations which Integral and Zapperz posted earlier

dimensional analysis and "number getting" are two different goals
 
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