Constant Acceleration Equation - physicist please!

  • Thread starter Gregor
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  • #1
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I need a proper Physicist to settle an argument (and a bet)
between myself and a friend over the correct equation for calculating constant acceleration
(from rest) from velocity over time.

my friend insists that A = V*T

and I say A = V*T²

in plain english,

my friend claims that at 1 g acceleration

9.80665 (m/s) / s = 9.80665 (mm/ms) / ms

and I say 9.80665 (m/s) / s = 9.80665 (µm/ms) / ms

who is right? :uhh:
 

Answers and Replies

  • #2
Integral
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Send me the bet. Neither are correct.

V = at + v
d= at2 + vt + x

d = distance traveled
a= acceleration
v = velocity
x = initial postion.
V= final velocity
 
  • #3
vanesch
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Gregor said:
I need a proper Physicist to settle an argument (and a bet)
between myself and a friend over the correct equation for calculating constant acceleration
(from rest) from velocity over time.

my friend insists that A = V*T

and I say A = V*T²

I agree with Integral to tell you that this is wrong!

However,...

in plain english,

my friend claims that at 1 g acceleration

9.80665 (m/s) / s = 9.80665 (mm/ms) / ms

and I say 9.80665 (m/s) / s = 9.80665 (µm/ms) / ms

who is right? :uhh:

Here, the last expression is correct:

micro divided by (milli x milli) = 1
 
  • #4
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Doh! :yuck:


well at least I can tell my friend he was wrong o:)

thanks!!! Integral / vanesch
I really appreciate your help
 
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  • #5
ZapperZ
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Gregor said:
I need a proper Physicist to settle an argument (and a bet)
between myself and a friend over the correct equation for calculating constant acceleration
(from rest) from velocity over time.

my friend insists that A = V*T

and I say A = V*T²

in plain english,

my friend claims that at 1 g acceleration

9.80665 (m/s) / s = 9.80665 (mm/ms) / ms

and I say 9.80665 (m/s) / s = 9.80665 (µm/ms) / ms

who is right? :uhh:

You are both wrong.

First of all, you can easily check this by simply using dimensional analysis. Acceleration has dimensions of Length/(Time)^2. Both of those answers are wrong.

Secondly, for any time of motion, acceleration is defined as

[tex]a=\frac{dv}{dt}[/tex]

For a constant acceleration, you can integrate this very easily to obtain

v = u + at

where "u" is the initial velocity at time t=0. Even if you let u=0, you would end up with

v = at or a = v/t.

(note: this "t" is the time at THAT instant, not the change in time).

This is not the same as what you wrote for both. So both of you lost your bet and you should donate them to a literacy charity.

Zz.

Edit: obviously, during my typing, a slew of answers came out that essentially say the same thing. However, I don't think Integral is a charity case. :)
 
  • #6
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So both of you lost your bet and you should donate them to a literacy charity.

ouch!!! :-(
 
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  • #7
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thanks ZapperZ

it looks as though my friend and I have unwittingly entered a strawman building contest :rolleyes:

at least i took first prize!! :blushing:
 
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  • #8
Integral
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ZapperZ said:
You are both wrong.

...snip...

Edit: obviously, during my typing, a slew of answers came out that essentially say the same thing. However, I don't think Integral is a charity case. :)

LOL, thats why I tried to keep my response short and sweet!

Whose asking for charity, I thought it was for services rendered! :rofl:
 
  • #9
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you could just look at units.
the first one says:
(m/s^2)=(m/s)(s) or (m)
m is not equal to (m/s^2) so it is wrong.

The second equation says:
(m/s^2)=(m/s)(s^2) or (m)(s)
(m)(s) is not equal to (m/s^2) so that equation is also wrong

* m=meters and s=seconds in the examples above.
 
  • #10
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Another crisis avoided...
 
  • #11
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For a constant acceleration, you can integrate this very easily to obtain

v = u + at

where "u" is the initial velocity at time t=0. Even if you let u=0, you would end up with

v = at or a = v/t.

so if we redefine the meter as 0.980665 m (the velocity at 0.1 s)
then the new value for 1 g = 10 m/s² exactly, correct?

but if we also redefine the second as 0.1 s

and a = v/t

then a = 1 m/s² since v = 1 and t = 1

therefore 1 = 1/1 or 1 = 1*1 or 1 = 1*1² or 1 = 1*1³

so a = v/t or a = v*t or a = v*t² or a = v*t³

what's wrong with this picture? :-/
 
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  • #12
ZapperZ
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Er... this is getting sillier by the minute. Even if the value of something is "1", it has DIMENSIONS. 1 m/s is not equal to 1 potato. But in your scheme of things, it is, because you mix and match the dimensions no matter what.

If you ignore the dimensions of the things you are manipulating, then you can get all kinds of nonsense. That is why this is physics, and not mindless mathematics.

Zz.
 
  • #13
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please humour me and my potato for a moment..

so if we redefine the meter as 0.980665 m (the velocity at 0.1 s)
then the new value for 1 g = 10 m/s² exactly, correct?

this statement agrees with dimenional analysis - correct?

therefore the equation holds true only as long as the base unit of time remains unchanged

and any change to the second would require a revised equation

otherwise we might was well say a = p / ud

where p = a potato, u = an untidy cat, and d = a picture of a happy dog



the original equation that I posted

A = V*T²

essentially yealds the correct result

since 9.80665 µm/ms² (A) = 0.00980665 m/ms (V at T^1) x 0.001 s (T²)

I simply forgot to define V as Velocity at T^1

algebraic definition makes all the difference :wink:
 
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  • #14
ZapperZ
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Gregor said:
please humour me and my potato for a moment..



this statement agrees with dimenional analysis - correct?

therefore the equation holds true only as long as the base unit of time remains unchanged

so any change to the second would require a revised equation

otherwise we might was well say a = p / ud

where p = a potato, u = an untidy cat, and d = a picture of a happy dog

This is why we teach kids in school to solve the problems algebraically and NOT plug in numbers into the problem until the very end.

The units for an 'acceleration' is defined as Length/(Time)^2. So now, let's look at what you did:

then a = 1 m/s² since v = 1 and t = 1

therefore 1 = 1/1 or 1 = 1*1 or 1 = 1*1² or 1 = 1*1³

so a = v/t or a = v*t or a = v*t² or a = v*t³

What is "1 = 1*1"? Did you do v = at? Fine, but 1 = 1*1^2? If what you did here is

v = a*t*t, then the equality NO LONGER HOLDS! Again, this is blind mathematics. These symbols have PHYSICAL MEANINGS that you simply cannot multiply at will. This is why I said why we teach students to do physics problems algebraically. By saying v = at^2, you have already commited yourself to changing the definition of what a "velocity" is that isn't the same as before.

This is not the way to do physics.

Zz.
 
  • #15
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I think that you are working backwards, Gregor. You cannot base a physical claim on a mathematical derivation that does not look at a system from a physical point of view. Zapperz is quite right in saying that there is no physical meaning in what has been described. By writing your new equations involving t by the observation of the fact that the "equality holds true" from a mathematical standpoint when multiplied by 1, you are changing the units of your system that no longer mean anything to what you are trying to prove (velocity). Indeed, this is an important difference between math and physics. For example, math would probably describe sqrt(2) as the irrational number that gives 2 when squared (gross simplification and/or generality, but lets keep this simple). In math, sqrt(2) can even be negative! Physics would probably see sqrt(2) as the measure of the hypotenuse of a right triangle whose adjacent sides measure 1. The important word here is MEASURE. This implies units. In math, sqrt(2) has no units. Dimensional analysis pretty much seperates math and physics and one can easily poison either discipline by forgetting that. In a way, perhaps i should ask that you avoid doing this as you would to divide by 0 (gross simplification). Allow me to divide by 0, and i can prove anything you wish (it just wont make any sense!) I am not saying that you cannot use math in physics, quite the opposite! But you must be wary of your logic. Physics requires a certain thought process that is similar (but not identical) to mathematics.
 
  • #16
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I understand your points completely, and of course you and Zapperz are absolutely right.

since I normally approach problems mathematically, I tend to deal with them in a reductive, rather than analytical way, being satisfied with correct solutions, with as few operations as possible.

my friend and I were making a thought experiment..

what would happen to g if the meter and/or the second were redefined.
so we got into an argument over the correct algorithm for deriving the new value for g.

In physics you're not allowed to toy with base units,
but in maths - everything is open game. :rolleyes:
 
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  • #17
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anyway, I think it's quite interesting to explore the algebraic effects of new units

for example - if the meter were redefined as 299 792 458 m
then the equation E = mc² becomes E = M

so what happens to the equation a = v/t
with 1 meter redefined as 0.980665 m and 1 second redefined as 0.1 s (?)
 
  • #18
ZapperZ
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Gregor said:
I understand your points completely, and of course you and Zapperz are absolutely right.

since I normally approach problems mathematically, I tend to deal with them in a reductive, rather than analytical way, being satisfied with correct solutions, with as few operations as possible.

my friend and I were making a thought experiment..

what would happen to g if the meter and/or the second were redefined.
so we got into an argument over the correct algorithm for deriving the new value for g.

In physics you're not allowed to toy with base units,
but in maths - everything is open game. :rolleyes:

I disagree. Try doing this in the mathematics forum that we have, and you'll find a few mathematicians objecting to it.

[example: 1 = 1 = sqrt(1) = sqrt(-1*-1) = sqrt(-1)*sqrt(-1) = i*i = -1 - nothing wrong with that?]

There is a difference in doing pure number theory and manipulation versus something with a physical quantity. If you wish to do the latter, you shouldn't have done it in the PHYSICS section of the forum.

And just so you know, in many areas of physics, h=k=c=1 by "definition". But this isn't a new definition. It is for "simplicity" and people in the field know what is being done so there's no confusion. It is why you see people often quoting the mass of a particle simply in terms of MeV rather than MeV/c^2. We all know how to convert those back to their true units.

Moral of the story: please do mathematics in the mathematics forum, especially if you wish to neglect the physical dimension of the quantities you are manipulating. If you wish to carry the discussion in the physics forums, then all the physics rules apply.

Zz.
 
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  • #19
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agreed

but surely you're not suggesting that SI units are universal constants,
carved in stone. in many cases equations are only approximations.
(such as mc²)

if a unit changes then it's correlation to other units also changes
and this effects the values in dimensional analysis.

in the example...

what happens to the equation a = v/t
with 1 meter redefined as 0.980665 m and 1 second redefined as 0.1 s (?)

does a = v/t still hold true through time in dimensional analysis?

(assuming that tomorrow the BIPM decided to redefine the meter and second)

this is not a problem for the maths forum
it's a physics problem

I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.
 
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  • #20
ZapperZ
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Gregor said:
agreed

but surely you're not suggesting that SI units are universal constants,
carved in stone. in many cases equations are only approximations.
(such as mc²)

It has nothing to do with "units". It has everything to do with dimensions. But even then, you cannot add something with units of meters to something with units of microns. You'll get nonsense.

if a unit changes then it's correlation to other units also changes
and this effects the values in dimensional analysis.

in the example...



does a = v/t still hold true through time in dimensional analysis?

(assuming that tomorrow the BIPM decided to redefine the meter and second)

this is not a problem for the maths forum
it's a physics problem

I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.

a = v/t is the DEFINITION when a is a constant. It is how it is DEFINED regardless of the units involved. The DIMENSION remains no matter what.

And note that you were the one who said that you were more interested in the manipulation of the mathematics. If you notice, you did something that was physically nonsensical, but you justified it simply by the fact that you're not doing anything wrong numerically. That is what prompted me to request that if you wish only to care about the numerical manipulation of numbers with utter disregard to the physical significance of that number, then the physics forum is NOT the place to do it. I'm only going by YOUR intentions.

Zz.
 
  • #21
Doc Al
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Gregor said:
but surely you're not suggesting that SI units are universal constants, carved in stone.
Huh? Who suggested any such thing? It's dimensions that count, not particular unit systems.


in many cases equations are only approximations.
(such as mc²)
Any equation, to make any physical sense at all, must have consistent units and dimensions.

if a unit changes then it's correlation to other units also changes
and this effects the values in dimensional analysis.
Dimensional analysis deals with dimensions, not particular systems of units.

so what happens to the equation a = v/t
with 1 meter redefined as 0.980665 m and 1 second redefined as 0.1 s (?)
All you've done is redefined your units: 1 "wacky meter" = 0.980665 m and 1 "wacky second" = 0.1 s. So what? Now acceleration has units of "wacky meters" per "wacky second"^2. So what?
this is not a problem for the maths forum
it's a physics problem
Where's the physics?

I would hope this question gets more than a sweeping dismissal, and lengthy explaination as to why such questions are not acceptable in Physics.
Because they are trivial? If done correctly, at least. "Equations" written with wanton disregard for dimensional consistency have no physical content and are just numerology.
 
  • #22
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It has nothing to do with "units". It has everything to do with dimensions. But even then, you cannot add something with units of meters to something with units of microns. You'll get nonsense.

the calculations were only done in meters

I simply converted the result to µm

and I did get the correct result (at least) :smile:


Huh? Who suggested any such thing?

sorry, it was the voices again :rolleyes: (what's that you say lucifer??) :devil:

All you've done is redefined your units: 1 "wacky meter" = 0.980665 m and 1 "wacky second" = 0.1 s. So what? Now acceleration has units of "wacky meters" per "wacky second"^2. So what?

so now 1 g = 1 wackymeter/wackysecond²
rather than 9.80665 m/s²

neat-o huh!

ok, don't hurt me :uhh:

Where's the physics?

it got stollen by uh... other kids, or something

Because they are trivial? If done correctly, at least. "Equations" written with wanton disregard for dimensional consistency have no physical content and are just numerology.

Oh Yea? well my my Dad has a PhD in Numerology and my Dad can beat up your Dad...
so nnnnn!!

and you know what? I'm not going to ride my bike past your Synchotron anymore! :cry:
 
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  • #23
ZapperZ
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Gregor said:
the calculations were only done in meters

I simply converted the result to µm

and I did get the correct result (at least) :smile:

What "correct" results? Show me the "calculations".

And note, I said, if you read carefully, that one cannot ADD (not convert) a length in meters to a length in micrometer and expect a meaningful answer. Do you think 3 +2 = 5 if 3 = 3 meter and 2 = 2 microns? Puhleeze! But that is what you are trying to do, and even worse, you multiply things just because they all result in "1".

Zz.
 
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  • #24
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I do enjoy that you are trying to think outside the box, Gregor, but this is turning to nonsense quite quickly and we are turning in circles.
 
  • #25
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What "correct" results? Show me the "calculations".

as you wish...

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


do you see any units here other than meters and seconds?


naturally 0.00000980665 m = 9.80665 µm and 0.001 s = 1 ms

but these were converted after the operation

the point is that the answer is correct

a at 1 g = 9.80665 µm / ms / ms

and I derived it with the calculation above, which works for any time interval

shall we try 0.5 s ?

sure!

9.80665 m x 0.5 s x 0.5 s = 2.4516625 m

so a = 2.4516625 m / 0.5 s / 0.5 s

feel free to tell me this result is incorrect
 
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