Can Absolute Velocity be Measured?

[pervect]

Yes I am somewhat familiar with the equations of motion and their application in this case, that was never a point of contention ,just the interpretation of the results of that math.
Correct me if I am in error but:

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude..

I haven't been following the rest of this thread, but since you asked for corrections...

I don't see why you say this. Accelerating particles have velocities. There's nothing wrong or ill-defined about an instantaneous velocity, just as there is nothing wrong with taking the derivative of a function, or finding a vector that's tangent to a curve at a point. In fact, all of these ideas are closely related.

 

[Austin0]

Yes I am somewhat familiar with the equations of motion and their application in this case, that was never a point of contention ,just the interpretation of the results of that math.
Correct me if I am in error but:

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude..

I haven't been following the rest of this thread, but since you asked for corrections...

I don't see why you say this. Accelerating particles have velocities. There's nothing wrong or ill-defined about an instantaneous velocity, just as there is nothing wrong with taking the derivative of a function, or finding a vector that's tangent to a curve at a point. In fact, all of these ideas are closely related.

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity
is only well defined or exact for an instant (zero duration) of time. So you can calculate an accurate momentum with this value which will be exact if there happens to be a collision at that same instant but it is a transitory value.
That there is a difference between velocity and instantaneous velocity.
You may disagree and be correct or think that I am splitting hairs [or infinitesimals ;-)) but if you look at the question at hand that is what it comes down to.

Thanks for your input. I don't enjoy being mistaken but I am always happy to be put straight if that is the case.

 

[GeorgeDishman]

1) Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations

Mathematicians would say they are not approximations.

Are you really sure about this assertion?

The statement is correct, a derivative is not an approximation.

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude.

Strictly speaking, velocity is a derivative and is exact at a particular instant.

2) Acceleration is a constant state of changing motion with a dynamic vector, varying in either; direction, magnitude or both.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivalent to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

3) Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's.
But it is not an actual velocity, which I think is demonstrated if you take the derived value and apply the math for velocity... dx=v x dt for some finite dt.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

This will not produce an exact prediction for the position of the particle. For that you need to use the applicable math for acceleration.

To predict the position, you cannot use simple multiplication in general, you have to integrate the velocity (the exception of course is when the velocity is constant) and if acceleration isn't constant then you need to integrate the jerk, etc..

Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

4) As far as I know that is exactly the definition of velocity.

"yuiop" is correct, velocity is the slope of the tangent to the line, not the ratio of finite deltas.

 

[russ_watters]

I'd take that one step further: only the instantaneous velocity as obtained by a derivative is exact. A velocity taken using change in position over a delta-t is an average over that delta-t and therefore only an approximation if used as a proxy for an instantaneous velocity.

Austin0 has the issue backwards.

And IMO, Xeno's paradox is little more than a silly riddle for non-scientists today even if it was a profound problem 2000 years ago. It isn't a difficult to figure out and doesn't even require much math, much less a discussion of the quantization of space/time.

 

[harrylin]

Welcome to PF!

Without reading your post, I can tell you this: the question posed in the title is misguided in that it assumes that there exists such a thing as absolute velocity. Current theory holds that there is no such thing, so it is meaningless/wrong to ask if it can be measured.

Also without reading that post, and while obviously it cannot be measured according to current theory, clearly your answer is equally misguided: current theory treats questions about things that in theory cannot be measured as belonging to philosophy; modern theories of physics make no statements about the existence or non-existence of such things.
Nevertheless, a modern interpretation of relativity as affected by quantum theory (based on Bell's theorem) does hold such a thing as a plausible model of reality - but that's another topic!

 

[harrylin]

[..] According to all known physics there is simply no such thing as absolute velocity - period; no qualifications or dancing around it. Any speculations otherwise should be pursued in a forum outside physicsforums (e.g. a religious forum).

Actually, such models (as well as block universe, many worlds interpretations etc.) are discussed in the QM forum as well as in this forum; physical models and possible explanations have always belonged to physics even though they are philosophical in the sense that they cannot be directly verified.

 

[harrylin]

Just a silly idea i have, please debunk it since i cannot figure it out: [..]

I have other things to do today and there is an infinite number of paradoxes that one can (and does!) invent (and some of which I did solve) - so, my excuses for not reading it. As I came late to this thread, has the error been pointed out to your satisfaction? and how good is your knowledge of relativity?

 

[Austin0]

Would you agree that many of the tricks of calculus ,like limits , provide exceedingly useful and accurate approximations?

Mathematicians would say they are not approximations.

The statement is correct, a derivative is not an approximation.

Wiki et al.

The derivative of a function at a chosen input value describes the best linear approximation of the function near that input value.

so that "f(x) becomes arbitrarily close to L" means that f(x) eventually lies in the interval (L - ε, L + ε),

\lim_{x \to c}f(x) = L
the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.
means that f(x) can be made to be as close to L as desired by making x sufficiently close to c.

Are these not correct?

SO I guess the first question is the meanings of exact and approximate.

ap·prox·i·ma·tion
n.
2. Mathematics An inexact result adequate for a given purpose.


approximation [əˌprɒksɪˈmeɪʃən]
n 4. (Mathematics) Maths
a. an estimate of the value of some quantity to a desired degree of accuracy
approximate
adj [əˈprɒksɪmɪt]
1. almost accurate or exact

Considering a particle under constant acceleration : to the limit \Deltat ---> \infty would you say the derived value was exactly c or approximately c?

1) I would say that strictly speaking the term velocity and its math only applies to inertial particles. With a constant motion , unchanging vector in both direction and magnitude.

Strictly speaking, velocity is a derivative and is exact at a particular instant

.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout.
I was comparing that with picking a durationless point on an accelerating path and saying it was motionless (not changing velocity) at that point.

This will not produce an exact prediction for the position of the particle. For that you need the applicable math for acceleration.

To predict the position, you cannot use simple multiplication in general, you have to integrate the velocity (the exception of course is when the velocity is constant) and if acceleration isn't constant then you need to integrate the jerk, etc .

If you will read what I wrote I think you will find that what you are saying here is exactly what I said. That I was comparing the case of constant Velocity (inertial motion) with Instantaneous Velocity (accelerated motion)
In the first case you simply apply the value dx/dt to predict position,, in the second case you could not and needed to apply other maths,integration or in this case the equation for constant acceleration.

3) Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's. But it is not an actual velocity which I think is demonstrated if you take the derived value and apply the math for velocity dx = v(dt) for some finite dt.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

There seems to be a miscommunication. You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process.The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt

Am I communicating this distinction I am talking about clearly??

Isn't the fundamental definition of velocity; rate of change of position as a function of time??
The difference in position with respect to the difference in time?

Isn't this validly expressed using the fundamental definition of delta as v=dx/dt
In the case of constant linnear motion , this function v=dx/dt. produces a value (dx/dt) which completely and exactly describes the motion of the particle over time without exception. correct??
Would you agree it is strictly applicable without restriction or qualification to a inertial particle??

Calculus does not require velocity to be defined in terms of a finite distance over a finite non zero time interval.

4) As far as I know that is exactly the definition of velocity. You are using the value v from the equation for acceleration and presenting it as a general definition. But it is only an actual velocity if it is final. If acceleration ceases at that point. Otherwise it is an instantaneous velocity and has a different meaning

"yuiop" is correct, velocity is the slope of the tangent to the line, not the ratio of finite deltas.

As you can see I was referring to the meaning and definition of the derived value

If the velocity is the slope of the tangent, then what is that but a simple ratio?
Expressed as a ratio of finite deltas? Eg. 2mm/sec

The tangent as a line, is indefinitely extended in space and time.
Isn't it true that it only exactly describes the motion of a world line or segment of a world line that is congruent or parallel ??
The instantaneous co-moving inertial frame , yes?.
That it is only congruent with the accelerated world line at a mathematical point. I.e. dimensionless in space and time.

Obviously this value (dx/dt) is incredibly useful as an input value for a variable in other equations but what is it's kinematic meaning as describing or predicting the motion of this accelerating particle in the real world other than an approximation??

I hope you now understand the mix up in the subject of our definitions and would like to know if you still think I was incorrect on any point.
Thanks

 

[Austin0]

Hi yuiop I don't know if you remember, but this debate actually started a long time ago in another thread.
We didn't conclude it then because it was off topic.
This discussion seems to have run into semantic snarls and I have let it sit for awhile. But I hope that those questions where you feel I am incorrect can be clarified so we might continue the discussion.

Put it another way. If an object had a velocity of 7 m/s and after accelerating it had a velocity of 9 m/s would you agree that at some point its velocity must have been 8 m/s? If your answer to that is yes, then you would also have to agree that if an object had a velocity of +1 m/s and later had a velocity of -1 m/s that its velocity at some point in between must have been zero.

Yes I would agree that it had to pass through a point where the calcuated velocity would be zero for an abstract dimensionless point of time .
Theb question is dos this imply motionlessnss??

Would I be wrong to say that the equation for determining instantaneous acceleration as applied to the apogee would be
a= limit as dt --->0 of dv/dt =0 also ?

Would you say this was a meaningful description of the acceleration of the particle at this point?

Or would it be more correct to say it was undergoing constant (non-zero) acceleration the whole time?


Would you agree that time, motion (change of position) and acceleration (change of velocity) are continua , not series of discrete steps??

 

[Austin0]

I'd take that one step further: only the instantaneous velocity as obtained by a derivative is exact. A velocity taken using change in position over a delta-t is an average over that delta-t and therefore only an approximation if used as a proxy for an instantaneous velocity.

Austin0 has the issue backwards..

A velocity taken using change in position over a delta-t with regard to an inertial particle is quite exact and that is the context that I was talking about.Not only is the value produced exact, but that value itself provides an eact description of the motion of the particle over time.
In contrast to an instantaneous value which when applied to an accelerating particle provides only an approximate description of the motion over a very limited range.Which was my point.

And IMO, Xeno's paradox is little more than a silly riddle for non-scientists today even if it was a profound problem 2000 years ago. It isn't a difficult to figure out and doesn't even require much math, much less a discussion of the quantization of space/time.

I essentialy agree with you. I only directly mentioned it in this discussion in one two word sentence;
"Zeno's Arrow"
I entered this not as an argument but rather as an illustration of a fallacy (IMO) which I assumed was generally known and understood. In this I may have been in error ..
Thanks for your response

 

[Dale]

Wiki et al.

The derivative is exact. The Wiki quote is referring to a Taylor series expansion. The series expansion uses a derivative to approximate a function. The series expansion is an approximation, not the derivative.

 

[Austin0]

The derivative is exact. The Wiki quote is referring to a Taylor series expansion. The series expansion uses a derivative to approximate a function. The series expansion is an approximation, not the derivative.

All the sources I found seemed to agree , not just Wiki. If a function is taken to a limit isn't it a derivative? If the result is within a range (L+infinitesimal,L-infinitesimal) isn't this an exceedingly close approximation?

SO does the Taylor expansion use the same symbolic representation?

So maybe I should ask what you mean by exact?

Considering a particle under constant acceleration : to the limit Δt --->\infty

v= a(dt) = ? ...would you say the derived value v was exactly c or approximately c?

You didn't answer this question.

Thanks

 

[yuiop]

Yes I would agree that it had to pass through a point where the calcuated velocity would be zero for an abstract dimensionless point of time.

I would agree that it was an abstract notion.

Theb question is dos this imply motionlessnss??

Similarly there are other points where the velocity is non-zero for a similarly abstract dimensionless point of time and I am sure you would use those abstract points to show that there is motion. A lot depends on how "motionlessness" is defined. If we define it as a change in location over a finite non zero time interval, the zero result for dr/dt at the apogee does not prove by itself, that the particle is "motionless at the apogee. However, since you made it clear that you prefer to work with non abstract intervals of non zero time intervals, then it easy to show that the average velocity of the particle averaged from just before the apogee to just after the apogee, is zero. You might then claim that is only an average, but its instantaneous velocity at any instant is non zero during that averaging period and we would be back to discussing abstract dimensionless points of time again. (No win situation )

Would I be wrong to say that the equation for determining instantaneous acceleration as applied to the apogee would be
a= limit as dt --->0 of dv/dt =0 also ?

No, we have defined the accleration as constant in this case and so the acceleration remains non zero even at the apogee. dr/dt=0 does not imply dv/dt=0. Again, if dismiss abstract instantaneous points of time (which are very handy mathematically) then we can work in terms of average velocity over a finite time interval that is as small as you like and we can then demonstrate that the average velocity over that tiny time interval is zero and the acceleration is non zero, if we zoom into a tinier (but non zero) time scale.

Would you say this was a meaningful description of the acceleration of the particle at this point?

Or would it be more correct to say it was undergoing constant (non-zero) acceleration the whole time?

The latter.

Would you agree that time, motion (change of position) and acceleration (change of velocity) are continua , not series of discrete steps??

Well I won't comit myself to that one. I have Zenoist sympathies so I am uncommitted on what happens physically at the quantum scale, but my hunch is that any ultimate TOE or theory of quantum gravity will involve discrete quantities rather than continua. Certainly Planck demonstrated that the only way to resolve the "ultraviolet catastrophe" is to assume the energy comes in discrete quantities and Einstein also demonstrated that in his Nobel prize winning analysis of the photoelectric effect and quantum theory by its very nature involves discrete quantities.

 

[Dale]

All the sources I found seemed to agree , not just Wiki.

You misunderstood Wiki, you probably misunderstood the other sources also. We can go through them one by one if you like.

If a function is taken to a limit isn't it a derivative?

Not necessarily, a derivative is a limit, but not all limits are derivatives.

SO does the Taylor expansion use the same symbolic representation?

Usually a Taylor series expansion is written explicitly as an approximation using "big O" notation. E.g.
f(x)=f(0)+f'(0)x+O(x^2)
Where the O term represents an error of the order of x^2 or smaller.

So maybe I should ask what you mean by exact?

I mean
f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact. There is no ≈ nor any O. If the limit exists then the derivative is exactly that value, by definition.

Considering a particle under constant acceleration : to the limit Δt --->\infty

v= a(dt) = ? ...would you say the derived value v was exactly c or approximately c?

You didn't answer this question.

That is not a derivative, it is a limit, but it is equal to c.

 

[and9]

You want to measure absolute velocity. The problem is all of your measurements are determining velocity RELATIVE to another object in motion (CMB, spaceship, etc). What do you consider the frame for which your velocity is absolute?

I like the time you put into the OP but the premise is flawed since you are attempting to disprove special relativity within the framework of special relativity.

 

[GeorgeDishman]

the derivative of the position of a moving object with respect to time is the object's instantaneous velocity.

Are these not correct?

They are correct but I think you already said you accept that:

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity is only well defined or exact for an instant (zero duration) of time.

That is correct, where you go wrong (and it may only be terminology) is when you say:

That there is a difference between velocity and instantaneous velocity.

In physics, the unqualified "velocity" means the instantaneous value while the value over a finite period might be called the "average velocity", thus you might see a statement such as "If a car travels at a velocity of 10k/hr east for 3 hours then north at the same speed for 4 hours, the average velocity is 7.1km/hr on a heading of 37 degrees."

You may disagree and be correct or think that I am splitting hairs [or infinitesimals ;-))

IMHO, the definition in the form of a function is more useful and may remove your reservation:

http://en.wikipedia.org/wiki/Derivative#The_derivative_as_a_function

but if you look at the question at hand that is what it comes down to.

I looked at the original question but can't see the connection, the answer to your question is that the graph of the passing particles' "relativistic mass" has a minimum when they are at rest relative to the craft.

Considering a particle under constant acceleration : to the limit \Deltat ---> \infty would you say the derived value was exactly c or approximately c?

I would say the velocity is asymptotic to the exact value c.

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space.

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity.

Look at your first statement, I've highlighted the key terms. If the particle is "inertial, its velocity is unchanging. If it is "motionless" at anyone instant then it follows that it is "motionless" at all times. By definition, the frame in which a particle is "motionless at all times" is its rest frame. Do you see what I was saying? I think you were trying to make some other point, such as considering an accelerating particle rather than an "inertial particle". You can't equate that to an accelerating particle because it is not inertial so I'm not sure what you were trying to say.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout.

The velocity has an exact value at any instant but that value varies with time.

A 'particle in continuous motion' has an exact value for its location at any instant but that value varies with time (ignoring QM for the moment).

The same applies to acceleration if the jerk is non-zero and so on. I see no difference in those descriptions.

If you will read what I wrote I think you will find that what you are saying here is exactly what I said. That I was comparing the case of constant Velocity (inertial motion) with Instantaneous Velocity (accelerated motion)
In the first case you simply apply the value dx/dt to predict position,, in the second case you could not and needed to apply other maths,integration or in this case the equation for constant acceleration.

You can apply the operation d/dt to either the function x(t) to get the function v(t) or to v(t) to get a(t), there is no difference. If x(t) is proportional to t, v(t) is a constant but it is only a special case of the more general function.

That is not the definition of velocity. v=dx/dt is a derivative, not a ratio. The value of the derivative is not dx divided by dt, it it the single unique limit value which that ratio approaches as dt -> 0.

There seems to be a miscommunication.

Yes.

You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process.

It is the other way round. The definition of the result is the exact value of the limit found by applying the operation known as differentiation and shown as "d/dt". You are confusing that with the approximate value obtained by process of dividing infinitesimals as a means of finding the exact limit.

The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt

It is not a ratio, it is the exact value of the limit towards which the ratio is an asymptotic approximation.

Am I communicating this distinction I am talking about clearly??

I think you did, you just have the role of the limit and the ratio reversed.

I hope you now understand the mix up in the subject of our definitions and would like to know if you still think I was incorrect on any point.
Thanks

I've snipped a bit here as I think it is already covered by the replies above, you should now see why I think you have it the wrong way round.

I was going to dig out another reference from a book by Penrose that goes farther but I can't lay my hands on it at the moment, sorry for the delay.

 

[Austin0]

Only that instantaneous velocity is only well defined or exact for an instant (zero duration) of time..

That is correct, where you go wrong (and it may only be terminology) is when you say:.

That there is a difference between velocity and instantaneous velocity..

In physics, the unqualified "velocity" means the instantaneous value while the value over a finite period might be called the "average velocity", .".

Yes , previous to this discussion I was unaware of the convention to consider velocity synonymous with instantaneous velocity..
While I might think the term more appropriately applied to inertial motion and then average and instantaneous applying to accelerating motion I certainly won't argue with convention.

Believe me I am well aware of the differences and confusion on that point has had nothing to do with the miscommunication.

IMHO, the definition in the form of a function is more useful and may remove your reservation:

http://en.wikipedia.org/wiki/Derivative#The_derivative_as_a_function.

On the samee page right under the fundamental expression of derivative :

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+h)-f(x)}{\Delta h}

was this:
which has the intuitive interpretation (see Figure 1) that the tangent line to f at a gives the best linear approximation to f near a (i.e., for small h).

Considering a particle under constant acceleration : to the limit Δ t ---> \infty would you say the derived value was exactly c or approximately c?.

I would say the velocity is asymptotic to the exact value c..

I agree completely ,which of course was my point. Asymptotic means never reaching the point of exact value. Only exceedingly close = proximate = approximate. Agreed??

To pick ANY durationless point on this continuum and declare there is no motion along the continuum is exactly equivelant to picking a durationless point on the path of an inertial particle and saying it is motionless in space..

I don't think you said what you meant. An inertial particle is motionless in it's rest frame at all times by definition.

There was no reference to the rest frame of the particle.It was an inertial test particle and an accelerating particle relative to one observation frame
If we pick a point of the inertial path and say: this point is an instant and having no duration, there can be no change of location in this instant. So therefore the particle is motionless in space at this point. I.e. Has no velocity..

Look at your first statement, I've highlighted the key terms. If the particle is "inertial, its velocity is unchanging. If it is "motionless" at anyone instant then it follows that it is "motionless" at all times. By definition, the frame in which a particle is "motionless at all times" is its rest frame. Do you see what I was saying? I think you were trying to make some other point, such as considering an accelerating particle rather than an "inertial particle". You can't equate that to an accelerating particle because it is not inertial so I'm not sure what you were trying to say..

I was describing a fallacy (IMO) In fact the initial premise of Zeno's arrow. SO just substitute arrow for inertial particle and it might make more sense.
I was not saying it actually was motionless but rather, that to say it was motionless was nonsense.
If you look at the following sentence below it is clear that the arrow was actually in continuous motion, never actually motionless . The point was the correspondence between this case and picking a similar durationless point on the path of an accelerating object and saying it was not accelerating.
In one case you are saying it is not changing position and in the second you are saying it is not changing velocity.

DO you think this is a valid or meaningful logical conclusion or description of reality?

I certainly don't.
It is trivially true but meaningless. An abstraction divorced from reality and falsified by that reality, wherein the particle is in continuous motion throughout..

Is it clear yet?

You are defining velocity in terms of a process for its determination.
I was talking about the definition of the term as applied to the result of that process..

It is the other way round. The definition of the result is the exact value of the limit found by applying the operation known as differentiation and shown as "d/dt". You are confusing that with the approximate value obtained by process of dividing infinitesimals as a means of finding the exact limit..

Isn't this what we are talking about??

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact.

If this is it then all the descriptions I have encountered including the page you referenced above described the process in terms of infinitesimals.

ANother source:

inputThe derivative of a function at a chosen input value describes the best linear approximation of the function near that value.
so that "f(x) becomes arbitrarily close to L" means that f(x) eventually lies in the interval (L - ε, L + ε),

Infinitesimals yes?

The meaning of the actual value thus derived and called velocity.
Which in all cases is in fact a ratio...correct? some dx/dt.

It is not a ratio, it is the exact value of the limit towards which the ratio is an asymptotic approximation..

Once again you are talking about the process and I the value.

We drop a ball. After 2 seconds we can calculate the instantaneous velocity is 20m/sec. This is the value of the velocity. It is a ratio. A dx/dt

dx=20,dt=1s correct?

The meaning of this value is a change of 1 sec in time results in a 20m change in position Right?

How do you propose to explicitely express a velocity value for a real world state of motion in a form which is not a dx/dt ?

Thanks for your help

 

[Austin0]

The derivative is exact.

So maybe I should ask what you mean by exact?

I mean
f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact. There is no ≈ nor any O. If the limit exists then the derivative is exactly that value, by definition.

Well I can't argue with that perfect self referential symmetry.

So:
Given that an instantaneous velocity of an accelerating particle is exact at that point:
That point being a mathematical point , it would follw that it was only exact within a restricted region of zero dimension, both spatially and temporally.
Therefore any finite interval of time must inevitably fall outside that region and be inexact.
Since any explicit value for that velocity would necessarily have a time term of finite duration
it would necessarily be approximate as applied to the motion of the particle in the real world..
SO would you say it was only abstractly exact?

Considering a particle under constant acceleration : to the LIMIT Δ t ---> \infty would you say the derived value was exactly c or approximately c?.

That is not a derivative, it is a limit, but it is equal to c.

So would this be a case where "Calculus says" something that does not have meaning or truth in the physics of the real world??
Entirely not the fault of calculus of course

Thanks

 

[Dale]

which has the intuitive interpretation (see Figure 1) that the tangent line to f at a gives the best linear approximation to f near a (i.e., for small h)

Note that this quote says that the tangent line is an approximation to f. That is not at all the same as saying that the derivative is an approximation. The tangent line is the first order Taylor series expansion, so this confirms my previous point.

Given that an instantaneous velocity of an accelerating particle is exact at that point:
That point being a mathematical point , it would follw that it was only exact within a restricted region of zero dimension, both spatially and temporally.

That doesn't follow. That would follow if the instantaneous velocity were exact only at that point, but that is not the case. The instantaneous velocity is exact at other points also (i.e. plug in a new value for x, take another limit, and you have the exact velocity at another point). If f is smooth (i.e. no teleportation) then the velocity function is defined at all points in the domain of f.

SO would you say it was only abstractly exact?

Although I don't agree with your logic, I do agree with your conclusion, but not for the reason you stated. The reason that I would agree is that in the abstract you can deal with the exact calculated values, but in reality you deal with measured values. Even classically, measured values always include some error, so they are not exact. However, the exact values of abstract theory are accurate predictors of our best inexact measurements, so the use of an instantaneous derivative is experimentally justified.

So would this be a case where "Calculus says" something that does not have meaning or truth in the physics of the real world??

No, calculus agrees in this case with the physics of the real world, as long as you understand what is meant by a limit as t->∞. However, I would emphasize again, that this limit is not a derivative and has nothing to do with the instantaneous velocity discussion.

 

[russ_watters]

Note that later in that same wiki, it does indeed affirm that the derivative at a point is the exact slope.

 

[GeorgeDishman]

Yes , previous to this discussion I was unaware of the convention to consider velocity synonymous with instantaneous velocity.

I suspect that clears it all up but I'll reply to the key points you raise.

On the samee page right under the fundamental expression of derivative :

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+h)-f(x)}{\Delta h}

I would say the velocity is asymptotic to the exact value c.

I agree completely ,which of course was my point. Asymptotic means never reaching the point of exact value. Only exceedingly close = proximate = approximate. Agreed??

Right, so the process of approaching the limit is always approximate while the limit itself is exact. Now look at the definition of f' above. The process involves the ratio of small but finite values while the derivative is defined as the limit which, as you note next, is exact.

[I'll skip the bit about Zeno, I think the rest of my reply makes it redundant.]

Isn't this what we are talking about??

f'(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}
This quantity is the definition of the derivative, and it is exact.

Yes, and velocity is defined as a derivative so it is exact, not an approximation.

If this is it then all the descriptions I have encountered including the page you referenced above described the process in terms of infinitesimals.

Yes, one process of finding the derivative is defined in terms of infinitesimals but the derivative is the exact limit.

It is not a ratio, it is the exact value of the limit towards which the ratio is an asymptotic approximation.

Once again you are talking about the process and I the value.

No, see above, the definition of the value is the limit, the process involves the ratio.

We drop a ball. After 2 seconds we can calculate the instantaneous velocity is 20m/sec. This is the value of the velocity. It is a ratio. A dx/dt

dx=20,dt=1s correct?

No, at the exact instant identified by t=1s, the rate at which the ball is falling is the exact value 20m/s.

The distance it has fallen is 10m so the ratio is the average speed over that 1s period and is only 10m/s.

The meaning of this value is a change of 1 sec in time results in a 20m change in position Right?

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics.

How do you propose to explicitely express a velocity value for a real world state of motion in a form which is not a dx/dt ?

"dx/dt" is a notation that indicates use of the derivative applied to the variable 'x'. The derivative is the limit at a particular instant as you have quoted several times above and it is exact, not an approximation. The ratio of infinitesimals is an approximation but that is only a process for finding the limit. The definitions you quote above make that clear so I'm unclear as to where you think there is a problem.

 

[Austin0]

I would really like to try and clear up some of the confusion that has developed in this discussion.

post #32

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity
is only well defined or exact for an instant (zero duration) of time.

Back at post 32 I clearly expressed that the question was not about calculus per se, and stated that an instantaneous velocity was exact.

Given that an instantaneous velocity of an accelerating particle is exact at that point:
That point being a mathematical point , it would follw that it was only exact within a restricted region of zero dimension, both spatially and temporally.
Therefore any finite interval of time must inevitably fall outside that region and be inexact.
Since any explicit value for that velocity would necessarily have a time term of finite duration
it would necessarily be approximate as applied to the motion of the particle in the real world..

That doesn't follow. That would follow if the instantaneous velocity were exact only at that point, but that is not the case. The instantaneous velocity is exact at other points also (i.e. plug in a new value for x, take another limit, and you have the exact velocity at another point). If f is smooth (i.e. no teleportation) then the velocity function is defined at all points in the domain of f.

Here you have questioned my logic. Fair enough , it may be flawed.
So let's examine it:
Simply parsing the engish it seems clear that I am referring to an (singular) instantaneous velocity at that (singular , specific) point. I make no statement explicit or implied about the velocity function itself or values derived from it at other points and am clearly referring to a singular specific quantitative value derived from that function.

Now looking at your response:

if the (general) instantaneous velocity were exact only at that point

My statements have been interpreted to become;

Only the instantaneous velocity derived at a single point on the path is exact.
The velocity function is only exact at a single point on the path of an accelerating particle.

1) Since these statements are so obviously boneheaded it was not difficult to refute them.
But as there was no foundation whatever for these interpretations in my actual words. that refutation is itself, a form of invalid argument we are all familiar with.

2) This argument also implicitely suggests that I might be such a bonehead, which may well be true but is not sufficiently proven in this instance.

3) I think if you examine this closely you will find that what you said here

[" The instantaneous velocity is exact at other points also (i.e. plug in a new value for x, take another limit, and you have the exact velocity at another point)" ]

actually supports my point.

That being;
if you want an exact value for a point outside the stated dimensionless point, it is necessary to input a different value of x or t into the function and derive a new and different quantitative value.

Would you disagree with this?

Although I don't agree with your logic, I do agree with your conclusion, but not for the reason you stated. The reason that I would agree is that in the abstract you can deal with the exact calculated values, but in reality you deal with measured values. Even classically, measured values always include some error, so they are not exact. However, the exact values of abstract theory are accurate predictors of our best inexact measurements, so the use of an instantaneous derivative is experimentally justified.

Obviously this value (dx/dt) is incredibly useful as an input value for a variable in other equations

Yes we can calculate an instantaneous velocity and this is a useful abstraction; eg. ICMIF's

The usefulness and validity has never been in question.
But I do have a question. If there are a pair of real world measurements (time-location)
would the derivative of this value differ from the average velocity directly indicated from the measurement?

No, calculus agrees in this case with the physics of the real world, as long as you understand what is meant by a limit as t->8. However, I would emphasize again, that this limit is not a derivative and has nothing to do with the instantaneous velocity discussion.

Are you changing your mind about the value being c?
I am unclear as to why it is not a derivative?
Is it because the function is a constant , a simple derivative??

I think it is germane to the underlying question.
The relationship of abstract values to the reality they describe.
Calculus is a fantastic tool but is still simply a complex set of algorithms. It produces values but does not interpret them. That requires intelligence and understanding.
As this exsample ; taken literally the value c does not describe the motion or the predicted outcome of real world measurements or conform to the physics of that world as we now understand it. That value must be interpreted as you said.
All I am saying is that an explicit velocity of an accelerating particle , no matter how useful or exact it may be abstractly , as a desription of the motion in the real world it cannot simply be taken literally but must be interpreted.

I hope all this may clear up some of the confusion.
thanks for your input. I know semantics is not our favorite subject either.

 

[Mentz114]

I am unclear as to why it is not a derivative

A derivative is a specific limit. The derivative of x² can be found by working out
<br /> \frac{(x+\delta x)^2-x^2}{\delta x} = \frac{x^2 + 2x\delta x + (\delta x)^2 - x^2}{\delta x}=2 x + \delta x<br />
and if we let δx = 0 the answer is exactly 2x. You might find it amusing to work out the derivative of xⁿ with this method.

In general a limit is just the value of any function when one of its arguments goes to some other value.

 

[Austin0]

A derivative is a specific limit. The derivative of x² can be found by working out
<br /> \frac{(x+\delta x)^2-x^2}{\delta x} = \frac{x^2 + 2x\delta x + (\delta x)^2 - x^2}{\delta x}=2 x + \delta x<br />
and if we let δx = 0 the answer is exactly 2x. You might find it amusing to work out the derivative of xⁿ with this method.

In general a limit is just the value of any function when one of its arguments goes to some other value.

Hi Mentz just looking at the 3rd power the handwriting on the wall seems to indicate nx^n-1 ?

I am still not sure why the equation for constant acceleration taken to the limit \Deltat ---->\infty is not a derivative? because of the infinity?

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??
In a case with actual inputs this represents the quantitative output value??

Thanks

thinking about it more, would the explicit answer in terms of x be: I don't know the notation but something like the series x(n-k)(n-k)... where k = the incrementing integers from 0 ,1,2...(n-1) ?

 

[Mentz114]

Hi Austin, you're right about nx^n-1.

I am still not sure why the equation for constant acceleration taken to the limit Δt ---->∞ is not a derivative? because of the infinity

I'm don't know which formula you're referring to, but a limit as Δt -> ∞ cannot be a derivative, as you conjecture. In the definition of the derivative the δ term is an infinitesimal which is taken to zero.

In the velocity equation
...
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??

uh, yes, that's the definition of dx/dt

In a case with actual inputs this represents the quantitative output value??

Not sure what you mean. An example. Suppose x(t) = x₀ + ht, where h is a constant, then the velocity is

dx(t)/dt = h.

I have to say that this is probably not the place for coaching in calculus. I was hoping you could see how straightforward it is and get a textbook ...

 

[GeorgeDishman]

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
am I correct in thinking the final dx/dt to the right of the equals sign is the actual result??
In a case with actual inputs this represents the quantitative output value??

uh, yes, that's the definition of dx/dt

I may be over-analysing but it may avoid some confusion if I check something here.

Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is

v=\lim_{\Delta t\to 0} \, \ ...

The notation "d/dt" is a way of writing that limit so as Menzt114 says, that is the definition of dx/dt.

 

[Dale]

Sorry I have been delayed. I was traveling for business very heavily. Some of your questions may have already been resolved, so I apologize for any redundancy.

Here you have questioned my logic. Fair enough , it may be flawed.
So let's examine it:
Simply parsing the engish it seems clear that I am referring to an (singular) instantaneous velocity at that (singular , specific) point. I make no statement explicit or implied about the velocity function itself or values derived from it at other points and am clearly referring to a singular specific quantitative value derived from that function.

Now looking at your response:

if the (general) instantaneous velocity were exact only at that point

My statements have been interpreted to become;

Only the instantaneous velocity derived at a single point on the path is exact.
The velocity function is only exact at a single point on the path of an accelerating particle.

Yes, upon reading your statements that is exactly how I had understood them. Even now, I have a hard time interpreting them any other way. But I am glad you have clarified.

1) Since these statements are so obviously boneheaded it was not difficult to refute them.
But as there was no foundation whatever for these interpretations in my actual words. that refutation is itself, a form of invalid argument we are all familiar with.

2) This argument also implicitely suggests that I might be such a bonehead, which may well be true but is not sufficiently proven in this instance.

Austin0, I never said nor implied anything about you being a bonehead. This site is for people who are trying to learn and people trying to learn make honest mistakes. It isn't boneheaded for someone to make an incorrect conclusion about a brand new concept and I understand that. So please don't get offended and imagine unintended insults.

My whole involvement in this thread was to correct the misunderstanding that the velocity is an approximation i.e. inexact. Your statements indicated to me that you still felt that it was inexace because it was a derivative. So I explained. That is all, no insult was provided nor implied.

3) I think if you examine this closely you will find that what you said here

[" The instantaneous velocity is exact at other points also (i.e. plug in a new value for x, take another limit, and you have the exact velocity at another point)" ]

actually supports my point.

That being;
if you want an exact value for a point outside the stated dimensionless point, it is necessary to input a different value of x or t into the function and derive a new and different quantitative value.

Would you disagree with this?

I agree. Mathematically f'(x)≠f'(x+Δx) in general, for Δx≠0. But both f'(x) and f'(x+Δx) are exact if f is smooth.

But I do have a question. If there are a pair of real world measurements (time-location)
would the derivative of this value differ from the average velocity directly indicated from the measurement?

In general the velocity does not equal the average velocity.

Are you changing your mind about the value being c?

No. The limit is exactly c.

I am unclear as to why it is not a derivative?
Is it because the function is a constant , a simple derivative??

No, it is not a derivative because it doesn't fit the form of a derivative. Recall the definition of a derivative:

f&#039;(x)=\lim_{\Delta x\to 0} \, \frac{f(\Delta x+x)-f(x)}{\Delta x}

Now, look at the limit you are trying to evaluate:

\lim_{t \to \infty}\,v(t)

There is no Δt which is approaching 0, so the limit cannot be expressed in the form of the above limit. Therefore it is not a derivative. Now, v(t) itself is a derivative v(t)=x'(t) so we can write the limit as

\lim_{t \to \infty}\, \left( \lim_{\Delta t\to 0} \, \frac{x(\Delta t+t)-x(t)}{\Delta t} \right)

This makes it clear that the overall expression is not a derivative, but a limit of a derivative. One of the key things that makes it not a derivative is the fact that the limit is taken to infinity. In a derivative the limit is always taken to 0.

This is important, because a limit to infinity is fundamentally a different thing than a limit to some finite value. A limit to some finite value means "as I get arbitrarily close to X ..." but you cannot get arbitrarily close to infinity. For instance, the real number .0001 is within .001 of 0. Can you give me a real number which is within .001 of infinity?

So a limit to infinity is a fundamentally different thing, it means what does f(x) tend to asymptotically as x increases without bound. In this case, that number is c. So the limit is c meaning that as t increases without bound v asymptotically approaches c.

I think it is germane to the underlying question.
The relationship of abstract values to the reality they describe.
Calculus is a fantastic tool but is still simply a complex set of algorithms. It produces values but does not interpret them. That requires intelligence and understanding.

I agree. I think you have the required intelligence, and hopefully you now have the required understanding also.

 

[Austin0]

Hi GeorgeDishman Thank you for your patience.There has been much confussion in this discussion and I have no doubt a certain frustration on both sides.

In the velocity equation v=\lim_{\Delta t\to 0} \, \frac{ x(t+\Delta t )- x(t)}{\Delta t}=dx/dt
the final dx/dt is the result?

I may be over-analysing but it may avoid some confusion if I check something here.

Austin0, if you are asking if the final "dx/dt" represents the division of a value dx by another value dt, the answer is no. The bit that gives you the value of v is


[\QUOTE]

I thought the final dx/dt represented the answer .A ratio like 20m/s not an operation of division to obtain the answer. Is this incorrect?

"dx/dt" is a notation that indicates use of the derivative applied to the variable 'x'. The derivative is the limit at a particular instant as you have quoted several times above and it is exact, not an approximation. .

There seems to be a question regarding the meaning and use of delta.
The term delta itself seems to have another intrinsic meaning , that being; interval.

Wrt dynamic systems it is found by subtracting one value of some variable at a point from the value at some other point to get a difference , an interval ..Correct?

Now delta may have a more specific meaning as an operator in the context of derivation but it is certainy widely used outside that context as meaning interval with no implied reference to the operation of derivation,limit etc.

Example

I'd take that one step further: only the instantaneous velocity as obtained by a derivative is exact. A velocity taken using change in position over a delta-t is an average over that delta-t and therefore only an approximation if used as a proxy for an instantaneous velocity.

DO you agree that this is in common usage??
Is there some distinction between small d delta and capitol D delta you are referring to?


Isn't the fundamental definition of velocity; rate of change of position as a function of time??
The difference in position with respect to the difference in time?

Are you saying this definition is invalid or has no relevance to a velocity as a description of motion in the world.??

Isn't this essential meaning validly expressed with symbols using the fundamental definition of delta as v=dx/dt ?

So any explicit velocity value , independent of the method of arriving at it , is expressed as a ratio of intervals of change eg. 20m/s

This value has an explicit literal meaning under the fundamental definition of velocity which is: the ratio of a specific interval/change of position with respect to a specific interval/change of time. Eg. 20m/s

I am not talking about the interpretation here but the literal meaning.
Would you agree?

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics. [\QUOTE]
I was of , course, aware of this as it is the point of this discussion but:

Here you are talking about interpretation which is fine.. So what is your interpretation of this value as a description of the real world motion of the particle??

Thanks

 

[Austin0]

Yes, upon reading your statements that is exactly how I had understood them. Even now, I have a hard time interpreting them any other way. But I am glad you have clarified..

Hi . Why is it hard to interpret them as written? Is my english not clear?? Actually this very thing seems to have repeatedly occurred throughout this discussion, not just with you but generally.
Everything i have said has been reinterpreted into a statement or argument directed to the proposition that a derivative or instantaneous velocity it not exact. This is in spite of the fact that right from the beginning and repeatedly after, I have clearly stated this was not the question , that it was given that a velocity was exact at an instant.
The result is that my communication has not been understood and just as in this case , my arguments and questions have not ever been addressed.

Austin0, I never said nor implied anything about you being a bonehead. It isn't boneheaded for someone to make an incorrect conclusion about a brand new concept and I understand that. So please don't get offended and imagine unintended insults..

I thought that the inclusion of a cartoon face would make it clear enough I was not serious.
But on the other hand your response here almost has me wondering if it was I and not you that made the incorrect conclusion about a brand new concept ;-)
But seriously, you have been nothing but helpful as long as I have been in the forum and I consider you ,with DrGreg, as being among the most considered and considerate members, so I never assumed any negative intent on your part whatever.

post 32

Hi I never meant to suggest there was anything wrong with instantaneous velocity, or derivatives or integration, I am well aware of their power. Only that instantaneous velocity is only well defined or exact for an instant (zero duration) of time...

Your statements indicated to me that you still felt that it was inexace because it was a derivative. ..

The post above is #32 we are now up to #57 and it is apparently still unclear that I never felt or said the value of an instantaneous velocity was inexact , I have been trying to talk about the meaning and interpretation of the derived value itself, independent of its method of determination. What that value meant as a description of the motion of a particle in the real world. But somehow this is not possible and it always, circularly, ends up back focused on derivatives.

Obviously this value (dx/dt) is incredibly useful as an input value for a variable in other equations but what is it's kinematic meaning as describing or predicting the motion of this accelerating particle in the real world other than an approximation??..

SO it is in this sense that I feel that no matter how exact a velocity may be abstractly , as a description of the motion in the real world it is an approximation confined to a small region of application.

This is also a rational interpretation of the statement ;" the tangent at a point is a linear approximation of the values of the curve near to that point of exact value."
As the curve represents the motion in the real world this means within a short distance on the actual path.

If you disagree with this interpretation of the tangent could you explain your alternative?

However, the exact values of abstract theory are accurate predictors of our best inexact measurements, so the use of an instantaneous derivative is experimentally justified.

.

But I do have a question. If there are a pair of real world measurements (time-location) would the derivative of this value differ from the average velocity directly indicated from the measurement?

In general the velocity does not equal the average velocity.

Since you didn't answer the question I will take a stab at it.
Given two measurements, the accuracy of the derived value is inversely dependent on the spatial separation between the two events. Increasing as the interval decreases. In this case derivation is no more accurate at interpolation than the simple dx/dt
and would return the same value . Both are approximations in this circumstance. Is this correct??
This is irrelevant to the initial question but there is another question.
Given an exact value at a point (ideally close measurements) what would you say about this value as a predictor of measurements of position or time at other points?

So what is your interpretation of an instantaneous velocity as a description of the motion of an accelerating particle.?

Thanks for your patience and help
.[/QUOTE]

 

[GeorgeDishman]

Hi GeorgeDishman Thank you for your patience.There has been much confussion in this discussion and I have no doubt a certain frustration on both sides.

No problem, sometimes it takes a few mails to get to the core of a problem.

I thought the final dx/dt represented the answer .A ratio like 20m/s not an operation of division to obtain the answer. Is this incorrect?

Yes, that is incorrect.

There seems to be a question regarding the meaning and use of delta.

The problem is that you are reading "dx/dt" as the ratio of two deltas. Instead you need to read it as an operation "d/dt" applied to a variable "x" which is a function of "t".

The definition of that operation is the derivative.

The term delta itself seems to have another intrinsic meaning, that being; interval.

There is a dual meaning, a "delta" is also used to mean a small finite step but we are not interested in that meaning here, that is a large part of the confusion.

Isn't the fundamental definition of velocity; rate of change of position as a function of time??

Yes, exactly. In your previous example some posts back, you had a velocity of 20m/s which was the rate of change of position at a specific instant.

The difference in position with respect to the difference in time?

No, in your example the position had only changed by 10m in 1s.

Are you saying this definition is invalid or has no relevance to a velocity as a description of motion in the world.??

We are all telling you that the first definition is correct for "velocity" while the second is "average velocity" and they are different things. Both are valid as you can see from the example but they are not the same thing (the first had the value 20m/s while the second was 10m/s).

Isn't this essential meaning validly expressed with symbols using the fundamental definition of delta as v=dx/dt ?

The fundamental definition of velocity is "rate of change of position" which is a derivative, not a ratio.

So any explicit velocity value , independent of the method of arriving at it , is expressed as a ratio of intervals of change eg. 20m/s

No, it is explicity the derivative.

This value has an explicit literal meaning under the fundamental definition of velocity which is: the ratio of a specific interval/change of position with respect to a specific interval/change of time. Eg. 20m/s

The change of position in 1s was 10m, the velocity after 1s was 20m/s downwards.

I am not talking about the interpretation here but the literal meaning.
Would you agree?

I already answered that:

No, the meaning is that, at that instant, the rate at which the distance is changing is a slope of 20m/s but since the acceleration is constant, it has that value only at that instant. That is what is called velocity in physics.

I was of , course, aware of this as it is the point of this discussion but:

So what is your interpretation of this value as a description of the real world motion of the particle??

The interpretation is that velocity is the instantaneous value of the rate at which the particle's position is changing, the velocity vector is the derivative of the position vector.