I When does the instantaneous velocity exist?

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Instantaneous velocity is defined as the derivative of the motion function f(t) and differs from average velocity over a time interval. When a force is applied to an object at rest, it requires time to change its motion status, and during this transition, the instantaneous velocity can be considered constant within a small interval. The discussion highlights the mathematical modeling of motion, suggesting that while trajectories can be approximated by short straight segments, a smoothly curving trajectory also exists. The Euler method is mentioned as a basic numerical approach to simulate motion, but more sophisticated methods can reduce bias. Ultimately, the continuous mathematical functions used in physics are assumed to accurately reflect physical reality, despite the theoretical possibility of an underlying discrete model.
Clockclocle
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The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of object even though we exert enough force it won't move if we do not stay exert on it enough time then the object must have a small interval of time to change their status of motion. In reality we can't make things change "instantaneous", so I guess in that small interval of time when the object change their status of motion the velocity of the object is the same on all the interval and equal to f'(a)
 
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Clockclocle said:
In reality we can't make things change "instantaneous", so I guess in that small interval of time when the object change their status of motion the velocity of the object is the same on all the interval and equal to f'(a)
I don't see it that way. At the moment force F is applied to mass m at the origin
  1. the instantaneous position of the object is zero
  2. the instantaneous velocity of the object is zero
  3. the acceleration of the object is a = F/m
This doesn't mean that, at the moment the force is applied, the instantaneous velocity does not exist. In this particular case, the non-zero acceleration guarantees that at any moment after the force is applied the instantaneous velocity will change from zero to some non-zero value in the direction of the acceleration, i.e. the object will be moving.

On Edit: Fixed typo thanks to @berkeman
 
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Clockclocle said:
The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of object even though we exert enough force it won't move if we do not stay exert on it enough time then the object must have a small interval of time to change their status of motion. In reality we can't make things change "instantaneous", so I guess in that small interval of time when the object change their status of motion the velocity of the object is the same on all the interval and equal to f'(a)
This is more of a mathematics question than a physics question. Or, perhaps, more about how we join physics with mathematics in a "model". There is no way to do justice to the subject matter in a brief post.

The model you seem to propose is that trajectories are composed of lots of little straight line segments. That once enough acceleration has built up, the particle suddenly snaps from one straight line to the next. The idea is that if we make the straight line segments short enough, this model is indistinguishable from reality.

I agree. This model works. It is pretty much the basis for Riemann integration or for the epsilon delta definition of a derivative. It is also how we run numerical simulations with the Euler method.

[You should know that the Euler method is about as primitive as it gets. You quickly want to improve it -- perhaps by pretending that the particle gets half of its delta v at the beginning of the interval and the other half at the end, thus eliminating some bias]

In mathematics, we phrase things differently. Instead of talking about "short enough", we get down to the nitty gritty about tolerances (the epsilon) and how short you need an arc to be to attain that tolerance (the delta). As a result, we can talk about the "limit" that is approached to within every tolerance you can choose if you can look as closely as you please.

Often, we can solve differential equations to obtain an exact formula for the motion of a particle under a particular force law. Sometimes we cannot solve the equations. But we can still prove that an exact and continuously differentiable trajectory exists -- one which does not take the form of a bunch of finite straight-line segments.

In summary: Yes, the short straight segment model works. Approximately. But a smoothly curving mathematically perfect trajectory also exists, even if we cannot always write down a formula for it or measure exactly enough to distinguish between the approximation and the mathematical ideal. No, we do not believe that particles actually follow short straight-line trajectories, even though experiment is silent on the question.
 
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Approaching it intuitively, say a ball starts at rest and is accelerated to 100 m/s. It starts at 0 m/s and ends at 100 m/s. And in between is at values between them.

The math works as a continuous function with the derivative indistinguishable from a small interval approximation. But, you are asking if there is an underlying stair-step function, or a continuous function IN THE PHYSICAL REALITY. The easiest assumption is that the mathematical continuous function reflects the physical reality. And experimentally, that is what you find. But the limits of experimental measurement leave it possible for an underlying stair-step function to still exist.

I'm in favor of assuming the classical physics interpretation that the continuous functions and calculus we use are a perfect description of the underlying reality.
 
"Officer, I can't have been driving 90 miles per hour. I've only been driving 20 minutes!"

(see the similarity?)
 
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Thread 'Question about pressure of a liquid'
I am looking at pressure in liquids and I am testing my idea. The vertical tube is 100m, the contraption is filled with water. The vertical tube is very thin(maybe 1mm^2 cross section). The area of the base is ~100m^2. Will he top half be launched in the air if suddenly it cracked?- assuming its light enough. I want to test my idea that if I had a thin long ruber tube that I lifted up, then the pressure at "red lines" will be high and that the $force = pressure * area$ would be massive...
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