# When does the instantaneous velocity exist?

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• Clockclocle
In summary, the instantaneous velocity at time a is defined as the derivative of the motion function f(t) and is not similar to average velocity in an interval of time. According to Newton's law, if an object is at rest, we must exert a force to make it move, but even if we exert enough force, it won't move unless we maintain the force for a long enough time. In reality, we cannot make things change instantaneously, so there must be a small interval of time during which the object changes its status of motion. However, this does not mean that the object follows a series of short straight-line segments, as a smoothly curving trajectory also exists and is mathematically perfect, even if it cannot always
Clockclocle
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)

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

Last edited:
berkeman and Lnewqban
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.

nasu and Lnewqban
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.

jbriggs444
"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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