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This question already has an answer here:

So in order for two things $A$ and $B$ to move apart, for example, relative to each other, $B$ can be set into motion away from $A$. This means that we have to increase $B$'s velocity and therefore the acceleration has to be positive. If the acceleration has to be positive and $A$ and $B$ were stationary before, that means that $B$'s acceleration has to increase from zero and therefore the third derivative of motion with respect to time has to be positive. The third derivative has to increase from zero and etc. etc. Is this just another way to state Zeno's paradox (and thus a dumb question) or does motion really involve an increase of the magnitude of infinitely many derivatives of velocity?

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marked as duplicate by ACuriousMind, Kyle Kanos, user10851, John Rennie, Danu Mar 28 '15 at 15:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Possible duplicate: physics.stackexchange.com/q/111251/2451 $\endgroup$ – Qmechanic Mar 25 '15 at 7:34
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    $\begingroup$ Zeno's paradox is not a dumb question. The ancient world had some brilliant mathematicians, but they did not understand infinity. This and a few other issues held back mathematics for more than 2000 years. Numbers generated intractable paradoxes. Geometry could be made rigorous. So math was geometry. Archimedes essentially discovered calculus, but did not consider his work a proof. In modern times, Newton invented calculus but it took another century to make the math rigorous. $\endgroup$ – mmesser314 Mar 25 '15 at 12:58
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    $\begingroup$ @mmesser314: I respectfully disagree. Zeno's paradox is most definitely a dumb question. The ancient Greeks may not have had a proper concept of infinity or of zero, but they did have atomism, the philosophical concept that at some microscopic level, there is a "resolution" (to borrow a modern computer term) to physical dimensions. It was known from the beginning that atomism provides a handy resolution to Zeno's paradox, but atomism was rejected because it contradicted Aristotle's philosophies, which was essentially math-heresy. (And heresy heresy too, once the Church accepted Aristotle.) $\endgroup$ – Mason Wheeler Mar 25 '15 at 20:43
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    $\begingroup$ @MasonWheeler - I agree with you (and disagree with mmesser314). Zeno's paradox was disproved at the moment it was introduced when smarter Greek philosophers than Zeno stood up and walked out of the room. $\endgroup$ – David Hammen Mar 26 '15 at 0:46
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    $\begingroup$ @DavidHammen - You have it backwards. Zeno walked out of the room too. He did not mathematically prove that reality was impossible. He showed that there was a problem with mathematics. But you are right. The Greeks sometimes took math-heresey way too seriously. Numbers were thought of as the ratio of integers. The man who discovered that no integers yield $\sqrt{2}$ was put to death. $\endgroup$ – mmesser314 Mar 26 '15 at 4:23
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Yes, if an object is stationary, then starts moving in the positive direction, all derivatives of the position can become positive when the object starts moving.

But so what? It's not a paradox. It's just a mathematical statement. All the derivatives are positive. No big deal.

It's hard to tell why you find this confusing. You might be thinking that there should be some cause for each derivative to change. E.g. "What is making the third derivative change? And if you can answer that, fine, but what's making the fourth derivative change?", etc. The answer is that there is no physics in the higher derivatives. Newton's second law $F=ma$ refers to the second derivative. Higher derivatives have no physics in them, they just follow mathematically from whatever the derivatives of the applied force are, so there is no need for an infinite series of physical causes being applied to all the derivatives.

Another possibility is that you've noticed that the motion is not analytic. For analytic functions, knowing all the derivatives at a point means you know the function everywhere. That means that if we want to model an object as being perfectly stationary, then starting to move, its position as a function of time cannot be an analytic function. But again, oh well. It's not analytic. So what?

If any of that is disturbing, it might help to keep in mind that describing the motion of an box that's sitting still on a table and then starts moving is not an attempt at some sort of perfect ultimate description of what's going on. Real boxes are made from some $10^{24}$ molecules all jiggling around and stuff. You can't possibly describe everything perfectly accurately. Saying that the position of the box is some function $x(t)$ is just rough description. As such it's not worth worrying about fine mathematical details such as infinite derivatives of the position; the function $x(t)$ isn't true enough to reality for, say, the 14,515th derivative to mean anything.

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    $\begingroup$ "all derivatives of the position will be positive when the object starts moving"... I don't think this makes sense. For example, if the object has v=0 for t<0 and v=t^5 for t>=0. Then the "object starts moving" at t=0, but clearly v=0 at t=0 as are the first 5 derivatives of v... although the sixth is a delta function... $\endgroup$ – hft Mar 25 '15 at 8:26
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    $\begingroup$ The phrase "when the object starts moving" is a colloquial one, it doesn't have to mean the same thing as t=0. In your example, all the derivatives are positive at some time arbitrarily close to t=0. $\endgroup$ – Mark Eichenlaub Mar 25 '15 at 10:09
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    $\begingroup$ ...except those higher than six, of course, so the statement in my answer assumes the function is smooth. But worry about details like that is missing the point. $\endgroup$ – Mark Eichenlaub Mar 25 '15 at 11:43
  • $\begingroup$ You could also just replace "positive" with "non-negative" and cover the $v = t^n$ case without changing the meaning of your answer. $\endgroup$ – tpg2114 Mar 25 '15 at 17:05
  • $\begingroup$ Good point. I decided to change it to "can become positive" because I think the mere possibility is what was motivating the question. $\endgroup$ – Mark Eichenlaub Mar 25 '15 at 17:57
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Who says all derivatives have to change continuosly? With a suitable setup, acceleration can jump from zero to something. Location and velocity can not change discontinuously, but acceleration can.

No, there are no "infinitely many derivatives of velocity". In your question, you write "acceleration has to increase from zero and therefore the third derivative of motion with respect to time has to be positive." This assumes that there is a third derivative, which implies that the second derivative is continuous. But the acceleration, the second derivative, may not be continuous. If it is not, there is no defined third derivative, the assumption is false and thus the conclusion does not hold.

Regarding Zeno's Paradox, I would not compare it with this. The paradox arises when time is not taken into account. In my view, once you take time into account and speak of derivatives, it is no longer Zeno's paradox.

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Have a look here: https://en.wikipedia.org/wiki/Non-analytic_smooth_function There are functions which are identically zero for negative arguments, non-zero for positive arguments and smooth everywhere.

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    $\begingroup$ Good point. However you don't really need such functions to resolve the paradox, since, in reality, it's not possible to start from exactly zero for all negative $t$ anyway. $\endgroup$ – leftaroundabout Mar 25 '15 at 11:36

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