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Change in velocity comes from acceleration, but that acceleration was also a leap that started from nothing. It was zero, then positive, and then something else. So it's application must have been pushed by some other, more interior, and higher-derivative force. But the force that pushed this positive acceleration into existence, it also arose from zero. So it must have a higher-order parent that produced it. And so on.

So is any nth time derivative of position ever really zero? Doesn’t any change in constant velocity rest finally on an infinite time derivative of position?

user739960
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    Do you allow for "mathematically perfect" motion? For example motion that is exactly $1~\frac{\text{m}}{\text{s}}$ forward? That has quite many derivatives that are exactly zero ... – Matti P. Jan 07 '20 at 06:10
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    You might want to consider the derivatives of $e^{-1/t^2}$ at $t=0$ – Henry Jan 07 '20 at 06:16

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