If we have, say, a material point with a zero velocity at the time $t=0$, and this point starts moving at a time $t>0$ , then we look at the force impressed on the point by inspecting the second derivative of the motion of the point. If our point follows a motion law of $e^{-\frac{1}{t^2}}$, then we have that all derivatives at $t=0$ are $0$. So my question is, why doesn’t this go against Newton's second law? To me it is like there is a correspondence between successive derivatives and the force impressed.
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2$\begingroup$ You are unconflicted about $x=t^4$? $\endgroup$– Cosmas ZachosCommented Dec 9, 2017 at 1:26
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1$\begingroup$ Possible duplicates: physics.stackexchange.com/q/111251/2451 , physics.stackexchange.com/q/60480/2451 , physics.stackexchange.com/q/172207/2451 , physics.stackexchange.com/q/190564/2451 and links therein. $\endgroup$– Qmechanic ♦Commented Dec 9, 2017 at 3:03
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$\begingroup$ What is that function defining? When you say "motion law," what does that mean? Is it time, position, velocity? Is $x$ referring to position? Without an equation, that function is meaningless. $\endgroup$– Bill NCommented Dec 9, 2017 at 4:45
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$\begingroup$ @CosmasZachos With $x=t^4$, at $t=0.1$, $\ddot{x}=0.12$. However, with $x=e^{-t^{-2}}$, at $t=0.1$, $\ddot{x}=1.5\cdot 10^{-37}$, and at $t=0.05$, $\ddot{x}=5\cdot 10^{-166}$. See the difference? The equilibrium at zero is so stable that there is no classical force (aside from quantum fluctuations) to move the body off zero. $\endgroup$– safesphereCommented Dec 9, 2017 at 5:30
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3$\begingroup$ -1 Not clear what you are asking. How does this violate $F=ma$? $a=\ddot x=0$ at $t=0$ but we are not told that $F \ne 0$. So where is the contradiction? $\endgroup$– sammy gerbilCommented Dec 9, 2017 at 10:05
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I have edited your question by changing $x$ to $t$, because you referred to your formula as the "motion law". If instead you meant the potential depending on $x$, it would lead to the same outcome anyway.
Your law of motion essentially implies an equilibrium at zero, because the force generated by a deviation from zero is astronomically small, smaller than quantum effects. In reality, for the point to start moving, the equilibrium must be broken. Normally this means that a deviation from zero generates an increasing force and the point follows this force with an increasing acceleration.
However, in your case, a deviation from zero does not generate an increasing force of a meaningful magnitude. For all practical reasons, the force remains a practical zero and therefore the point will not start moving.
So the catch in your question is in the formulation that the point will start moving according to the law you have prescribed. However, there is a difference between mathematical formulas and the real world. To move according to your formula, there must be a tangible reason for the point to do so, such as a force capable of moving this material point. If there is no such a force present, the point would have no reason to follow the law of motion you have artificially prescribed to it.
The bottom line here is that there is no contradiction with the Newton's second law. This law states that acceleration is proportional to force. In other words, if you apply a force to your point, it would indeed get moving. In your case, you are not applying any force. The numbers your formula generates are less than the influence of the moons of Jupiter.
Stating this differently, there is no real world scenario, in which a material point would follow this hypothetical law of motion, because it is not realistic. However, if you want to consider a purely hypothetical thought experiment, then sure, first, the initial equilibrium must be broken by a spontaneous symmetry breakdown and then the point would feel a vanishingly small, but non-zero force, and would indeed follow your equation. The only clarification is that this may not happen exactly at the time zero, because the moment of the initial symmetry breakdown is random, unless you give the point a slight push in a certain direction.