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This question already has an answer here:

Suppose we have a material point. If it is moving from position $X_0$ with initial velocity $V_0$ and constant acceleration $A$, then from elementary physics course I remember that its movement is described by the equation

$$X(t) = X_0 + V_0t + At^2/2.$$

Now, my question is, what is the equation of the movement of the material point if its acceleration is an arbitrary function of $t$: $A(t)$. Is it simply:

$$X(t) = X_0 + V_0t + A(t)t^2/2,$$

or is it more complicated than that? From the looks of $At^2/2$ I have a suspicion that integrals may be involved.

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marked as duplicate by jinawee, David Z Apr 16 '14 at 15:39

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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It's not as simple as that. You'll have to obtain velocity and displacement by integrating your given acceleration and using correct boundary conditions.

For example:

Suppose the acceleration is given by A(t) = 2t [m/s²] and the problem states that the particle starts its movement from rest and from the origin of your coordinate system, so that X(t=0)=0 and V(t=0)=0.

The velocity of that particle would be an integral in time of the acceleration, that is V(t) = t² + C [m/s], where C is a constant of integration.

Now, you know that V(0) = 0, so C = 0 is the only possible value that satisfies your movement.

Integrating velocity in time you´ll obtain the displacement, that is

X(t) = t³/3 + B [m], where, again, B is a constant of integration. Since X(0)=0 , B = 0.

Sometimes boundary conditions are imbued within text, so you gotta pay attention to some details, but the method of obtaining the equation of movement is the same for every problem.

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  • $\begingroup$ Hi Friquinho, this is perfect for a comment but it isn't an answer. $\endgroup$ – Brandon Enright Apr 16 '14 at 15:57
  • $\begingroup$ @BrandonEnright - I dare say that your comment may be no longer valid after the considerable edit that was made to this answer... Take a look. $\endgroup$ – Floris Apr 16 '14 at 17:51

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