# Movement with non-constant acceleration [duplicate]

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.

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.

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• @BrandonEnright - I dare say that your comment may be no longer valid after the considerable edit that was made to this answer...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