I have an equation that says $$a(t)= \alpha * s(t)+\beta$$ where $a$ is the acceleration, $\alpha$ and $\beta$ are constants, $s$ is position and $t$ the time. My question is how to get the resulting position function $p(t)$ that gives the position in dependence to the time.

I fail to understand how to get this equation since the acceleration depends on the position and vice versa this seems like a circular reference to me.

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    $\begingroup$ Have you taken a course on differential equations? What you have here is: $$\frac{d^2s}{dt^2} = \alpha s + \beta$$. $\endgroup$ – tmwilson26 Nov 13 '15 at 14:10

It's not really a circular reference, it's an ordinary differential equation: $$ \frac{d^2y}{dx^2}=f(y) $$ In particular, it is a general 2nd order ODE, which, for some functions $f(y)$, has known analytic solutions.

One way to think of this is a mass on a spring:

(source: Julius O Smith, Standorf MUS420/EE367A Supplementary: The Laplace Transform)

If you watch the motion of the spring, then you'll note that at each point in time $t$, the direction of the motion depends on where it was just beforehand. So the velocity, $\dot{x}=dx/dt$, is pointing to the right in the image, but after compressing all the way, the velocity would go to the left. The rate of change of the velocity (acceleration, $a=dv/dt$) also depends on where it was.

This is what Hooke's law says: the force, $F(t)$, depends on the position, $x(t)$: $$ F(t)=-kx(t) $$ but we also know that $F=m\,(d^2x/dt^2)$, so $$ \frac{d^2x}{dt^2}=\frac{k}{m}x(t) $$ which is similar to your equation.

There are numerous methods of solving these types of equations (e.g., trial functions or integration), but it takes practice & experience to know when to use which.


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