# How to convert acceleration as a funtion of position to a function of time?

1. Oct 16, 2011

### WindScars

Suppose I have acceleration defined as a function of position, "a(x)". How to convert it into a function of time "a(t)"? Please give an example for the case a(x)= x/s²

2. Oct 16, 2011

### elegysix

not sure what s is supposed to be, if its a function of t or x, this won't work, but anything else is fine.

a(x)=x/s^2 = F/m

x'' - mx/s^2 = 0

has the general solution x = c*e^(sqrt(m)*t/s)+ d*e^(-sqrt(m)*t/s)

differentiating this twice gives you
a(t) = (mc/s^2)*e^(sqrt(m)*t/s) + (md/s^2)*e^(-sqrt(m)*t/s)

where c and d are constants you can find from boundary conditions or initial values

3. Oct 16, 2011

### Ken G

That's correct except drop the "m", it doesn't belong. Also, there's a general way to do this-- write a(x) = v*dv/dx, use that to derive the work-energy theorem that the integral of a(x) over dx equals the change in v2/2. Solve that for v(x), and say t = integral of v(x) over dx. That gives you t(x), which you can invert to x(t), then plug into a(x(t)). This requires that t(x) be invertible, but it's the best you can do in general.