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

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To convert acceleration defined as a function of position, a(x), into a function of time, a(t), one can use the relationship between velocity and position. Starting with a(x) = x/s², the general solution for position is x = c*e^(sqrt(m)*t/s) + d*e^(-sqrt(m)*t/s). By differentiating this solution twice, one can derive a(t) as a function of time. An alternative method involves using the work-energy theorem, where a(x) = v*dv/dx, to find velocity as a function of position, then integrating to express time as a function of position, t(x), which can be inverted to find x(t). This approach allows for the substitution back into a(x(t)) to obtain the desired acceleration function in terms of time.
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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
 
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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
 
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.
 
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