Acceleration as a function of position, and velocity.

[Axecutioner]

If acceleration is a function of position, can you express velocity as a function of position, or would there have to be a time variable? For example, if F = kx = ma, and v₀ is given, then a = (k/m)x, so what would the function for v be?

I would think that, if you know the acceleration at every point along a given path and you know initial velocity, you would also know velocity at every point as well, without having to know what time is.

My actual problem is more complicated than a spring, but it comes down to ∫ f(x) d? = ∫ a d? = v, but what would the integral be? dx or dt?

Thanks!

 

[AlephZero]

Yes, you can do this. For example you can use the chain rule

dv/dt = (dv/dx)(dx/dt)

but dx/dt = v, so

a = dv/dt = v dv/dx.

 

[Ken G]

I would think that, if you know the acceleration at every point along a given path and you know initial velocity, you would also know velocity at every point as well, without having to know what time is.

That's correct, if you take what AlephZero said, and write it as 0.5*d(v²)/dx = F(x), then integrate over x, you get the concept of kinetic energy on the RHS and potential energy on the LHS, which let's you figure out v(x) without ever knowing t.

My actual problem is more complicated than a spring, but it comes down to ∫ f(x) d? = ∫ a d? = v, but what would the integral be? dx or dt?

It's over dt, to get the second integral to =v. But it's not useful, since F is a function of x not t. So get v(x) with conservation of energy, and if you want to know what t is doing, say v(x)=dx/dt so t is the integral of dx/v(x).