## velocity from acceleration if acceleration is a function of space...

Hello Forum,

usually the acceleration a is a function of time: a(t)= d v(t) /dt

to find v(t) se simply integrate v(t)= integral a(t) dt

What if the acceleration was a function of space, i.e. a(x)?

what would we get by doing integral a(x) dx? The velocity as a function of space, v(x)?

but a(x) is not defined to be v(x)/dx or is it? maybe some chain rule is involved..

thanks,

fisico30

Recognitions:
Homework Help
 Quote by fisico30 What would we get by doing integral a(x) dx? ... chain rule ...
You use chain rule

a(x) = dv/dt = f(x)

multiply dv/dt by dx/dx:

(dv/dt)(dx/dx) = (dx/dt)(dv/dx) = v dv/dx = f(x)

multiply both sides by dx

v dv = f(x) dx

If f(x) can be integrated and the integral of f(x) = g(x), then

1/2 v2 = g(x) + c

This results in an equation that relates velocity and position.

v = sqrt(2 (g(x) + c))

v = dx/dt = sqrt(2 (g(x) + c))

and then integrate

dx/(sqrt(2 (g(x) + c))) = dt

assuming h(x) is the integral of dx/(sqrt(2 (g(x) + c))), you get

t = h(x) + d

where d is constant of integration. It may be possible to solve this equation to get x = some function of t.

One simple case is a(x) = -x, which results in x(t) = e(sin(t+f)), where e and f are constants.
 Thanks rgcldr! great explanation and example! fisico30

## velocity from acceleration if acceleration is a function of space...

Hi rcgldr,

one question:

when you get v dv = f(x) dx, is v a function of t or of x, i.e. is v equal to v(t) or v(x)?

It looks like it would be a function of x, v(x), since 1/2 v^2 = g(x) + c results in v(x) = sqrt(2 (g(x) + c)).

I am confused because when we write v=dx/dt I always assume that v must be a function of time t, i.e. v(t).
When in your first step you write a(x) = dv/dt I tend to think that the v must be function of t since dv/dt is a derivative with respect to time.

So which one is correct? a(x)= dv(x)/dt or a(t)=dv(t)/dt. Acceleration is always the time derivative of the velocity vector but the velocity vector can be a function of any dependent variable: v(t), v(x), v(F), where F is force, etc....

Thanks,
fisico30
 I guess my point is: if a function, like a, is defined as a time derivative of another function, dv/dt, does the differentiated function need to be a function if time?

Recognitions:
Homework Help
 Quote by fisico30 when you get v dv = f(x) dx, is v a function of t or of x, i.e. is v equal to v(t) or v(x)? It looks like it would be a function of x, v(x), since 1/2 v^2 = g(x) + c results in v(x) = sqrt(2 (g(x) + c)).
At this point, it's just a relationship betwen v and x.

 I am confused because when we write v=dx/dt I always assume that v must be a function of time t, i.e. v(t).
It is, but I was only able to take advantage of that in the last step where I find t as a function of x.

 When in your first step you write a(x) = dv/dt I tend to think that the v must be function of t since dv/dt is a derivative with respect to time.
True, but I used chain rule to get rid of the dt and end up with v dv = f(x) dx.

 So which one is correct? a(x)= dv(x)/dt or a(t)=dv(t)/dt.
Both are correct, but if acceleration is defined as a function of x, then it may not be possible to find an equation for acceleration as a function of time. In the simple example I gave, a(x) = -x, you will be able to find both a(x) and a(t). I ended up finding x(t) = e sin(t + f). You can take the derivative of this to find v(t), and the derivative of v(t) to find a(t), so a(x) = -x, and a(t) = -e sin(t+f). For other situations, you may not be able to solve for x(t), v(t), or a(t), in which case numerical integration will be required.
 So it is mathematically legal and ok to write a derivative of a function even if the independent variable a of the function and the differentiation variable b are not the same: f(a) and db to create df(a)/db... I guess that is all the chain rule is about....