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

by fisico30
Tags: acceleration, function, space, velocity
 P: 374 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
HW Helper
P: 7,108
 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.
 P: 374 Thanks rgcldr! great explanation and example! fisico30
 P: 374 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
 P: 374 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?
HW Helper
P: 7,108
 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.
 P: 374 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....

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