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

by fisico30
Tags: acceleration, function, space, velocity
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fisico30
#1
Feb22-12, 03:26 PM
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
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rcgldr
#2
Feb22-12, 07:17 PM
HW Helper
P: 7,179
Quote Quote by fisico30 View Post
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))

To get time versus position, you start with

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.
fisico30
#3
Feb23-12, 07:42 AM
P: 374
Thanks rgcldr!

great explanation and example!

fisico30

fisico30
#4
Mar2-12, 10:03 AM
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
fisico30
#5
Mar2-12, 10:12 AM
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?
rcgldr
#6
Mar2-12, 11:03 AM
HW Helper
P: 7,179
Quote Quote by fisico30 View Post
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.
fisico30
#7
Mar3-12, 08:03 AM
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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