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Velocity from acceleration if acceleration is a function of space... 
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#1
Feb2212, 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 


#2
Feb2212, 07:17 PM

HW Helper
P: 7,033

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 v^{2} = 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. 


#3
Feb2312, 07:42 AM

P: 374

Thanks rgcldr!
great explanation and example! fisico30 


#4
Mar212, 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 


#5
Mar212, 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?



#6
Mar212, 11:03 AM

HW Helper
P: 7,033




#7
Mar312, 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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