Chain rule and Kinematic quantities x,v,a

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fog37
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Hello Forum,

I have a couple of kinematics questions.

The position of a point object is given by the position vector x(t). Speed is v(t)=dx(t)/dt and the acceleration a(t)= dv(t)/dt. What if we wanted to know the velocity and/or the acceleration as a function of position, i.e v(x) or a(x)?

For example, given x(t)=3t^2+2, how would we find v(x) and a(x)? I think the chain rule is needed. The chain rule is a a formula for computing the derivative of the composition of two or more functions.

What if we were given v(x) from the beginning, for example v(x) = 3x +x^2, and wanted to know the position x(t) or a(t) for the object?

A generic force F can be a function of several independent variables, F(x, t, v):
  • distance x
  • time t
  • speed v
Are these three independent variables associated to a hypothetical object to which the force F is applied? Since F=ma, then the acceleration a(x, t , v)

If the acceleration varies with time t, then the acceleration should also automatically be a function of distance x, since the object occupies different positions x at different times t. So does it make sense to write a(x,t) or should we just write a(t) or a(x)? Same goes for v(x,t): if the object occupies different positions at different times it also means its speed is a function of time...

Thanks!
 
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fog37 said:
What if we wanted to know the velocity and/or the acceleration as a function of position, i.e v(x) or a(x)?
Obviously, if x(t) fulfills some constraints (for example that you do not drive in circles), you can invert it and find t=f(x). Then v(t) = v(f(x)) etc. But v(t) is still dx(t)/dt, so v(f(x)) = [itex]\frac{1}{\frac{dt}{dx}}[/itex] - which gives you another clue: If you stand still, you cannot create t=f(x).
 
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thanks!

So a function is usually invertible when it has some contraints?