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Chain rule and Kinematic quantities x,v,a

  1. Sep 28, 2015 #1
    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!
     
  2. jcsd
  3. Sep 28, 2015 #2

    Svein

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    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).
     
  4. Sep 28, 2015 #3
    thanks!

    So a function is usually invertible when it has some contraints?
     
  5. Sep 28, 2015 #4

    Svein

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