1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

A very simple question about velocity and acceleration

  1. Dec 8, 2007 #1
    [SOLVED] A very simple question about velocity and acceleration

    1. The problem statement, all variables and given/known data
    I should be way over this by now (I took elementary mechanics a year ago), but occasionally I find out there's unacceptably much I don't understand about very elementary physics (perhaps is it a sign to be taken seriosly about the future of my physics career...). And this question in particular is ridiculously easy, I guess.

    In Fowles's and Cassiday's Analytical mechanics it says:
    It goes on to say that
    [tex]\ddot{x}=\frac{d\dot{x}}{dt}=\frac{dx}{dt}\frac{d\dot{x}}{dx}=v\frac{dv}{dx}[/tex]

    Now what I don't understand is that if F is a function of x, then acceleration (a) should be a function of x also. However
    [tex]\ddot{x}=\frac{d^2x}{dt^2}=a=\frac{dx}{dt}\frac{d\dot{x}}{dx}=v\frac{dv}{dx} \not= va[/tex]
    although if F=F(x) and a=a(x) then shouldn't
    [tex]\frac{dv}{dx}=a[/tex]?

    On the other hand the book states that
    [tex]F(x)=mv\frac{dv}{dx}[/tex]
    and if I divide by m I get:
    [tex]\frac{F}{m}=a=v\frac{dv}{dx}[/tex]

    So dv/dx is not a in this case but dv/dx times v. But if the force is independent of time, then how can velocity still be dx/dt? How can the position derivated with respect to time give out velocity in full, if there's an acceleration affected by the position and therefore a velocity affected by the position?

    So I guess I'm having trouble understanding even the initial quotation. I know how it is, but I don't understand why it is.
     
  2. jcsd
  3. Dec 8, 2007 #2

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Okay, first of all:

    WHY?

    a = dv/dt by definition

    Therefore:

    [tex]\frac{dv}{dx}=\frac{dv}{dt} \frac{dt}{dx} = a\frac{1}{v}[/tex]
     
  4. Dec 8, 2007 #3

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Who says the force is independent of time? The particle moves along a trajectory x(t), so everywhere you see x, you can replace it with x(t):

    F(x) = F(x(t)), which, if we substitute in the time dependence of x, becomes F(t).

    Does this help?

    Take the example of simple harmonic motion (F = -kx). Yes, the acceleration of a mass on a spring does depend on WHERE it is, as does its velocity. But that doesn't change the fact that WHERE it is (its position) is a well-defined function of time x(t) = Acoswt (for example) so that:

    v(t) = -Awsinwt

    a(t) = -Aw^2coswt (so that a(x) = -w^2x)
     
    Last edited: Dec 8, 2007
  5. Dec 8, 2007 #4
    Yes, I know. But if a function is not dependend on a variable, then shouldn't derivating the function with respect to that variable return zero? If f(x)=ax and neither a nor x is dependend on y, then df/dy=0?

    Edit: Well you posted that latter post and for some reason I didn't see it though I refreshed just before posting this.
     
  6. Dec 8, 2007 #5

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Soo...are you saying that that cleared it up then? Your question is answered?
     
  7. Dec 8, 2007 #6
    Well the book says... "If the force is independent of -- time, then..."

    Possibly. I don't have the time to think about it thoroughly right now, but I'll ponder it later and report back.
     
  8. Dec 8, 2007 #7

    cepheid

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    They could be saying that the force has no EXPLICIT time dependence. I.e. the force depends on time ONLY because the force depends explicitly on position, which in turn depends on time. Mathematically, the lack of explicit time dependence is expressed as follows:

    Say the force is a function of position and time (in general):

    F = F(x,t)

    then:

    [tex]\frac{d}{dt} F(x,t) = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial t} [/tex]

    The statement that there is no explicit time dependence is a statement that

    [tex] \frac{\partial F}{\partial t} = 0 [/tex]

    There really is only one independent variable. And if the function x(t) has a unique inverse t(x) (which is not the case for SHM), then it doesn't matter which you pick:

    F(x,t) = F(x(t), t) = f(t)

    OR

    F(x,t) = F(x, t(x)) = g(x)

    If you write F(x,t) as f(t), then that latter function, of course, does have an explicit time dependence, but it is a time dependence that is determined solely by the dependence of the force on the position and of the position on time.
     
    Last edited: Dec 8, 2007
  9. Dec 9, 2007 #8
    Yes, of course. Whenever an object is at some explicit position, it is always at some explicit moment of time, too, even though their relation wouldn't be explicitly defined. Classically, at least. And when and object moves, time flows also. There is no change of position if there is no passing of time.

    Thanks for your help. I guess I'm beginning to understand this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A very simple question about velocity and acceleration
Loading...