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Confusion regarding notation of instantaneous velocity wrt something.

  1. Jun 17, 2013 #1
    I know that v(t)=dx/dt

    Then what is v(x) and how?

    Is it also dx/dt or something else?
     
  2. jcsd
  3. Jun 17, 2013 #2
    Take the simplest example, motion in one dimension with constant acceleration. Write down the two equations describing v(t), x(t) and try to eliminate the parameter t from the two equations to get v(x).
     
    Last edited: Jun 17, 2013
  4. Jun 17, 2013 #3
    First of all thanks for replying.

    I took the equation v(t)=2t^2 and differentiated it to get x(t)=2t.
    But again the same confusion cropped up. If v(t)=dx/dt then what is x(t)?
     
  5. Jun 17, 2013 #4

    jtbell

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    To get v(t) from x(t) you differentiate x(t).

    To get x(t) from v(t) you use the operation that "undoes" differentiation. Which operation is that?
     
  6. Jun 18, 2013 #5
    integration
     
  7. Jun 18, 2013 #6

    HallsofIvy

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    Yes. And what is the integral (also called "anti-differentiation) of [itex]2t^2[/itex]? That is, what function has [itex]2t^2[/itex] as its derivative?
     
  8. Jun 18, 2013 #7
    You mean ##v(x)## not ##x(t)##, right?
     
  9. Jun 18, 2013 #8
    You remember back in algebra you learned that rate times time equals distance: rt = d

    Well in this case, x replaces the distance d, and v replaces the rate r. If the velocity is changing with time, you need to express the relationship in terms of differentials: v dt = dx. But, otherwise, it is basically the same thing you learned in Algebra I.
     
  10. Jul 1, 2013 #9
    If v(t)= dx/dt (change in position wrt time)
    then what is v(x) (change in position wrt position?)
    =dx/dx=1???
     
  11. Jul 1, 2013 #10

    NascentOxygen

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    You need to think carefully before writing your questions, because a poorly considered question won't encourage the help you may be hoping for.

    v(t) is velocity as it varies with time, and v(t)=dx/dt

    But more precisely, v(t)=d x(t) / dt

    x(t) being displacement as it varies with time

    Following the usual convention, v(x) must be velocity as it varies with displacement

    Exercise: If x(t)=t^2, and you already showed v(t)= dx(t)/dt = 2t, then determine v(x).
     
  12. Jul 1, 2013 #11

    jtbell

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    More conceptually, imagine that you are driving a car down the road. At 1:00 PM you look at the instrument panel and observe that the speedometer reads 50 km/hr and the odometer reads 23765.8 km. Then v(t) = 50 km/hr for t = 1:00 PM, and v(x) = 50 km/hr for x = 23765.8 km.
     
  13. Jul 1, 2013 #12

    NascentOxygen

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    No you didn't. You took the equation v(t)=t^2 and differentiated it to get

    a(t) = 2t because dv/dt gives acceleration.
     
  14. Jul 2, 2013 #13
    If v(t)=dx/dt
    v(x)= d?/d?
     
  15. Jul 2, 2013 #14
    Have you read the previous posts? If you are still confused, ask a follow up question. Please don't just repeat the question. When I first saw this thread I was excited because I have seen a lot of students struggle with this, but you must actually make an attempt to learn
     
  16. Jul 2, 2013 #15
    Please go back to your advanced algebra textbook and look up what the notation f(x) means.

    f(x) means "the parameter f expressed as a function of the parameter x"

    f(t) means "the parameter f expressed as a function of the parameter t"

    So, v(t) represents the velocity v expressed as a function of the time t

    v(x) represents the velocity v expressed as a function of the distance x

    In both these cases, v is the same parameter.
     
  17. Jul 2, 2013 #16
    Okay now I understand.I finally worked it out. Thank you very much for talking some sense into me.
    v(t) and v(x) are both equal to dx/dt but both the functions define velocity wrt different parameters while dx/dt simply means change in x in very small interval of time.
    Am I right?
     
  18. Jul 2, 2013 #17
    YES. Perfect!!!!
     
  19. Jul 2, 2013 #18

    NascentOxygen

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    This is a good question, and has been addressed. But is there an alternative approach?

    Normally, we are given x as a function of time, viz., x(t), and to determine d##x(t)##/dt we differentiate w.r.t. time. But you could take that x(t) and re-arrange it into time as a function of x, t(x) (at least, you can for some functions x(t), otherwise, use a restricted range.)

    Now, given t(x), how to determine velocity, dx/dt?

    v(x) = 1/(d##t(x)##/dx)

    I.e., v(x) = the reciprocal of d##t(x)##/dx

    You can see that this has units of m/sec, i.e., velocity.
     
  20. Jul 3, 2013 #19
    Thank You. This made things more clear.
     
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