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Motion problem -- Finding velocity and acceleration with derivatives [title edited by Mentor]

  1. Jul 10, 2017 #1
    1. The problem statement, all variables and given/known data
    Coordinates of a particle which moves on a xy coordinate system given with:
    x=-(5m)sinωt
    y=(4m)-(5m)cosωt
    In these equlations t's unit given as second, and ω's unit
    second^-1. A-) Found velocity and acceleration components when t=0 B-) Write equlations for position and acceleration vektors when t>0C-)Define particle's way
    on a xy graph.

    2. Relevant equations
    V=dx/dt a=dV/dt


    3. The attempt at a solution
    I tried to take deridatives of equlations, but I couldn't.
    Can someone explain deridativation method clearly:D
     
    Last edited by a moderator: Jul 10, 2017
  2. jcsd
  3. Jul 10, 2017 #2
    Are you sure you didn't mean derivation? Derivation can allow you to find the motion of a particle at a specific point in time.
     
  4. Jul 10, 2017 #3
    Yes my English is not well so there can be some writing mistakes. I mean I actually cannot derivate the equlations which given in that problem:smile:
     
  5. Jul 10, 2017 #4
    Do you know how to find the derivatives of trig functions? Also, where is ##t## in the functions, and what does ##m## indicate?
     
  6. Jul 10, 2017 #5
    Okay I am so sorry this is my first post. In the equlations y's at right side are actually t I know how to derivate trig functions
     
  7. Jul 10, 2017 #6
    I think I fixed it
     
  8. Jul 10, 2017 #7
    So is ##x=-5m\sin(\frac 1 t) t##? and ##y=4m-5m\cos(\frac 1 t) t##? I'm guessing ##m## is a constant. By the way, if you want to write equations more clearly here's the page which shows you how: https://www.physicsforums.com/help/latexhelp/. I found it pretty simple.
     
    Last edited: Jul 10, 2017
  9. Jul 10, 2017 #8
    Yes m is meter unit. Equlations are correct
     
  10. Jul 10, 2017 #9
    Alright. So if you find the derivative with respect to time you will get the velocity. Do it again and you'll get acceleration. Knowing the derivative of trig functions and the chain rule should make it manageable.
     
  11. Jul 10, 2017 #10
    I cannot derivate it still :/ can you explain me step by step
     
  12. Jul 10, 2017 #11
    Do you know the chain rule?
     
  13. Jul 10, 2017 #12
    Yes it is df(g(x))/dx=g'(x)f'(g(x)) am I wrong?
     
  14. Jul 10, 2017 #13

    scottdave

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    Originally you have this:
    x=-(5m)sinωt
    y=(4m)-(5m)cosωt

    In these equations t's unit given as second, and ω's unit
    second^-1.

    Then somehow you have x = (-5 meters)sin(1/t)t, but where did the extra t come from? You need to keep ω in there, which has dimension of (1 / time) which cancels out the time [the argument of sine or cosine need to be dimensionless].
     
  15. Jul 10, 2017 #14
    Yes when you dimensionally analyze the problem this problem occurs. The problem is about problem then ?
     
  16. Jul 10, 2017 #15

    scottdave

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    Let's say you have f(θ), and you want to take the derivative with respect to t, and θ is a function of t, then you have df/dt = (df/dθ)*(dθ/dt).
    So for sin(θ) you have (d/dt) of sin(θ) = cos(θ)*(dθ/dt). Since θ = ωt, then dθ/dt = ω
     
  17. Jul 10, 2017 #16

    scottdave

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    Looking at x(t) = (-5 meters)*sin(ωt), taking the first derivative gives (ω)*(-5 meters)*cos(ωt) {trig function and chain rule}. Note that this now has the dimension of {length / time} which is velocity. Taking the derivative of velocity {second derivative of x} gives a = x'' = (ω2)*(5 meters)*sin(ωt), which has the dimension of {length / time2} which is acceleration. Do similar for the y component.
     
  18. Jul 10, 2017 #17
    I unterstood until last equlation but why θ=ωt? Did I missunderstood something in question?
     
  19. Jul 10, 2017 #18

    scottdave

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    This was given in the problem statement: x=-(5m)sinωt. It probably should have been clarified as x=-(5m)sin(ωt). Omega is the angular frequency, and multiply that by time gives you a dimensionless quantity as the argument for sine.
     
  20. Jul 10, 2017 #19
    Okay now I understand that I completely miss understood that question I assumed that x= -5m sin(ω)t
    not x= -5sin(ωt) Thanks for all your help
     
  21. Jul 11, 2017 #20

    Ray Vickson

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    PF helpers are not allowed to do that; read the posting rules, which state that YOU must first do some work on the problem before receiving help.
     
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