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Kinematics proble

  1. Nov 5, 2008 #1
    a particle move according to the vector
    [tex]\vec{r}[/tex]=3cos[tex]^{2}[/tex](6t)[tex]\hat{x}[/tex]-5sin([tex]\frac{PI}{6}[/tex]t)[tex]\hat{y}[/tex]

    what is the formula for the route it takes, y(x)? when will the particle stop and how many times will it stop in an hour??

    i said, the vectors in the direction x and y
    x(t)=3cos[tex]^{2}[/tex]6t=3(1-sin[tex]^{2}[/tex]6t)
    y(t)=-2sin6t
    -------------
    sin6t=[tex]\frac{-y}{2}[/tex]

    x=3(1-([tex]\frac{-y}{2}[/tex])[tex]^{2}[/tex])
    x(y)=3-[tex]\frac{3}{4}[/tex]y[tex]^{2}[/tex]

    this course is a parabula lying on its side, (minus infinity) on its x axis with its maximum at Max(3,0)

    to find out when the particle stops i say
    the particle will stop when the parabula peaks-- [tex]\frac{dx}{dy}[/tex]=0
    x(y)=3-0.75y[tex]^{2}[/tex]
    [tex]\frac{dx}{dy}[/tex]=-1.5y
    y=0
    the particle will stop every time y=0
    y(t)=-2sin(6t)=0
    sin(6t)=0
    6t=[tex]PI[/tex]*K
    t=[tex]\frac{PI}{6}[/tex]*K (K being a positive whole number)

    does this mean that the particle will stop every [tex]\frac{PI}{6}[/tex] seconds??

    how can this be, since the route the particle moves on is a parabula and doesnt peak more than once, i realise that where i have misunderstood something is in the transition from the trigonometric equations x(t) and x(y) which repeat themselves to the parabula x(y).
     
  2. jcsd
  3. Nov 5, 2008 #2

    Hootenanny

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    Careful
     
  4. Nov 5, 2008 #3
    sorry, the -5sin...was a typo, , meant to be -2sin(6t)y like i wrote in all the calculations
     
  5. Nov 5, 2008 #4

    Hootenanny

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    What about the [itex]\pi/6[/itex] in the question? Was that a typo as well?
     
  6. Nov 5, 2008 #5
    correct equation is r=3cos^2(6t)x-2sin(6t)y
    but that doesnt really make a diffeerence to what im asking, its more of a technical question, the end result as a number isnt of that much importance,,,
    after reading over what i wrote i now think that i was wrong, and that the particle will stop when |v|=0 meaning when cos(6t)=0
    6t=PI/2 +PI*K
    t=PI/12 + (PI/6)*K

    from this how do i know how many times it will stop in an hour??
     
  7. Nov 5, 2008 #6

    Hootenanny

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    The particle doesn't stop when

    [tex]\frac{dy}{dx}=0[/tex]

    Rather, it stops when

    [tex]\frac{d\bold{r}}{dt}=0[/tex]
     
  8. Nov 5, 2008 #7
    thats essentially what i did by saying Vx=0 and Vy=0

    but then i get to t=PI/12 + (PI/6)*K how do i translate that into how many times it will stop in an hour
     
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