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Homework Help: Vector question

  1. Jan 16, 2014 #1
    I was given an example in a lecture of a vector question which is typical of this module.


    The position of mass m is described as a vector r, from an origin such that r= 7t i + (4t-3t^2) j metres.

    Find the magnitude and directions of r and dr/dt when dy/dt=0

    Now I have worked through this and know what to do but when I looked at the model answer I noticed what to me looks like a mistake?

    I've attached the solution done by my lecturer but he deduces when 4t-3t^2 = 0

    That t=2/3 which it can't be it has to be 4/3?

    I'm not sure if I am mistaken that's why I'm asking and if I am wrong I would like to know why:) thanks guys:) ImageUploadedByPhysics Forums1389878760.572193.jpg
  2. jcsd
  3. Jan 16, 2014 #2
    I don't know whether these notes are in your own handwriting or in your lecturer's handwriting, but whoever wrote them got the equation for dy/dt wrong. The equation for dy/dt is

    dy/dt =4-6t

    This component of velocity is zero when t = 2/3.
  4. Jan 16, 2014 #3
    It was the lecturers handwriting
  5. Jan 16, 2014 #4
    So then the second part where he solves for the magnitude of r must be wrong as the t is added back in?
  6. Jan 16, 2014 #5


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    No, that part is okay. Everything looks okay to me in the attachment except for one thing.

    The attachment has in it:

    [tex] \vec r = 7 t \ \hat \imath + (4t - 3t^2) \ \hat \jmath [/tex]
    [tex] x = 7 t [/tex]
    [tex] y = 4t - 3 t^2 [/tex]
    [tex] \frac{dy}{dt} = 4 t - 3 t^2 = 0 [/tex]
    That last line has the problem. I think your instructor meant to write:

    [tex] \frac{dy}{dt} = \frac{d}{d t} \{ 4 t - 3 t^2 \} = 0 [/tex]
    but neglected to write in the d/dt operator.

    Everything looks okay after that though.

    If you take the derivative of y with respect to t and set it equal to zero, you'll find that t = 2/3, which is shown on the attachment correctly.
  7. Aug 16, 2015 #6
    Yeah, as collinsmark said, it might just be a prob of not writing d/dx as a step,
    The ans is still supposed to be 2/3 seconds.
  8. Aug 16, 2015 #7
    The last previous post on this thread was over a year and a half ago. I'm closing this thread.

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