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Acceleration problem

  1. Jul 22, 2009 #1
    The tangential acceleration and normal acceleration is given as follows:
    [tex]\vec{a_{T}} = \frac{4cos(2t)sin(2t)}{1 + 2sin^{2}(2t)} \cdot (\hat{x} - sin(2t)\hat{y} + sin(2t)\hat{z}),
    \vec{a_{N}} = - \frac{2cos(2t)}{1 + 2sin^{2}(2t)} \cdot (2sin(2t)\hat{x} + \hat{y} - \hat{z})[/tex] (#0)

    Find the magnitude of the tangential and normal acceleration components (given above).

    The solution for the tangential and normal magnitudes are given by the following,
    [tex] \vec{a_{T}} = \frac{4cos(2t)sin(2t) }{\sqrt{1 + 2sin^{2}(2t)}},
    \vec{a_{N}} = \frac{2\sqrt{2}cos(2t)}{\sqrt{1 + 2sin^{2}(2t)} }[/tex] (#1)

    I mean if we had to solve for the magnitude of some vector [tex]a = 3\hat{x} - 33\hat{z},[/tex] then we say,
    [tex]|a| = \sqrt{3^{2} + (-33)^{2}}[/tex]. How would we apply this to our equation (#0) to get equation (#1)?


    Last edited: Jul 22, 2009
  2. jcsd
  3. Jul 22, 2009 #2


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    Homework Helper

    Re: Acceleration

    [tex]|a\hat{x} + b\hat{y} + c\hat{z}| = \sqrt{a^2+b^x+c^2}[/tex]

    that's all there is to it.
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