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inkyvoyd

"the position of an object is described by the parametric equations x=ln t and y=5t^2. What is the accerleration of the object in m/sec^2 when t=2?"

  • one year ago
  • one year ago

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  1. inkyvoyd
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    Now, what I did was calculuate d^2y/dt^2 and d^2x/dt^2. then I calculated \(\sqrt{a_x^2+a_y^2}\), for t=2...

    • one year ago
  2. inkyvoyd
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    But, my sister tells me I'm supposed to find d^2y/dx^2... what am I supposed to do?

    • one year ago
  3. inkyvoyd
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    @amistre64

    • one year ago
  4. inkyvoyd
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    http://chat.stackexchange.com/transcript/message/6955366#6955366

    • one year ago
  5. inkyvoyd
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    Question answered by people at math.stackexchange :) http://math.stackexchange.com/questions/241312/acceleration-of-a-particle-described-by-parametric-equations

    • one year ago
  6. phi
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    accerleration of the object is the change in velocity of the object with time velocity is the change in position of the object with time The position of the object is the vector (x y) its velocity is the vector (dx/dt dy/dt) its acceleration is the vector (d^2 x/dt^2 d^2 y/dt^2) You found the magnitude of the acceleration, which may or may not be what the question is asking for.

    • one year ago
  7. inkyvoyd
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    @phi , the problem is that the units they want are in m/sec^2 - so direction wouldn't really do anything in here :S

    • one year ago
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