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inkyvoyd

  • 2 years ago

"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?"

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  1. inkyvoyd
    • 2 years ago
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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...

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

  3. inkyvoyd
    • 2 years ago
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    @amistre64

  4. inkyvoyd
    • 2 years ago
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    http://chat.stackexchange.com/transcript/message/6955366#6955366

  5. inkyvoyd
    • 2 years ago
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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

  6. phi
    • 2 years ago
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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.

  7. inkyvoyd
    • 2 years ago
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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

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