ksaimouli
describe the motion of the particle
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ksaimouli
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|dw:1352072429317:dw|
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t >=0
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@zepdrix
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what i did is for velocity =0 to find the critical points
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|dw:1352072565266:dw|
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so from (-inf,2) left and (2,inft) right
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is this right
Algebraic!
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no, it's a particle that's moving left initially, and accelerating to the right...
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Vi =-4
a=2
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so how to approach it then
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you're right that it has V=0 at t=2, but other than that, I'm not sure what your trying to say.
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that's it.
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initially moving left, accelerating right, 'turning around' at t=2, then heading right (and accelerating) forever.
ksaimouli
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the answer is it begins at a position 3 moving in a negative direction.it moves to position -1 when t=2 and then changes direction moving in a positive direction thereafter
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i dont know how they got it
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@Algebraic!
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you left something out then, there's nothing about position in your statement.
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what's that second line supposed to be? I guessed it was a=2...
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is it supposed to be d(0)=2 or something?
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a particle is moving along the x axis with position functionx(t) find the (a) velocity (b) acceleration (c)describe the motion of the particle at t>=0
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|dw:1352073305950:dw|
ksaimouli
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this was the question
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...
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when t=0 x=3
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when V= 0, t=2 and x=?