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what i did is for velocity =0 to find the critical points
so from (-inf,2) left and (2,inft) right
is this right
no, it's a particle that's moving left initially, and accelerating to the right...
Vi =-4 a=2
so how to approach it then
you're right that it has V=0 at t=2, but other than that, I'm not sure what your trying to say.
initially moving left, accelerating right, 'turning around' at t=2, then heading right (and accelerating) forever.
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
i dont know how they got it
you left something out then, there's nothing about position in your statement.
what's that second line supposed to be? I guessed it was a=2...
is it supposed to be d(0)=2 or something?
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
this was the question
when t=0 x=3
when V= 0, t=2 and x=?