instantaneous velocity and speed problem

1. The problem statement, all variables and given/known data
the position function x(t) of a particle moving along an x axis is x=4.0-6.0t^2, with x meters and t in seconds. (a) at what time and (b) where does the particle (momentarily)stop?

x=4.0-6.0t^2

and i made the equation equal this t=sqrt(4.0/6 = .8164995809
i put it into the equation and x=0

is this correct. I'm assuming that the particle is zero meters when it stops
but what do i know i'm not an expert
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 Quote by afcwestwarrior I'm assuming that the particle is zero meters when it stops but what do i know i'm not an expert
You can't just assume the answer. Instead, figure it out. Hint: Given the position as a function of time, find the velocity as a function of time.
 So are you saying that I have to find the velocity as a function of time. I'm not sure by what you mean by this. Do you mean that I use V=change inx/ change in t

instantaneous velocity and speed problem

Are you saying that I have to find the derivative.

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 Quote by afcwestwarrior Are you saying that I have to find the derivative.
That would be most wise.
 Ok, so would it be like this x=4.0-6.0t^2 x=-(2)(6)(t) x=-12(t) so x=-12t then t=x/-12
 then i plug 12 into the equation and i get 4.0-6(12)^2=-860m

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 Quote by afcwestwarrior Ok, so would it be like this x=4.0-6.0t^2 x=-(2)(6)(t) x=-12(t)
That gives you the velocity as a function of t (not x). v = dx/dt.
 So how do I do that.
 hmm. would it be -12=x/t

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 Quote by afcwestwarrior So how do I do that.
Do what? You already found the velocity, you just mislabeled it "x" instead of "v".
 oh ok lol, so that's all i do, v=-12t

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 Quote by afcwestwarrior oh ok lol, so that's all i do, v=-12t
Good. Now go back and answer part (a).
 Ok so where do I plug in the -12t, x=4.0-6.0t^2 would it be 4-6(-12)^2=-860

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 Quote by afcwestwarrior Ok so where do I plug in the -12t,
You don't plug it in anywhere, you use that equation for speed to answer question (a). What's the time when the particle momentarily stops? (What speed must it have when it stops?)
 the speed would have to be zero right.
 -12(0)= it has to be zero