instantaneous velocity and speed problem

afcwestwarrior

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

i made x=0
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

Doc Al

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.

afcwestwarrior

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

afcwestwarrior

Are you saying that I have to find the derivative.

Doc Al

That would be most wise.

afcwestwarrior

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

afcwestwarrior

then i plug 12 into the equation and i get 4.0-6(12)^2=-860m

Doc Al

That gives you the velocity as a function of t (not x). v = dx/dt.

afcwestwarrior

So how do I do that.

afcwestwarrior

hmm. would it be -12=x/t

Doc Al

Do what? You already found the velocity, you just mislabeled it "x" instead of "v".

afcwestwarrior

oh ok lol, so that's all i do, v=-12t

Doc Al

Good. Now go back and answer part (a).

afcwestwarrior

Ok so where do I plug in the -12t,
x=4.0-6.0t^2

would it be 4-6(-12)^2=-860

Doc Al

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?)

afcwestwarrior

the speed would have to be zero right.

afcwestwarrior

-12(0)= it has to be zero