# Instantaneous velocity and speed problem

## Homework Statement

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

Doc Al
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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

Are you saying that I have to find the derivative.

Doc Al
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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

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

Doc Al
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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

Doc Al
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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

Doc Al
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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

Doc Al
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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

Doc Al
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the speed would have to be zero right.
Of course. What value of t gives you a speed of zero?

zero

That means that the particle stops at 4 meters right.

Doc Al
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zero
Good. So (a) is done. Now answer part (b).

Doc Al
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That means that the particle stops at 4 meters right.
Right!

Lol. I feel like an idiot thanks man.

now c and d are negative and positive times the particle passes through the origin how do i figure out this

Doc Al
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now c and d are negative and positive times the particle passes through the origin how do i figure out this
You are given the formula for the particle's position (x) as a function of time. Solve for t when x = ?