Instantaneous velocity and speed problem

AI Thread Summary
The position function of a particle is given by x(t) = 4.0 - 6.0t^2. To determine when the particle momentarily stops, the velocity must be found by taking the derivative, resulting in v(t) = -12t. The particle stops when v(t) = 0, which occurs at t = 0 seconds, indicating it is at the position x = 4.0 meters. The discussion also touches on finding the times when the particle passes through the origin by solving for t when x = 0. Overall, the key takeaway is that the particle stops at 4 meters at time t = 0 seconds.
afcwestwarrior
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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

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
 
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afcwestwarrior said:
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.
 
afcwestwarrior said:
Are you saying that I have to find the derivative.
That would be most wise. :wink:
 
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
 
Last edited:
then i plug 12 into the equation and i get 4.0-6(12)^2=-860m
 
afcwestwarrior said:
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.
 
  • #10
hmm. would it be -12=x/t
 
  • #11
afcwestwarrior said:
So how do I do that.
Do what? You already found the velocity, you just mislabeled it "x" instead of "v".
 
  • #12
oh ok lol, so that's all i do, v=-12t
 
  • #13
afcwestwarrior said:
oh ok lol, so that's all i do, v=-12t
Good. Now go back and answer part (a).
 
  • #14
Ok so where do I plug in the -12t,
x=4.0-6.0t^2

would it be 4-6(-12)^2=-860
 
  • #15
afcwestwarrior said:
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?)
 
  • #16
the speed would have to be zero right.
 
  • #17
-12(0)= it has to be zero
 
  • #18
afcwestwarrior said:
the speed would have to be zero right.
Of course. What value of t gives you a speed of zero?
 
  • #19
zero
 
  • #20
That means that the particle stops at 4 meters right.
 
  • #21
afcwestwarrior said:
zero
Good. So (a) is done. Now answer part (b).
 
  • #22
afcwestwarrior said:
That means that the particle stops at 4 meters right.
Right!
 
  • #23
Lol. I feel like an idiot thanks man.
 
  • #24
now c and d are negative and positive times the particle passes through the origin how do i figure out this
 
  • #25
afcwestwarrior said:
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 = ?
 
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