anonymous
  • anonymous
The velocity of a particle moving along the x-axis is v(t) = t2 – 2t, with t measured in minutes and v(t) measured in feet per minute. To the nearest foot find the total distance travelled by the particle from t = 0 to t = 3 minutes.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
@zepdrix pretty sure I just integrate, just wanted to confirm
zepdrix
  • zepdrix
Ummm
anonymous
  • anonymous
https://gyazo.com/1ad2f2f534c3e4c134c1e809420a3e7a

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More answers

welshfella
  • welshfella
yes integrate between t = 0 and t = 3
zepdrix
  • zepdrix
I mean, yes, integrating will get you your position function, good. The issue is, if the particle moved backwards and then also forwards during those first 3 minutes, then we might have an issue. The s(t) is going to tell us about displacement, not total distance. So maybe we need to look at critical points first.
zepdrix
  • zepdrix
Maybe I'm mistaken hehe, rusty
anonymous
  • anonymous
It says move along so I doubt it means back and forth.
anonymous
  • anonymous
Anyways I got 0.
welshfella
  • welshfella
yes it does move in 2 directions 0 is the displacement no the distance travelled
zepdrix
  • zepdrix
Setting v(t)=0 gives us:\[\large\rm 0=t^2-2t\qquad\to\qquad t=0,~t=2\] These are the times when the velocity changes from positive to negative. If you do a sign chart or something similar, you see that the particle was moving backwards from t=0 to t=2. So maybe what you need to do is break up the integral.
welshfella
  • welshfella
yes
zepdrix
  • zepdrix
\[\large\rm total~distance=\left|\int\limits_0^2 t^2-2t~dt\right|+\int_2^3 t^2-2t~dt\] Somethingggg like that.
welshfella
  • welshfella
yes
zepdrix
  • zepdrix
Maybe that's more complicated :) Just integrate,\[\large\rm s(t)=\frac{1}{3}t^3-t^2+c\] The particle is moving backwards from t=0 to t=2, so calculate the distance it covered in that interval,\[\large\rm s(2)-s(0)\]And then calculate the distance it traveled while moving forward,\[\large\rm s(3)-s(2)\]and add those quantities together. Same thing, just without the ugly integrals I guess.
anonymous
  • anonymous
Nah I like the first one better, actually how I remember doing it. I got 2.667
welshfella
  • welshfella
the graph looks something like below so you need to calculate the 2 areas |dw:1449618711512:dw|
welshfella
  • welshfella
- 2 integrals
anonymous
  • anonymous
-2 integrals?
zepdrix
  • zepdrix
Hmm I came up with positive 4 for the distance. I should check my work again though >.<
anonymous
  • anonymous
Not sure how you got 4 :/
zepdrix
  • zepdrix
Ah yes :) 2.667 my bad
anonymous
  • anonymous
:)
anonymous
  • anonymous
Thanks. Unfortunately, I didn't do so well on the ones I did on my own. Confused the graph of f' with the graph of f
zepdrix
  • zepdrix
Oo that's no good :o

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