anonymous
  • anonymous
PLEASE HELP a particle moves so that x, its distance from the origin at time t, t is greater then or equal to 0 is given by x(t)=cos^3(t). the first interval in which the particle is moving to the right it 1.)0< t
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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DLS
  • DLS
Differentiate x(t) wrt t to get v(t)
anonymous
  • anonymous
you mean to find the velocity?
DLS
  • DLS
yes. please write down the expression for the velocity for me.

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More answers

anonymous
  • anonymous
3(cos(t))^2(-sin(t))
DLS
  • DLS
If v(t) is +ve then the object is going forward(let's consider it right..since its 1D motion) and backwards if its negative. Evaluate your v(t) expression for all the given options.
anonymous
  • anonymous
i'm sorry because i really font know how to do this, would that mena i set the v(t) to equal zero to find the roots?
DLS
  • DLS
I'll do the first one for you. When t is from 0 to pi/2 So lets take t = pi/4. \[\Large v(\pi/4) = - 3 \cos^2(\frac{\pi}{4})\sin (\frac{\pi}{4}) <0\] So its not A.
anonymous
  • anonymous
OH! Aright one moment ill do the next one
DLS
  • DLS
sure :)
DLS
  • DLS
On a side note, you can do this manually too, that was just one way I told you. Notice v(t) more carefully. \[\Large v(t) = - 3 \cos^2(t) \sin(t)\] Observe that cos^2(t) is always +ve..so if we somehow get sin(t)<0 then v(t) would be +ve. So it boils down to ..in which quadrant is sine negative ?
anonymous
  • anonymous
\[3\cos(3\pi/4)^2(-\sin(3\pi/4))<0\]
DLS
  • DLS
After you get that answer, then you can go for the first method to cross check your answer..just pickup a number from the range and substitute to check whether its +ve or not.
anonymous
  • anonymous
sin is negative in three and four i think
anonymous
  • anonymous
i'm not sure what you mean
DLS
  • DLS
Yep..so " the first interval " phrase in the question will help your decide between the 3rd and the 4th quadrant
DLS
  • DLS
Since 3rd quadrant is the first interval where sine is negative, therefore t is from pi to 3pi/2. Now if you want to check your answer then choose a number between [pi,3pi/2] ..say 3pi/4. \[\Large v(\frac{3 \pi}{4}) = - 3 \cos^2(\frac{3\pi}{4})\sin(\frac{3\pi}{4}) = -3*(\frac{-1}{\sqrt2})*(\frac{1}{\sqrt2})>0\]
DLS
  • DLS
so our answer is correct and verified.
anonymous
  • anonymous
I'm not sure this is correct or even relevant but when i plug in v(t) into my graphing calculator then the first two waves are below the x axis in the third quadrant is that significant, or an i graphing the wrong equation?
DLS
  • DLS
you don't need to check it, it was an additional step. the question is just asking for the first interval where sine is negative (so that v(t) is positive) and its the 3rd quadrant.
anonymous
  • anonymous
ok and the first interval would be 3pi/2 < t <2pi
DLS
  • DLS
3rd quadrant means theta ranges between: \[\Huge [\pi,\frac{3\pi}{2}]\]
anonymous
  • anonymous
right, becasue when i plug in pi/2 into the v(t) equation the answer is 0
anonymous
  • anonymous
THANK YOU FOR YOUR TIME @DLS

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