butterflydreamer
  • butterflydreamer
Motion Question: A particle is moving along the x-axis. The distance of the particle from the origin O is given by the equation " x = 6t + e^(-4t) where 't' is time in seconds. Explain why the particle will never come to rest.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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butterflydreamer
  • butterflydreamer
So i've worked out the velocity but i'm not sure why the particle will never come to rest? |dw:1433156056342:dw|
IrishBoy123
  • IrishBoy123
\(x(t) = 6t + e^{-4t}\) the the second term will -> 0 as t -> infinity, but will always be > 0, and the first term is plain linear or solving \(\dot x(t) = 6 -4 e^{-4t}\) gives \(t = - \frac{1}{4} ln \frac{3}{2} < 0\)
butterflydreamer
  • butterflydreamer
ohh okay.. i kind of understand it.. But i'm still a bit confused to be honest

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butterflydreamer
  • butterflydreamer
What i did was i think by the solving method? I found that when v = 0 , t = -1/4 *ln(3/2) So because t<0 this means it will never come to rest?
IrishBoy123
  • IrishBoy123
yes, precisely! t > 0
butterflydreamer
  • butterflydreamer
okaaay! Wait LOL why did you write t>0 ?
Michele_Laino
  • Michele_Laino
the only time at which v(t)=0, is such that: \[\Large \exp \left( {4t} \right) = \frac{2}{3}\] which is the time at which the motion is started
butterflydreamer
  • butterflydreamer
ohhh riggghtt! That makes a bit more sense. Thank you both :))! @IrishBoy123 @Michele_Laino
Michele_Laino
  • Michele_Laino
:)
butterflydreamer
  • butterflydreamer
oh hang on! I think i finally completely understand this. Since i solved for t and t<0 when v=0 This can't occur since t>0 because time can't be negative LOL. And since velocity is greater than 0 (v>0) , the particle never stops! The world makes sense :')

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