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butterflydreamer
 one year ago
Motion Question:
A particle is moving along the xaxis. 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.
butterflydreamer
 one year ago
Motion Question: A particle is moving along the xaxis. 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.

This Question is Closed

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0So i've worked out the velocity but i'm not sure why the particle will never come to rest? dw:1433156056342:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(x(t) = 6t + e^{4t}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0ohh okay.. i kind of understand it.. But i'm still a bit confused to be honest

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0What 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
 one year ago
Best ResponseYou've already chosen the best response.1yes, precisely! t > 0

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0okaaay! Wait LOL why did you write t>0 ?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0ohhh riggghtt! That makes a bit more sense. Thank you both :))! @IrishBoy123 @Michele_Laino

butterflydreamer
 one year ago
Best ResponseYou've already chosen the best response.0oh 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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