butterflydreamer one year ago 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.

1. butterflydreamer

So i've worked out the velocity but i'm not sure why the particle will never come to rest? |dw:1433156056342:dw|

2. 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 $$t = - \frac{1}{4} ln \frac{3}{2} < 0$$

3. butterflydreamer

ohh okay.. i kind of understand it.. But i'm still a bit confused to be honest

4. 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?

5. IrishBoy123

yes, precisely! t > 0

6. butterflydreamer

okaaay! Wait LOL why did you write t>0 ?

7. 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

8. butterflydreamer

ohhh riggghtt! That makes a bit more sense. Thank you both :))! @IrishBoy123 @Michele_Laino

9. Michele_Laino

:)

10. 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 :')