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A particle moves along the x-axis so that its acceleration at any time $t\geq0$ is given by $a(t)=12t-4$. At time $t=1$, the velocity of the particle is $v(1)=7$ and its position is $x(1)=4$.

Alright, so part a said find the velocity equation. I got: $v(t)=6t^2-4t+5$.

But part b says: At what values of $t$ does the particle change direction? The particle has to be stopped to change direction, so I started to try and find where the particle stops ($v=0$). But when I set the velocity equal to zero, it ends up being imaginary.

Am I doing something wrong?

(Also: is the position equation $x(t)=2t^3-2t^2+5t-1$?)

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  • $\begingroup$ The position equation is definitely correct. But I find the velocity result odd as well; I'll solve it myself to be sure. $\endgroup$
    – Semiclassical
    Aug 27, 2014 at 21:26
  • $\begingroup$ The velocity result is valid well. To do a physical check on this, interpret them in a different way by replacing $a\to v$ and $v\to x$. Then $v(t)=12t-4$, so we have a constant acceleration of $a=12$. So now the situation is as though we were tossing a ball up and seeing if it reaches a certain height; in this case, the initial velocity is not enough to do so. In the original situation, we could have similarly said that the initial acceleration wasn't enough to bring the particle to a stop. $\endgroup$
    – Semiclassical
    Aug 27, 2014 at 21:36

1 Answer 1

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All your results are correct. Sometimes, questions do that $-$ they ask for something that does not even exist, and you just have to point it out like you just did. Your position and velocity functions are both correct and you are right that there is no real solution to $ v(t) = 0 $.

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  • $\begingroup$ So (out of curiosity) when finding the distance traveled from t=1 to t=3, the distance and displacement will be the same thing, since the particle doesn't stop? $\endgroup$
    – Hello
    Aug 27, 2014 at 21:39
  • $\begingroup$ @Hello Yes, good observation! By the way, do I get an accept? :D $\endgroup$
    – Ahaan S. Rungta
    Aug 27, 2014 at 21:45
  • $\begingroup$ Sure, by the way I posted a new question! $\endgroup$
    – Hello
    Aug 27, 2014 at 22:01

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