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 $x(t)=2t^3-2t^2+5t-1$?)
Best Answer
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 $.