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