If the position of a particle is given by $x(t) = 47t − 3t^3$, where $x$ is in meters and $t$ is in seconds, Graph $x(t), v(t),$ and $a(t).$
I'm not sure how to approach this. Should I differentiate the equation to get $v$, then graph that? But what would be the $x$ and $y$ points?
You have $v(t)=-9t^2+47$ and $a(t)=-18t$. Then you plot it in the same manner as $x(t)$. You plot for example $y=v(t)$ as a function of $t$.