The position $\vec r$ of a particle moving in an xy plane is given by $\vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $\vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
What is the angle between the positive direction of the x axis and a line tangent to the particle's path at $t = 2.10 s$? (counterclockwise from the +x-axis)
Is it asking for a degree based on $r$, $v$, or $a$?
Since it mentioned $s$, I thought it referred to $v$. So, I did:
$\frac{-333.4}{10.28}$ times tan inverse. That got $-88.1$, then I added $180$ degrees since it was in Q4, for $91.76$ degrees. It says it's wrong, though.
"Direction" refers to velocity. So calculate the derivative vector at time $t=2.1$ seconds.