Are there kinematic equations for the direction of a vector? I know there are basic kinematic equations for the motion of a particle. Given an initial velocity, an initial position, and a constant acceleration, a future position can be determined.
Are there equations like these for the direction of a vector?
Let's say I have the initial direction of a unit vector. I also know its initial angular velocity and angular acceleration. The angular acceleration is held constant. How can I calculate the new direction of this vector at a future time?
 A: If the angular velocity and (constant) angular acceleration are around 
the same axis, the formulas for angular velocity and angular displacement (i.e., direction) of the vector are basically the same as for motion along a straight line under constant linear acceleration, replacing velocity with angular velocity and replacing distance from the initial point with angle rotated from the initial direction.
If the angular velocity and (constant) angular acceleration are around
different axes then the problem is more complicated.
In general, the unit direction vector 
$\hat v(t)$ as a function of time
is a solution to the differential equation
$$
\frac{d}{dt} \hat v(t) = \omega(t) \times \hat v(t)    \tag1
$$
where $\omega(t)$
(a vector parallel to the axis of rotation)
is the angular velocity at time $t$.
If the angular acceleration $\alpha$ (also a vector) is constant, then
$$
\omega(t) = \omega(0) + t\alpha.   \tag2
$$
If $\alpha$ is parallel to the same axis as $\omega(0)$,
then Equation $(2)$ says that $\omega(t)$ is a scalar multiple of
$\omega(0)$ for all $t$; that is, the axis of rotation is fixed,
$\hat v(t)$ travels in a circular orbit around that axis,
and $\omega(t) \times \hat v(t)$ is always tangent to that circular orbit
at the position of $\hat v(t)$.
You can let the scalar $\theta(t)$ be the cumulative angle through which
$\hat v(t)$ has rotated up to time $t$, and treat
$\omega(t)$ and $\alpha$ as scalars; Equation $(1)$ then has the solution
$\theta(t) = \theta(0) \int_0^t \omega(\tau) \,d\tau
= \theta(0) + t \omega(0) + \frac12 t^2 \alpha$.
But here's the rub: if $\alpha$ and $\omega(0)$ are parallel to 
different axes, then the axis of rotation is constantly changing.
Equation $(1)$ is still true, but its solution is not as simple
as in the other case.
A: The equations of motion generalize to vector equations, e.g.
$$
m \ddot{x} = F(x)
$$
for vectors $\ddot{x}$ and $F$. Or
$$
L = \frac{1}{2}mv^2 - V(x)
$$
for vectors $v$ and $x$.
Your example is somewhat restricted, as it usually assumes that $u$ is constrained to a circle of radius $r$:
$$
u(t) 
= (r\cos \phi(t),r \sin \phi(t))
= r (\cos \phi(t),\sin \phi(t))
= r \, e_r
$$
for a unit vector it would have $r = 1$.
