A particle is speeding up whenever $v(t)$ and $a(t)$ share the same sign and is slowing down whenever they do not share the same signs. Since $0$ is neither positive nor negative, would the following be true? Whenever $v(t)=0$ and $a(t)=0$ the particle is at rest. When $v(t)=0$ and $a(t)\ne0 \ $or$ \ v(t)\ne0$ and $a(t)=0$ the particle is changing direction.
Here is the graph of the particle's position (red), velocity (blue), and acceleration (green) on the problem I worked on for reference.
An object being "at rest" means its velocity is zero.
So the first scenario when $v(t) = a(t) = 0$, the particle is considered to be at rest at time $t$, since its velocity is zero.
In the case that $v(t) \neq 0$ but $a(t) = 0$, the particle is not changing direction, as some acceleration (change in velocity) would be required for this to happen.
Now for the scenario when $v(t) = 0$ but $a(t) \neq 0$. If the motion is one-dimensional, the particle must be changing direction, which means that its velocity is changing sign. This is because if it were not changing directions, then its velocity would have the same sign before and after time $t$, implying that the velocity would achieve a turning point at time $t$, but this means that its acceleration (derivative of velocity) is zero, contrary to assumption.
This is also true in higher dimensions, in which you can apply the one dimensional argument.