According to Morin "Classical Mechanics", the derivative of a general vector $\vec A$ in an accelerating frame may be given as $$\frac{d\vec A}{dt}=\frac{\delta \vec A}{\delta t}+\vec \omega\times \vec A, $$ where $\frac{\delta \vec A}{\delta t}$ denotes the rate of change of $\vec A$ measured in the rotating frame. I understand the derivation, but I am somewhat confused about the "general vector" part. The derivation assumes you have position vector, or that's how I interpreted it, because then $\vec r=\vec R+\vec r'$, where $\vec r$ is the position in the inertial frame, $\vec R$ is the position of the origin of the accelerating frame in the inertial frame and $\vec r'$ is the position in the accelerating frame. This is all fine and good, but surely if I take any other vector, say the velocity $\vec v$, this rule no longer applies? So my question: what is different about this "general vector" $\vec A$, or what am I not quite grasping?

According to Morin "Classical Mechanics" (Section 10.1, page 459), the derivative of a general vector $\vec A$ in an accelerating frame may be given as $$\frac{d\vec A}{dt}=\frac{\delta \vec A}{\delta t}+\vec \omega\times \vec A, $$ where $\frac{\delta \vec A}{\delta t}$ denotes the rate of change of $\vec A$ measured in the rotating frame. I understand the derivation, but I am somewhat confused about the "general vector" part. The derivation assumes you have position vector, or that's how I interpreted it, because then $\vec r=\vec R+\vec r'$, where $\vec r$ is the position in the inertial frame, $\vec R$ is the position of the origin of the accelerating frame in the inertial frame and $\vec r'$ is the position in the accelerating frame. This is all fine and good, but surely if I take any other vector, say the velocity $\vec v$, this rule no longer applies? So my question: what is different about this "general vector" $\vec A$, or what am I not quite grasping?