# What does the derivative of unit vector of velocity with respect to time represent?

let an object move with a constant accelration a. in my book,the following derivatve is said to be non-constant(variable). $$\frac{d[\frac{v}{|v|}]}{dt}$$ what does this mean? as far as i can think,it should mean the rate of change of direction of velocity(as $$\frac{v}{|v|}$$ is a unit vector)if yes then what would its value be like

Bonus Question what would the graph of a body with negative accelration and initial positive velocity with respect to time look like?i think it should be a straight line on positive axis and then a sudden shift to a line on negative axis.

• the derivative of the unit tangent vector is actually the normal vector (normal to velocity & trajectory) Jun 15, 2021 at 18:41

Lets denote, $$\vec v = v~\hat e$$, so derivative of it with respect to time will be :
\tag 1 \begin{align} \frac{d}{dt}(v~\hat e) &=\\ v\frac{d~\hat e}{dt}+\hat e\frac{d~v}{dt}&=\\ v~\hat e_{\perp} + \hat e~a_{\parallel} &=\\ \vec a_{\perp} + \vec a_{\parallel} \end{align}
So, any acceleration can be decomposed into parallel and perpendicular parts wrt trajectory. If object moves linearly then there's no centripetal acceleration. If it moves in a circle,- then there's no parallel part of acceleration. Consequently, if object moves in a spiral (say elementary particle in a non-linear accelerator) then it has both $$\vec a_{\perp},\vec a_{\parallel}$$ acceleration components.