Confused about linear displacement in circular motion We know that velocity is the derivative of displacement w.r.t. time. Keeping this mind, suppose that we are traveling across a circle where we go from point A to point B. In textbooks, it is said that the linear displacement is the arc length $AB$ given as, $s=r\theta$ where $r$ is the radius and $\theta$ is the angular displacement.But isn't displacement the straight line distance from two points? So shouldn't $s$ be the length of the line segment $AB$ instead of arc length? I am really confused about this matter since all the laws in circular motion are derived taking arc length as displacement.
 A: The definition of velocity is: $$\mathbf v = \frac{d\mathbf r}{dt} = \lim_{\Delta t\to0}\frac{\mathbf {\Delta r}}{\Delta t}$$  In the case of circular motion, when $\mathbf {\Delta r}$ (whose modulus is the lenght of straight line between 2 points in the arc) is too small), that modulus can be approximated by the length of the arc $\Delta s$. So, the speed $v$, the modulus of $\mathbf v$ is: $$v = \frac{ds}{dt}$$
A: You are mixing two things: average velocity (that is a vector) and average speed (that is a number). They are the same only if the trajectory is a straight line, and you do not change direction (someone could consider "speed with sign", but here I consider speed as always positive.. sign is reminiscent of being a vector in 1D, and here I hant to maximise the difference between the two concepts). After all, I use as definition of speed the same definition used by the "speedometer". The average velocity is different, and can not be extracted by monitoring the speedometer only, unless you are on a straight line without inverting the direction of your motion.
The important point:: If $s$ is the vector from A to B, then you obtain the average velocity. If $s$ is just the lentgh of the trajectory (calculated by flollowing the path!), then you have the speed (just a number).
The second case (speed) is the one you use when you drive your car: you may turn many times, and even drive the same road in opposite directions.. but in the end (sone cars) tell you the total "distance" (which is in reality the lenght of the path you followed) and the  total time.
Driving down the same road two times in opposite directions cancels out when it comes to calculate the average velocity.
A final funny thing: since you are asking about the circular motion, assume  to have the usual uniform circular motion. The speed is constant (it is just "angular velocity times radius), but the instantaneous velocity changes (its direction changes). Also the average velocity up to a certain moment in time changes: in the first instants of the motion it will be almost equal to the instantaneous velocity, but after exactly one cycle (you are back to the initial position) it will be zero! When you start the second cycle, the average velocity won't be again infinitesimally close to the instantaneous velocity, but rather close to zero (by continuity). The average velocity oscillates, dropping to the "zero vector" any time you pass from the initial point, and the amplitude of the oscillations decreases in time. The speed is always constant.
