Let us consider a point A in a two-dimensional Cartesian co-ordinate system, the position of this point can be described by a specific single vector. This vector is the displacement of that point A with respect to the origin O of the co-ordinate system, as shown in the figure below.

This vector is known as the “position vector” of that point A with respect to the origin O.

Here the position vector is denoted by the \( \vec{OA}=\vec{r} \).

\( \vec{r} \) gives the magnitude as well as the direction of displacement.

If \( \hat{r} \) be the unit vector along \( \vec{r} \), then we can write,

\( \displaystyle{\hat{r}=\frac{\vec{r}}{|\vec{r}|}}\tag{1} \). where, \( |\vec{r}| \) is the magnitude of the vector \( \vec{r} \).

Expression of velocity:

Velocity \( \vec{v} \) is the derivative of the displacement \( \vec{r} \) with respect to time.

Therefore, \( \displaystyle{\vec{v}=\frac{d\vec{r}}{dt}} \)

or, \( \displaystyle{\vec{v}=\frac{d}{dt}(r\hat{r})} \) [using equation (1) ]

or, \( \displaystyle{\vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}} \)

\( or,\ \displaystyle{\vec{v}= \frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{d\theta}\frac{d\theta}{dt}}\tag{2} \)

We know that, \( \hat{r}=\cos\theta\ \hat{i}+\sin\theta\ \hat{j} \)

and, \( \hat{\theta}=-\sin\theta\ \hat{i}+\cos\theta\ \hat{j} \)

[ To know the derivations of expression of \( \hat{r} \) and \( \hat{\theta} \) in terms of the unit vectors \( \hat{i} \) and \( \hat{j} \), CLICK HERE ].

Now derivative of \( \hat{r} \) with respect to \( \theta \) is

\( \displaystyle{\frac{d\hat{r}}{d\theta}=-\sin\theta\ \hat{i}+\cos\theta\ \hat{j}} \)

\(or,\ \displaystyle{\frac{d\hat{r}}{d\theta}=\hat{\theta}}\tag{3} \)

Using equation (3) in equation (2), we get

\( \displaystyle{\vec{v}=\frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}} \)

\(or,\ \displaystyle{\vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta}}\tag{4} \)

where, \( \frac{dr}{dt} \) is denoted by \( \dot{r} \) and \( \frac{d\theta}{dt} \) is denoted by \( \dot{\theta} \).

Components of velocity:

Equation (4) can be written as,

\( \displaystyle{\vec{v}=\vec{v_r}+\vec{v_{\theta}}} \)

Here, radial component of the velocity is \( \displaystyle{\vec{v_r}=\dot{r}\hat{r}} \)

and the transverse component of the velocity is \( \displaystyle{\vec{v_{\theta}}=r\dot{\theta}\hat{\theta}} \).

The magnitude of radial velocity \( \displaystyle{|\vec{v_r}|=\dot{r}} \)

and the magnitude of the transverse velocity \( \displaystyle{|\vec{v_{\theta}}|=r\dot{\theta}} \).

Again the angular velocity \( \displaystyle{\omega=\frac{d\theta}{dt}=\dot{\theta}} \)

therefore, \( \displaystyle{|\vec{v_{\theta}}|=r\omega} \)

Now the magnitude of the velociy \( \vec{v} \) is given by,

\( \displaystyle{|\vec{v}|=\sqrt{[{\dot{r}}^2+{r}^2{\dot{\theta}}^2]}} \)