Position Vector, Displacement and Velocity

In this section we shall see how to describe motion in two dimensions using vectors.

Fig.1: (a) Position vector r

The position vector r of a particle P located in a plane with reference to the origin of an x-y reference frame (Fig. 4.12) is given by

$\mathbf{r}=x \mathbf{\hat{i}}+y \mathbf{\hat{j}}$

where x and y are components of r along x-, and y- axes or simply they are the coordinates of the object.

Fig. 1b: Displacement ∆r and average velocity v of a particle

Suppose a particle moves along the curve shown by the thick line and is at P at time t and P′ at time t′ [Fig. 4.12(b)]. Then, the displacement is :

$\Delta \mathbf{r}= \mathbf{r'}- \mathbf{r}$

and is directed from P to P′ .

We can write Eq. (4.25) in a component form:

$\Delta \mathbf{r}=\left (x' \mathbf{\hat{i}}+y' \mathbf{\hat{j}}  \right )-\left (x \mathbf{\hat{i}}+y \mathbf{\hat{j}}  \right )$

$\Delta \mathbf{r}=\Delta x \mathbf{\hat{i}}+\Delta y \mathbf{\hat{j}}$

where ∆x = x ′ – x, ∆y = y′ – y

VELOCITY

The average velocity ($\mathbf{\bar{v}}$) of an object is the ratio of the displacement and the corresponding time interval:

$\mathbf{\bar{v}}=\frac{\Delta \mathbf{r}}{\Delta t}=\mathbf{i} \frac{\Delta x}{\Delta t}+\mathbf{j} \frac{\Delta y}{\Delta t}$

Or, $\mathbf{\bar{v}}=\bar{v_x}\mathbf{i} +\bar{v_y} \mathbf{j}$

Since $\mathbf{\bar{v}}=\frac{\Delta \mathbf{r}}{\Delta t}$, the direction of the average velocity is the same as that of ∆r (Fig. 4.12). The velocity (instantaneous velocity) is given by the limiting value of the average velocity as the time interval approaches zero:

$\boldsymbol{v}=\lim_{\Delta t\rightarrow 0}\frac{\Delta \mathbf{r}}{\Delta t}=\frac{d\mathbf{r}}{dt}$

The meaning of the limiting process can be easily understood with the help of Fig 4.13(a) to (d). In these figures, the thick line represents the path of an object, which is at P at time t. P$_1$, P$_2$ and P$_3$ represent the positions of the object after times ∆t$_1$, ∆t$_2$, and ∆t$_3$. ∆r$_1$, ∆r$_2$, and ∆r$_3$ are the displacements of the object in times ∆t$_1$, ∆t$_2$, and ∆t$_3$, respectively. 

Fig.2 As the time interval ∆t approaches zero, the average velocity approaches the velocity v. The direction of v is parallel to the line tangent to the path

The direction of the average velocity $\mathbf{\bar{v}}$ is shown in figures (a), (b) and (c) for three decreasing values of ∆t, i.e. ∆t$_1$,∆t$_2$, and ∆t$_3$, (∆t$_1$ > ∆t$_2$ > ∆t$_3$). As ∆t → 0, ∆r → 0 and is along the tangent to the path [Fig. 4.13(d)]. Therefore, the direction of velocity at any point on the path of an object is tangential to the path at that point and is in the direction of motion.

We can express v in a component form:

$\mathbf{\bar{v}}=\frac{d \mathbf{r}}{dt}$

$\mathbf{v}=\lim_{\Delta t\rightarrow 0}\left (\mathbf{i} \frac{\Delta x}{\Delta t}+\mathbf{j} \frac{\Delta y}{\Delta t}  \right )$

     $= \mathbf{i}\lim_{\Delta t\rightarrow 0} \frac{\Delta x}{\Delta t}+\mathbf{j}\lim_{\Delta t\rightarrow 0} \frac{\Delta y}{\Delta t}$

$\mathbf{v}=\mathbf{i} \frac{d x}{dt}+\mathbf{j} \frac{dy}{dt}=v_x\mathbf{i}+v_y\mathbf{j}$

where $v_x=\frac{dx}{dt}$, $v_y=\frac{dy}{dt}$

So, if the expressions for the coordinates x and y are known as functions of time, we can use these equations to find v$_x$ and v$_y$.

The magnitude of v is then

$v=\sqrt{v_x^2+v_y^2}$

and the direction of v is given by the angle θ:

tan θ = $\frac{v_y}{v_x}$, θ = tan $^{-1}\left(\frac{v_y}{v_x}\right)$

v$_x$ , v$_y$ and angle θ are shown in Fig. 4.14 for a velocity vector v at point p.

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