# 1.14 Velocity  (Page 5/6)

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$\begin{array}{l}\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{dx}{dt}\mathbf{i}+\frac{dy}{dt}\mathbf{j}\\ \mathbf{v}={v}_{x}\mathbf{i}+{v}_{y}\mathbf{j}\\ v=|\mathbf{v}|=\surd \left({v}_{x}^{2}+{v}_{y}^{2}\right)\end{array}$

Similarly, one dimensional motion (For example : x – direction) is described by one of the components of velocity.

$\begin{array}{l}\mathbf{v}=\frac{d\mathbf{r}}{dt}=\frac{dx}{dt}\mathbf{i}\\ \mathbf{v}={v}_{x}\mathbf{i}\\ v=|\mathbf{v}|={v}_{x}\end{array}$

## Few words of caution

Study of kinematics usually brings about closely related concepts, terms and symbols. It is always desirable to be precise and specific in using these terms and symbols. Following list of the terms along with their meaning are given here to work as reminder :

1: Position vector : r : a vector specifying position and drawn from origin to the point occupied by point object

2: Distance : s : length of actual path : not treated as the magnitude of displacement

3: Displacement : AB or Δ r : a vector along the straight line joining end points A and B of the path : its magnitude, | AB | or |Δ r | is not equal to distance, s.

4: Difference of position vector : Δ r : equal to displacement, AB . Direction of Δ r is not same as that of position vector ( r ).

5: Magnitude of displacement : | AB | or |Δ r |: length of shortest path.

6: Average speed : ${v}_{a}$ : ratio of distance and time interval : not treated as the magnitude of average velocity

7: Speed : v : first differential of distance with respect to time : equal to the magnitude of velocity, | v |

8: Average velocity : ${\mathbf{v}}_{\mathbf{a}}$ : ratio of displacement and time interval : its magnitude, $|{\mathbf{v}}_{\mathbf{a}}|$ is not equal to average speed, ${v}_{a}$ .

9: Velocity : v : first differential of displacement or position vector with respect to time

## Summary

The paragraphs here are presented to highlight the similarities and differences between the two important concepts of speed and velocity with a view to summarize the discussion held so far.

1: Speed is measured without direction, whereas velocity is measured with direction. Speed and velocity both are calculated at a position or time instant. As such, both of them are independent of actual path. Most physical measurements, like speedometer of cars, determine instantaneous speed. Evidently, speed is the magnitude of velocity,

$\begin{array}{l}v=|\mathbf{v}|\end{array}$

2: Since, speed is a scalar quantity, it can be plotted on a single axis. For this reason, tangent to distance – time curve gives the speed at that point of the motion. As $ds=vXdt$ , the area under speed – time plot gives distance covered between two time instants.

3: On the other hand, velocity requires three axes to be represented on a plot. It means that a velocity – time plot would need 4 dimensions to be plotted, which is not possible on three dimensional Cartesian coordinate system. A two dimensional velocity and time plot is possible, but is highly complicated to be drawn.

4: One dimensional velocity can be treated as a scalar magnitude with appropriate sign to represent direction. It is, therefore, possible to draw one dimension velocity – time plot.

5: Average speed involves the length of path (distance), whereas average velocity involves shortest distance (displacement). As distance is either greater than or equal to the magnitude of displacement,

$\begin{array}{l}s\ge |\Delta \mathbf{r}|\mathrm{and}{v}_{a}\ge |{\mathbf{v}}_{\mathbf{a}}|\end{array}$

## Exercises

The position vector of a particle (in meters) is given as a function of time as :

$\begin{array}{l}\mathbf{r}=2t\mathbf{i}+2{t}^{2}\mathbf{j}\end{array}$

Determine the time rate of change of the angle “θ” made by the velocity vector with positive x-axis at time, t = 2 s.

Solution : It is a two dimensional motion. The figure below shows how velocity vector makes an angle "θ" with x-axis of the coordinate system. In order to find the time rate of change of this angle "θ", we need to express trigonometric ratio of the angle in terms of the components of velocity vector. From the figure :

$\begin{array}{l}\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}\end{array}$

As given by the expression of position vector, its component in coordinate directions are :

$\begin{array}{l}x=2t\phantom{\rule{4pt}{0ex}}\mathrm{and}\phantom{\rule{4pt}{0ex}}y=2{t}^{2}\end{array}$

We obtain expression of the components of velocity in two directions by differentiating "x" and "y" components of position vector with respect to time :

$\begin{array}{l}{v}_{x}=2\phantom{\rule{4pt}{0ex}}\mathrm{and}\phantom{\rule{4pt}{0ex}}{v}_{y}=4t\end{array}$

Putting in the trigonometric function, we have :

$\begin{array}{l}\mathrm{tan}\theta =\frac{{v}_{y}}{{v}_{x}}=\frac{4t}{2}=2t\end{array}$

Since we are required to know the time rate of the angle, we differentiate the above trigonometric ratio with respect to time as,

$\begin{array}{l}{\mathrm{sec}}^{2}\theta \frac{d\theta }{dt}=2\end{array}$

$\begin{array}{l}⇒\left(1+{\mathrm{tan}}^{2}\theta \right)\frac{d\theta }{dt}=2\\ ⇒\left(1+4{t}^{2}\right)\frac{d\theta }{dt}=2\\ ⇒\frac{d\theta }{dt}=\frac{2}{\left(1+4{t}^{2}\right)}\end{array}$

At t = 2 s,

$\begin{array}{l}⇒\frac{d\theta }{dt}=\frac{2}{\left(1+4\phantom{\rule{2pt}{0ex}}x\phantom{\rule{2pt}{0ex}}{2}^{2}\right)}=\frac{2}{17}\phantom{\rule{2pt}{0ex}}\mathrm{rad}/s\end{array}$

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