# Physics 9th: Motion | Graphical Representation of Motion

## Physics 9th: Motion | Graphical Representation of Motion

### Graphical Representation of Motion

A Graph represents the relation between two variable quantities in pictorial form. A graph is plotted between two variable quantities.

The quantity that is made to alter at will is called independent variable. The other quantity, which varies as a result of this change is called the dependent variable.

These graphs are used to calculate speed, acceleration and distance.

For the purpose of graphical representation speed and velocity are taken in the same meaning.

There are two types of graphs we draw

1.    Distance –Time graph

2.    Speed – Time graph or Velocity-Time graph

### DISTANCE –TIME GRAPH

(a) When we draw Distance –Time Graphs, time is always taken on the x-axis and distance on the y-axis.

In this graph AB is a straight line parallel to x-axis. In this graph, the distance travelled is not increasing with time. This is a graph of stationary object.

(b) The Distance-Time graph of a body moving at uniform speed is represented by a straight line OA as shown in figure. It can be used to calculate the speed of the body by finding the slope of the graph

$\displaystyle Speed=\frac{Dis\tan ce\,\,travelled}{Time\,\,taken}$

$\displaystyle Speed=\frac{AB}{OB}$

$\displaystyle \frac{AB}{OB}$ is also taken as the slope of the graph, which indicates the speed of the body.

(c) When the Speed – Time graph is a curve, it represents non-uniform motion. This curved line is called a PARABOLA

### SPEED-TIME GRAPH

#### When speed remains constant (zero Acceleration)

If the speed-time graph of a body is a straight line parallel to the time axis, then speed of the body is constant (uniform)

$\displaystyle \text{Speed}=\frac{\text{Distance}}{\text{Time}}$

$\displaystyle \text{Distance}=\text{Speed}\times \text{Time}$

= YZ ´ OZ

Therefore, the area enclosed by the speed-time graph and the x-axis(time-axis), represents the Distance travelled by the body.

#### When speed increases at uniform rate (uniform Acceleration)

The speed –time graph for uniformly changing speed will be a straight line.

Acceleration $\displaystyle =\frac{change\,\,in\,\,speed}{Time\,\,Taken}$

Acceleration $\displaystyle =\frac{BC}{OC}$

Therefore acceleration of a body is given by the slope of the graph $\displaystyle \left( i.e.\frac{BC}{OC} \right)$

The distance travelled = Area under DOBC = $\displaystyle \frac{1}{2}\times OC\times BC$

#### When speed decreases at uniform rate(Retardation)

In speed time graphs, a straight line sloping downwards indicates uniform retardation.

#### When initial speed of the body is not zero

When the initial speed of the body is not zero, the speed-time graph is represented as a straight line forming a trapezium with the time-axis.

Distance travelled     = Area of Trapezium OXYZ

= $\displaystyle \frac{1}{2}$ (sum of parallel sides) (Height)

$\displaystyle =\frac{1}{2}\left( OX+YZ \right)\left( OZ \right)$

$\displaystyle =\frac{1}{2}\left( u+v \right)\left( t \right)$

#### When speed changes at non-uniform rate

Speed-time graph of non-uniform speed (non-uniform acceleration) is a curved line.

# Physics 9th: Motion | Distance and Displacement

## 9th Physics | Motion | Distance and Displacement

### Distance

(a)    The distance travelled by a body is the actual length of the path covered by a moving body irrespective of the direction in which the body travels

(b)    It is a scalar quantity therefore it has only magnitude

### Displacement

When a body moves from one point to another, the shortest distance between the initial position and final position of the body, along with direction, is known as displacement.

It is vector quantity therefore it has magnitude as well as direction.

Distance travelled by a moving body cannot be zero but the displacement of a moving body can be zero.

With increase in time distance can not decrease but displacement can.

### Some Solved Questions Based on Distance and Displacement

#### (c)    ABCDA

Solution:

(a)    Distance travelled from A to B = 10 m

Displacement from A to B = 10 m

Therefore distance and displacement are equal if     the object travels in the same straight path

(b)    Distance travelled from A to C = AB + BC

= 10 + 10

= 20 m

Displacement from A to C     = AC = $\displaystyle 10\sqrt{2}$    m.

Therefore distance travelled is always more than the displacement if there is a change in direction of motion

(c)    Distance travelled from A, back to A = AB + BC + CD + DA

= 10 + 10 + 10 + 10 = 40 m

Displacement from A, back to A = 0 m

This is because the shortest distance between initial position (A) and final position     (A) is zero.

Therefore the displacement of a moving object may be zero but distance travelled cannot be zero.

#### Question:     Calculate the distance and displacement of a body moving in a complete circle.

Solution:

If we take a round trip and reach back at the starting point, then although we have travelled some distance, our final displacement will be zero. This is because the straight line distance between initial and final position is zero

#### Question:     A farmer moves along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of the farmer at the end of 2 minutes 20 seconds?

Solution:

In figure ABCD is a square field of side 10 m.

Time for one round = 40 s

Total time = 2 min 20 s

= (2 ´ 60 + 20) s

= 140 s

Number of rounds completed $\displaystyle =\frac{140}{40}=3.5$

If framer starts from A, it will complete 3 rounds $\displaystyle \left( A\to B\to C\to D\to A \right)$ at A. In the last 0.5 round starting from A, he will finish at C

\ Displacement $\displaystyle =AC$, where $\displaystyle AC=\sqrt{A{{B}^{2}}+B{{C}^{2}}}=\sqrt{{{10}^{2}}+{{10}^{2}}}=10\sqrt{2}$m