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Describing Motion Around Us Class 9 Science Notes

Describing Motion Around Us

Everything in our universe, from the largest galaxies to the smallest particles, is in a state of motion. To understand this complex world, scientists start by studying simpler, idealized forms of motion. The three basic types are:

Linear motion: Moving in a straight line.
Circular motion: Moving in a circle.
Oscillatory motion: Moving back and forth.

In this chapter, we will explore linear motion and uniform circular motion in detail. We'll build on your knowledge of distance, time, and speed, and introduce new concepts like displacement, velocity, and acceleration. We will learn to describe motion not just with words, but also with numbers, graphs, and equations.

Motion in a Straight Line

When an object moves along a straight path, we call it linear motion. This is the simplest type of motion to study. You can see it all around you: a car on a straight highway, a ball falling vertically, or a train on a straight track.

To describe an object's motion, we first need to know its position at different times.

Describing position

To describe the position of an object, you need a reference point, also called the origin. An object's position is its distance and direction from this fixed reference point.

An object is in motion if its position changes with time, relative to the reference point.
An object is at rest if its position does not change with time, relative to the reference point.

For motion in a straight line, there are only two directions: forward and backward. We can represent these directions using positive (+) and negative (-) signs. Typically, positions to the right of the origin are considered positive, and positions to the left are negative.

Note

An instant of time is a single point in time, like a single reading on a clock (e.g., 3:15 PM). A time interval is the duration between two instants of time (e.g., the 10 minutes between 3:15 PM and 3:25 PM).

Physical quantities can be classified as scalars or vectors.

Scalars are quantities that can be described by just a numerical value (magnitude), like distance or time.
Vectors are quantities that require both a magnitude and a direction to be fully described, like displacement.

Distance travelled and displacement

Let's understand two important quantities used to describe the overall motion of an object.

Distance travelled is the total length of the path covered by an object. It is a scalar quantity and only has a numerical value (magnitude).
Displacement is the net change in an object's position. It is the shortest distance between the initial and final positions. Displacement is a vector quantity, meaning it has both magnitude and direction.

The SI unit for both distance and displacement is the metre (m).

Example

An athlete starts at point O (origin), runs to point A (100 m mark), and then runs back to point B (40 m mark).

Distance Travelled: The athlete runs 100 m forward (O to A) and then 60 m back (A to B). Total distance = $100 \text{ m} + 60 \text{ m} = 160 \text{ m}$ .
Displacement: The athlete's starting position was O (0 m) and the final position is B (40 m). Displacement = Final position - Initial position = $40 \text{ m} - 0 \text{ m} = 40 \text{ m}$ in the positive direction.

In this case, the distance travelled (160 m) is not equal to the magnitude of the displacement (40 m).

Note

The total distance travelled and the magnitude of displacement are equal only if the object moves in a straight line in a single direction, without turning back. The magnitude of displacement is always less than or equal to the total distance travelled.

Average speed and average velocity

To describe how fast or slow an object is moving, we use the concepts of speed and velocity.

Average speed is the total distance travelled by an object divided by the total time interval. It tells us the average rate at which an object covers distance. Since distance has no direction, average speed also has no direction.

$\text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}}$

Average velocity is the displacement of an object divided by the total time interval. It tells us the rate at which an object's position changes, and in which direction. Since displacement has a direction, average velocity also has a direction.

$\text{average velocity} = \frac{\text{displacement}}{\text{time interval}}$

If we represent average velocity by $v_{av}$ , displacement by $s$ , and time interval by $t$ , the formula is:

$v_{av} = \frac{s}{t}$

The SI unit for both average speed and average velocity is metres per second ( $\text{m s}^{-1}$ or m/s). Another common unit is kilometres per hour ( $\text{km h}^{-1}$ ).

Uniform motion is when an object travels equal distances in equal intervals of time. In this case, the object moves at a constant speed.
Non-uniform motion is when an object travels unequal distances in equal intervals of time. This means its speed is changing.

Example

India's Scientific Contributions The concept of speed is ancient. The following problem is from the Ganitakaumudi (14th century CE). Consider two postmen. They start walking towards each other from a distance of 210 yojanas. One travels 9 yojanas per day and the other covers 5 yojanas per day. Can you determine in how many days they will meet each other?

Given

Total distance = 210 yojanas
Speed of first postman = 9 yojanas/day
Speed of second postman = 5 yojanas/day

To Find

The number of days until they meet.

Solution

First, find the combined distance they cover in one day. Combined distance per day = $9 \text{ yojanas} + 5 \text{ yojanas} = 14 \text{ yojanas}$

To meet, they must cover the total distance of 210 yojanas together. Time taken = $\frac{\text{Total distance}}{\text{Combined distance per day}}$

$\text{Time} = \frac{210}{14} = 15 \text{ days}$

Final Answer The postmen will meet each other after 15 days.

Example

Sarang takes 50 seconds to swim from one end of a 25 m pool to the other end and back. Find his average speed and average velocity.

Given

Length of pool = 25 m
Total time taken, $t = 50$ s

To Find

Average speed
Average velocity

Solution

First, let's determine the total distance and displacement. Sarang swims 25 m to one end and 25 m back. Total distance travelled = $25 \text{ m} + 25 \text{ m} = 50 \text{ m}$

Sarang starts and ends at the same point. Displacement = $0 \text{ m}$

Now, we can calculate the average speed and average velocity.

Average speed: $\text{average speed} = \frac{\text{total distance travelled}}{\text{time interval}} = \frac{50 \text{ m}}{50 \text{ s}} = 1 \text{ m s}^{-1}$

Average velocity: $\text{average velocity} = \frac{\text{displacement}}{\text{time interval}} = \frac{0 \text{ m}}{50 \text{ s}} = 0 \text{ m s}^{-1}$

Final Answer Sarang's average speed is $1 \text{ m s}^{-1}$ , while his average velocity is $0 \text{ m s}^{-1}$ .

Average acceleration

When the velocity of an object changes, we say it is accelerating. The jolts you feel when a car starts or stops are due to acceleration.

Average acceleration is the rate of change of velocity. It is calculated by dividing the change in velocity by the time interval over which the change occurs.

$\text{average acceleration} = \frac{\text{change in velocity}}{\text{time interval}} = \frac{\text{final velocity - initial velocity}}{\text{time interval}}$

If an object's velocity changes from an initial value $u$ at time $t_1$ to a final value $v$ at time $t_2$ , the average acceleration $a$ is:

$a = \frac{v-u}{t_2 - t_1}$

The SI unit of acceleration is metres per second squared ( $\text{m s}^{-2}$ or $\text{m/s}^2$ ).

Like velocity, acceleration is a vector quantity and has direction.

If an object's speed is increasing, the acceleration is in the same direction as the velocity.
If an object's speed is decreasing (deceleration), the acceleration is in the opposite direction to the velocity.

Example

A bus is moving on a highway at

36 \text{ km h}^{-1}

. The driver accelerates for 10 s, and the velocity increases to

54 \text{ km h}^{-1}

. Later, the driver applies the brakes, and the bus stops in 5 s. Find the average acceleration during (i) acceleration and (ii) braking.

Given

(i) Acceleration phase:

Initial velocity, $u = 36 \text{ km h}^{-1}$
Final velocity, $v = 54 \text{ km h}^{-1}$
Time interval, $t = 10$ s

(ii) Braking phase:

Initial velocity, $u = 54 \text{ km h}^{-1}$
Final velocity, $v = 0 \text{ m s}^{-1}$
Time interval, $t = 5$ s

To Find

(i) Average acceleration when the accelerator was pressed. (ii) Average acceleration when the brakes were pressed.

Solution

First, we must convert the velocities from $\text{km h}^{-1}$ to $\text{m s}^{-1}$ . To convert, we use the factor $\frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18}$ .

$36 \text{ km h}^{-1} = 36 \times \frac{1000}{3600} \text{ m s}^{-1} = 10 \text{ m s}^{-1}$
$54 \text{ km h}^{-1} = 54 \times \frac{1000}{3600} \text{ m s}^{-1} = 15 \text{ m s}^{-1}$

(i) When the driver presses the accelerator The variables are: $u = 10 \text{ m s}^{-1}$ , $v = 15 \text{ m s}^{-1}$ , $t = 10$ s.

$a = \frac{v - u}{t} = \frac{15 \text{ m s}^{-1} - 10 \text{ m s}^{-1}}{10 \text{ s}} = \frac{5 \text{ m s}^{-1}}{10 \text{ s}} = 0.5 \text{ m s}^{-2}$ The positive sign indicates acceleration is in the direction of velocity.

(ii) When the driver presses the brake The variables are: $u = 15 \text{ m s}^{-1}$ , $v = 0 \text{ m s}^{-1}$ , $t = 5$ s.

$a = \frac{v - u}{t} = \frac{0 \text{ m s}^{-1} - 15 \text{ m s}^{-1}}{5 \text{ s}} = \frac{-15 \text{ m s}^{-1}}{5 \text{ s}} = -3 \text{ m s}^{-2}$ The negative sign indicates the acceleration is opposite to the direction of velocity.

Example

When an object is dropped, its velocity increases as it falls. The table below shows the velocity of a dropped object at different times. Find the average acceleration for each one-second interval.

Time (s)	Velocity (m/s)
0	0
1	9.8
2	19.6
3	29.4
4	39.2

Solution

We calculate the acceleration for each interval using $a = \frac{\Delta v}{\Delta t}$ .

Between 0 s and 1 s: $a = \frac{(9.8 - 0) \text{ m s}^{-1}}{(1 - 0) \text{ s}} = 9.8 \text{ m s}^{-2}$
Between 1 s and 2 s: $a = \frac{(19.6 - 9.8) \text{ m s}^{-1}}{(2 - 1) \text{ s}} = 9.8 \text{ m s}^{-2}$
Between 2 s and 3 s: $a = \frac{(29.4 - 19.6) \text{ m s}^{-1}}{(3 - 2) \text{ s}} = 9.8 \text{ m s}^{-2}$
Between 3 s and 4 s: $a = \frac{(39.2 - 29.4) \text{ m s}^{-1}}{(4 - 3) \text{ s}} = 9.8 \text{ m s}^{-2}$

Final Answer The average acceleration is constant and is equal to $9.8 \text{ m s}^{-2}$ in every interval. This constant acceleration is due to Earth's gravity and is denoted by g. Since the velocity is increasing, the acceleration is in the direction of motion (downwards).

Graphical Representation of Motion

Graphs are powerful tools for visualizing and analyzing motion. They show how quantities like position and velocity change over time.

Note

For the graphs discussed here, we consider motion in a straight line in one direction only. In this special case, distance equals the magnitude of displacement, and speed equals the magnitude of velocity.

Position-time graphs

A position-time graph plots an object's position (on the y-axis) against time (on the x-axis).

Shape of the graph: The shape tells us about the object's motion.
- A straight line indicates motion with constant velocity.
- A curved line indicates accelerated motion (velocity is changing).
- A horizontal line (parallel to the time axis) indicates the object is at rest; its position is not changing.
Slope of the graph: The slope of a position-time graph gives the velocity of the object. A steeper slope means a higher velocity. $\text{Slope} = \frac{\text{Change in Position}}{\text{Change in Time}} = \text{Velocity}$

Example

For a vehicle starting from rest and speeding up, the position and time data are given. Plot the position-time graph.

Time (s)	Position (m)
0	0
2	1
4	4
6	9
8	16
10	25
12	36

Answer

When these points are plotted on a graph with Time on the X-axis and Position on the Y-axis, they do not form a straight line. Instead, they form a curve that gets steeper over time. This curved shape indicates that the vehicle is in accelerated motion—its velocity is increasing.

Example

What does the position-time graph of a vehicle that is stationary at 40 m from the origin look like?

Answer

The graph would be a horizontal straight line at the position value of 40 m. The position is 40 m at all times, so the line is parallel to the time axis. This indicates the vehicle is at rest.

Example

The position-time graphs for two objects, A and B, are shown as straight lines. Which object has a higher velocity?

Answer

By observing the graphs, the line for object B is steeper than the line for object A. The slope of a position-time graph represents velocity. A steeper slope means a greater velocity. Therefore, the velocity of object B is higher than that of object A.

Velocity-time graphs

A velocity-time graph plots an object's velocity (on the y-axis) against time (on the x-axis).

Shape of the graph:
- A horizontal line indicates motion with constant velocity (zero acceleration).
- A straight, sloped line indicates constant acceleration. If the slope is positive (upward), it's constant acceleration. If the slope is negative (downward), it's constant deceleration.
Slope of the graph: The slope of a velocity-time graph gives the acceleration of the object. $\text{Slope} = \frac{\text{Change in Velocity}}{\text{Change in Time}} = \text{Acceleration}$
Area under the graph: The area enclosed by the velocity-time graph and the time axis gives the displacement of the object during that time interval. $\text{Area} = \text{Velocity} \times \text{Time} = \text{Displacement}$

Kinematic Equations for Motion in a Straight Line with Constant Acceleration

For the special case of an object moving in a straight line with constant acceleration, we can use a set of three equations, known as the kinematic equations, to describe its motion.

Let:

$u$ = initial velocity
$v$ = final velocity
$a$ = constant acceleration
$t$ = time interval
$s$ = displacement

The three kinematic equations are:

Velocity-time relation: $v = u + at$
Position-time relation: $s = ut + \frac{1}{2}at^2$
Position-velocity relation: $v^2 = u^2 + 2as$

These equations allow you to calculate any of the five variables if you know the values of at least three of them.

Note

These equations are valid only when the acceleration is constant. When solving problems, be mindful of the signs (+ or -) for velocity, displacement, and acceleration, as they indicate direction.

Example

A car moving on a highway applies its brakes, which cause a constant acceleration of

-4 \text{ m s}^{-2}

. Calculate the stopping distance if the initial velocity was (i)

54 \text{ km h}^{-1}

and (ii)

108 \text{ km h}^{-1}

.

Given

Acceleration, $a = -4 \text{ m s}^{-2}$
Final velocity, $v = 0 \text{ m s}^{-1}$ (since the car comes to a stop)

To Find

The stopping distance, $s$ , for two different initial velocities.

Formula

We can use the position-velocity relation, as it connects $u, v, a$ , and $s$ . $v^2 = u^2 + 2as$

Solution

First, convert the initial velocities to $\text{m s}^{-1}$ . (i) $u = 54 \text{ km h}^{-1} = 15 \text{ m s}^{-1}$ (ii) $u = 108 \text{ km h}^{-1} = 30 \text{ m s}^{-1}$

Rearranging the formula to solve for $s$ : $0 = u^2 + 2as \implies -2as = u^2 \implies s = \frac{-u^2}{2a}$ Or more simply: $s = \frac{v^2 - u^2}{2a}$ $s = \frac{(0)^2 - u^2}{2 \times (-4)} = \frac{-u^2}{-8} = \frac{u^2}{8}$

(i) For $u = 15 \text{ m s}^{-1}$ $s = \frac{(15)^2}{8} = \frac{225}{8} = 28.125 \text{ m}$

(ii) For $u = 30 \text{ m s}^{-1}$ $s = \frac{(30)^2}{8} = \frac{900}{8} = 112.5 \text{ m}$

Final Answer The stopping distance is approximately (i) 28.1 m and (ii) 112.5 m. This shows that doubling the speed more than doubles the stopping distance, highlighting the importance of maintaining a safe speed.

Motion in a Plane

When an object moves in a plane, its motion is described in two dimensions. Examples include a car turning a corner or a satellite orbiting the Earth.

Uniform circular motion

When an object moves in a circular path, its motion is called circular motion.

A special case of this is uniform circular motion, which occurs when an object moves along a circular path at a constant speed.

Even though the speed is constant, the object is still accelerating. Why? Because velocity is a vector, with both magnitude (speed) and direction. In circular motion, the direction of the object's movement is constantly changing. At any point on the circle, the velocity is directed along the tangent to the circle at that point. Since the direction of velocity is continuously changing, the object is accelerating.

Note

Acceleration can be caused by a change in the magnitude of velocity (speed), a change in the direction of velocity, or both. In uniform circular motion, acceleration is due solely to the continuous change in direction.

For an object in uniform circular motion with radius $R$ that takes time $T$ to complete one revolution:

The distance travelled in one revolution is the circumference, $2\pi R$ .
The displacement after one revolution is $0$ , because it returns to the starting point.
The average speed is $v_{av} = \frac{2\pi R}{T}$ .
The average velocity over one revolution is $0$ , because the displacement is zero.

Describing Motion Around UsClass 9 Science Notes

Describing Motion Around Us

Motion in a Straight Line

Describing position

Distance travelled and displacement

Average speed and average velocity

Given

To Find

Solution

Given

To Find

Solution

Average acceleration

Given

To Find

Solution

Solution

Graphical Representation of Motion

Position-time graphs

Answer

Answer

Answer

Velocity-time graphs

Kinematic Equations for Motion in a Straight Line with Constant Acceleration

Given

To Find

Formula

Solution

Motion in a Plane

Uniform circular motion