# Class 9 Science Chapter 7 Motion NCERT Notes

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## Chapter 7 Motion Class 9 Science CBSE NCERT Notes

Motion: If an object changes its position with respect to a reference point with elapse of time, the object is said to be in motion.

Rest: When an object does not change its position with respect to a reference point with elapse of time, the object is said to be in rest.

Example: When a vehicle’s position with regard to an electric pole (a reference point) varies over time, the vehicle is said to be in motion. If the same vehicle doesn’t change its position in relation to that electric pole, it is said to be in the state of rest. Thus, two elements are required in order to observe the motion of any object: a reference point and time. In the above example, the object is a car, and the reference point is an electric pole. To observe the motion of an object, a building, tree, or any other static object may be used as a reference point.

### Motion along a straight line

When an object moves along a straight line, the motion of the object is called rectilinear motion. For example, motion of a vehicle along a straight road.

### Uniform and non Uniform motion

Uniform Motion: If a body travels equal distance in equal intervals of time then it is in uniform motion.

Non-uniform motion: If a body travels unequal distance in equal intervals of time then it is in uniform motion.

### Distance and Displacement

Distance is the length of path covered by a moving object in the given time irrespective of direction. Distance has only magnitude and no direction. SI unit of distance is meter (m).

Displacement is the shortest possible distance covered by a moving object from initial point in a particular direction. In other words, shortest distance between initial point and final point is called the displacement.

Displacement has both magnitude and direction while distance has only magnitude.

Suppose, a ball is rolling along a straight line.

Case 1: Suppose, the ball starts moving from point A and reaches at point B.

Thus, distance covered by ball = 10 m

Displacement of ball = 10 m towards west.

Case 2: Suppose, ball starts moving from point A and reaches to B. Again it returns on the same path from point B and reaches at A.

Thus, distance covered by the ball = distance from A to B + Distance from B to A = 10 m + 10 m = 20 m.

In this condition, distance covered by ball = 20 m.

Since, ball returns at point A, thus displacement of the ball = 0.

Case 3: Suppose, the ball starts moving from point A, reaches point B and returns back to point C.

Then, the distance covered by ball = distance from A to B + Distance from B to C = 10 m +7 m = 17 m

Displacement of ball = Distance of point C from A = 3 m towards west.

### Measuring the Rate of Motion

Different objects may take different amounts of time to cover a given distance. Some of them move fast and some move slowly. One of the ways of measuring the rate of motion of an object is to find out the distance travelled by the object in unit time. This quantity is referred to as speed.

Speed= Distance/Time

Average Speed = total distance travelled/Total time taken

#### Velocity

The rate of motion is more meaningful if we specify if we specify its direction of motion with speed, which is termed as velocity.

Velocity = Displacement/Time

Velocity is a vector quantity. Its value changes when either its magnitude or direction changes. It is also denoted by v.

For non-uniform motion in a given line, average velocity will be calculated in the same way as done in average speed.

Average velocity = Total displacement/Time

For uniformly changing velocity, the average velocity can be calculated

Average velocity = Initial velocity + Final velocity/2

V_{(avg)} = u+v/2

#### Acceleration

The rate of change in velocity is called acceleration. Acceleration is generally denoted by ‘a’.

Thus, acceleration = change in velocity/time taken

⇒ a = final velocity – initial velocity/time taken to change in velocity

⇒ a = v-u/t

A positive sign of the magnitude of acceleration shows increase in velocity and a negative sign show decrease in velocity. If there is decrease in acceleration, it is called Retardation. This means, rate of decrease in velocity is called Retardation.

Thus, SI unit of acceleration (a) = m/s/s = ms^{2} or ms^{-2}

Acceleration in the case of Uniform Velocity: In the case of uniform velocity, the speed or direction of a moving object is not changed and thus there is no change in acceleration. Therefore, in the case of uniform velocity acceleration will be zero.

### Graphical Representation of Motion

#### Distance–time graphs

The change in the position of a body with time can be represented on the distance time graph. In this graph distance is taken on the y-axis and time is taken on the x-axis.

The distance time graph for uniform speed is a straight line (linear). This is because in uniform speed a body travels equal distances in equal intervals of time.

We can determine the speed of the body from the distance – time graph.

For the speed of the body between the points A and B, distance is (s_{2} – s_{1}) and time is (t_{2}– t_{1}).

v = s/t

v = (s_{2} – s_{1})/(t_{2} – t_{1})

v = 20-10/10-5 = 10/5

2 ms^{-1}

ii) The distance-time graph for non uniform motion is non linear. This is because in non uniform speed a body travels unequal distances in equal intervals of time.

b) Velocity-time graphs: The change in the velocity of a body with time can be represented on the velocity time graph. In this graph velocity is taken on the y-axis and time is taken on the x-axis.

i) If a body moves with uniform velocity, the graph will be a straight line parallel to the x-axis. This is because the velocity does not change with time.

To determine the distance travelled by the body between the points A and B with velocity 20 km h-1.

v = s/t

s = v ✕ t

v = 20 kmh^{-1} = AC or BD

t = t_{2} – t_{1} = DC

= AC (t_{2} – t_{1})

s = AC ✕ CD

s = area of the rectangle ABDC

ii) If a body whose velocity is increasing with time, the graph is a straight line having an increasing slope. This is because the velocity increases by equal amounts with equal intervals of time.

The area under the velocity-time graph is the distance (magnitude of displacement) of the body.

The distance travelled by a body between the points A and E is the area ABCDE under the velocity-time graph.

s = area ABCDE = area of rectangle ABCD + area of triangle ADE

s = AB ✕ BC + 1/2 (AD ✕ DE)

iii) If a body whose velocity is decreasing with time, the graph is a straight line having an decreasing slope. This is because the velocity decreases by equal amounts with equal intervals of time.

iv) If a body whose velocity is non uniform, the graph shows different variations. This is because the velocity changes by unequal amounts in equal intervals of time.

### Equations of motions by graphical method

The motion of a body moving with uniform acceleration can be described with the help of three equations called equations of motion.

The equations of motion are:

v = u + at

s = ut + 1⁄2 at^{2}

2as = v^{2} – u^{2}

where u is the initial velocity

v is the final velocity

a is acceleration

t is the time

s is the distance travelled

#### Equation for velocity-time relation (v = u + at )

Consider a velocity – time graph for a body moving with uniform acceleration ‘a’. The initial velocity is u at A and final velocity is v at B in time t.

Perpendicular lines BC and BE are drawn from point B to the time and velocity axes so that the initial velocity is OA and final velocity is BC and time interval is OC. Draw AD parallel to OC.

We observe that

BC= BD + DC = BD + OA

Substituting BC= v and OA = u

We get, v = BD + u

or, BD = v-u

Acceleration = Change in velocity/time

a = BD/AD = BD/OC

or, a = v-u/t

v-u = at

or, v = u + at

#### Equation for position – time relation (s = ut + 1⁄2 at^{2})

Consider a velocity-time graph for a body moving with uniform acceleration ‘a’ travelled a distance s in time t.

The distance travelled by the body between the points A and B is the area OABC.

s = area OABC (which is a trapezium)

= area of rectangle OABC + area of triangle ABD

= OA ✕ OC + 1/2 (AD ✕ BD)

Substituting OA = u, OC = AD=t,

BD = v – u = at

We get, s = u ✕ t + 1/2 (t ✕ at)

or, s = ut + 1/2 at^{2}

#### Equation for position-velocity relation (2as = v^{2} -u^{2})

Consider a velocity-time graph for a body moving with uniform acceleration ‘a’ travelled a distance s in time t. The distance travelled by the body between the points A and B is the area OABC.

s = area of trapezium OABC

s = (OA+ BC) ✕ OC/2

Substituting OA = u, BC = v and OC = t

We get, s = (u + v) ✕ t/2

From velocity – time relation

t = (v-u)/a

(v+u) ✕ (v – u)/2a

or, 2as = v^{2} – u^{2}

### Motion along a circular path

Motion of an object along a circular path is called circular motion. Since, on a circular path the direction of the object is changing continuously to keep it on the path, the motion of the object is called accelerated motion.

Velocity in the case of circular motion.

If the radius of circle is ‘r’

Therefore, circumference = 2πг

Let time ‘t’ is taken to complete one rotation over a circular path by any object

Therefore, velocity (v) = Distance/Time

v = Circumference//t

v = 2𝝿r/t

Where, v = velocity, r = radius of circular path and t = time