Class 9 Exam  >  Class 9 Notes  >  Science Class 9  >  Chapter Notes: Motion

Motion Class 9 Notes Science Chapter 7

In our daily lives, objects such as birds, fish, and cars can be either at rest or in motion. Motion is observed when an object's position changes over time. Sometimes, motion is inferred indirectly, like noticing dust moving to deduce air movement. Perception of motion can vary: passengers in a moving bus see trees moving backward, while onlookers outside the bus see both the bus and its passengers in motion.

MotionMotion

Describing Motion

To describe an object's motion, we use a reference point, or origin, as a fixed location. For instance, if a school is 2 km north of a railway station, the railway station is the reference point. This origin helps us measure and describe the object's position relative to it.

What is Motion?

A body is said to be in a state of motion when its position changes continuously with reference to a point. Motion can be of different types depending upon the type of path by which the object is travelling through:

Types of MotionTypes of Motion

  • Circulatory motion/Circular motion: In a circular path.
  • Linear motion: In a straight-line path.
  • Oscillatory/Vibratory motion: To and fro path with respect to origin.

Motion Along a Straight Line

The simplest type of motion is along a straight line. 
Let's understand this with an example. Imagine an object moving along a straight path, starting from point O, which we use as the reference point.

Motion along a straight lineMotion along a straight line

Example Description

  • Initial Motion: The object starts at point O and moves to point A via points C and B.
  • Return Motion: The object then travels back from A to C through B.

Distance Covered

The total path length covered by the object is the sum of the distances traveled:

  • Distance from O to A: 60 km
  • Distance from A to C: 35 km
  • Total Distance from O to C = OA + AC = 60 km + 35 km = 95 km

Distance is a scalar quantity, meaning it only has magnitude and no direction.

  • Scalar quantity: It is the physical quantity having its own magnitude but no direction.
    Example: Distance, Speed.
  • Vector quantity: It is the physical quantity that requires both magnitude and direction.
    Example: Displacement, Velocity.

Distance and Displacement

1. Distance

The actual path of length travelled by an object during its journey from its initial position to its final position is called the distance.

  • Distance is a scalar quantity that requires only magnitude but no direction to explain it.
  • Example: Ramesh travelled 65 km. (Distance is measured by odometer in vehicles.)

Motion Class 9 Notes Science Chapter 7

2. Displacement

The shortest distance travelled by an object during its journey from its initial position to its final position is called displacement.

  • Displacement is a vector quantity requiring both magnitude and direction for its explanation.
  • Example: Ramesh travelled 65 km southwest from Clock Tower.
  • Displacement can be zero (when the initial point and final point of motion are the same)
    Example: Circular motion.

Example of Zero DisplacementExample of Zero Displacement

Question for Chapter Notes: Motion
Try yourself:
What is the definition of distance?
View Solution

Example 1: A body travels in a semicircular path of radius 10 m starting its motion from point ‘A’ to point ‘B’. Calculate the distance and displacement.

Sol. Given, π = 3.14, R = 10 mMotion Class 9 Notes Science Chapter 7

The distance is the length of the semicircular path.

Distance = Circumference of a circle ÷ 2 

Distance = 2πR ÷ 2 

Distance = πR 
= 3.14 × 10 = 31.4 m

Where as , 

Displacement = 2 × R = 2 × 10 = 20 m

Example 2: A body travels 4 km towards North then he turns to his right and travels another 4 km before coming to rest. Calculate 

(i) total distance travelled, 

(ii) total displacement.

Sol. The total distance is the sum of all the paths travelled:

Total DistanceTotal Distance = 4km (North) + 4km (Right) = 8km

Since displacement is the shortest straight-line distance between the starting point and the final point.

The path forms a right triangle, where:

  • One leg =  km4 \, \text{km}4 km (North direction),
  • Other leg = 4 km (Right direction).

Motion Class 9 Notes Science Chapter 7

Question for Chapter Notes: Motion
Try yourself:What is the difference between distance and displacement?
View Solution

Uniform and Non-uniform Motion

1. Uniform Motion

  • When a body travels equal distances in equal intervals of time, then the motion is said to be a uniform motion.Motion Class 9 Notes Science Chapter 7
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2. Non-uniform Motion

  • In this type of motion, the body travels unequal distances in equal intervals of time.
  • Two types of non-uniform motion:
    (i) Accelerated Motion: When the motion of a body increases with time.
    (ii) De-accelerated Motion: When the motion of a body decreases with time.

Motion Class 9 Notes Science Chapter 7

Measuring the Rate of Motion

The measurement of distance travelled by a body per unit of time is called speed.
i.e. Speed (v) = Distance Travelled/Time Taken = s/t

  • SI unit: m/s (meters/second)
  • If a body is executing uniform motion, then there will be a constant speed.
  • If a body is travelling with a non-uniform motion, then the speed will not remain uniform but have different values throughout the motion of such a body.
  • For non-uniform motion, the average speed will describe one single value of speed throughout the motion of the body.
    i.e. Average speed = Total distance travelled/Total time taken

Conversion Factor
Change from km/hr to m/s = 1000m/(60×60)s = 5/18 m/s

Example: What will be the speed of body in m/s and km/hr if it travels 40 km in 5 hrs?
Sol: Distance (s) = 40 km
Time (t)  = 5 hrs.
Speed (in km/hr) = Total distance/Total time = 40/5 = 8 km/hr
40 km = 40 × 1000 m = 40,000 m
5 hrs = 5 × 60 × 60 sec.
Speed (in m/s) = (40 × 1000)/(5×60 ×60) = 80/36 = 2.22 m/s

Speed with Direction

  • It is the speed of a body in a given direction.
  • The measurement of displacement travelled by a body per unit of time is called velocity.
    i.e. Velocity = Displacement/Time
  • SI unit of velocity: ms-1
  • Velocity is a vector quantity. Its value changes when either its magnitude or direction changes.
  • It can be positive (+ve), negative (-ve) or zero.
  • For non-uniform motion in a given line, average velocity will be calculated in the same way as done in average speed.
    i.e. Average velocity = Total displacement/Total time
  • For uniformly changing velocity, the average velocity can be calculated as follows:-

Avg. Velocity (vavg) = (Initial velocity + Final velocity)/2 = (u+v)/2
where,  u = initial velocity, v = final velocity 

Example 1: During the first half of a journey by a body it travels with a speed of 40 km/hr and in the next half it travels at a speed of 20 km/hr. Calculate the average speed of the whole journey.

Sol: The average speed for a journey where the distances are equal but the speeds are different is not simply the arithmetic mean 

When a body covers equal distances at different speeds, the correct formula for average speed is:
Motion Class 9 Notes Science Chapter 7Given:

  • Speed during the first half (v1) = 40 km/hr
  • Speed during the second half (v2v_2) = 20 km/hr

Now, use the correct formula:
Motion Class 9 Notes Science Chapter 7

Motion Class 9 Notes Science Chapter 7

Example 2: A car travels 20 km in first hour, 40 km in second hour and 30 km in third hour. Calculate the average speed of the vehicle.

Sol: Speed in 1st hour = 20 km/hr

Distance travelled during 1st hr = 1 × 20= 20 km

Speed in 2nd hour = 40 km/hr

Distance travelled during 2nd hr = 1 × 40= 40 km

Speed in 3rd hour = 30 km/hr

Distance travelled during 3rd hr = 1 × 30= 30 km

Average speed = Total distance travelled/Total time taken

= (20 + 40 + 30)/3 = 90/3 = 30 km/hr

Question for Chapter Notes: Motion
Try yourself:
What is the definition of uniform motion?
View Solution

Rate of Change of Velocity

  • Acceleration is seen in non-uniform motion and it can be defined as the rate of change of velocity with time.
    i.e. Acceleration (a) = Change in velocity/Time = (v-u)/t
    where, v = final velocity, u = initial velocity
  • Here, v > u, then ‘a’ will be positive (+ve). If v is greater than u, then acceleration (a) will be positive.

Example: A car speed increases from 40 km/hr to 60 km/hr in 5 sec. Calculate the acceleration of car.
Sol. u = 40km/hr = (40×5)/18 = 100/9 = 11.11 m/s
v  = 60 km/hr = (60×5)/18 = 150/9 = 16.66 m/s
t = 5 sec
a = (v-u)/t = (16.66 - 11.11)/5 = 5.55/5 = 1.11 ms-2

Question for Chapter Notes: Motion
Try yourself:Which quantity requires both magnitude and direction.
View Solution

Retardation/Deceleration

  • Deceleration is seen in non-uniform motion during decrease in velocity with time. It has same definition as acceleration.
    i.e. Deceleration (a') = Change in velocity/Time = (v-u)/t
  • Here, v < u, ‘a’ = negative (-ve).Motion Class 9 Notes Science Chapter 7


Example: A car travelling with a speed of 20 km/hr comes into rest in 0.5 hrs. What will be the value of its retardation?
Sol. v = 0 km/hr, u = 20 km/hr, t = 0.5 hrs
Retardation, a = (v-u)/t = (0-20)/0.5 = -200/5 = -40 km hr-2

Watch the animated video below to understand the concepts in an easy manner.

00:00
10:40

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Chapter Notes: Motion
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Graphical Representation of Motion

1. Distance-Time Graph (s/t graph)

(i) s/t graph for uniform motion:
Motion Class 9 Notes Science Chapter 7(ii) s/t graph for non-uniform motion:
Motion Class 9 Notes Science Chapter 7
(iii) s/t graph for a body at rest:
Motion Class 9 Notes Science Chapter 7v = (s2 - s1)/(t2 - t1)
But, s2 - s1
∴ v = 0/(t2 - t1) or v = 0

2. Velocity-Time Graph (v/t graph)

(i) v/t graph for uniform motion:
Motion Class 9 Notes Science Chapter 7a = (v2 - v1)/(t2 - t1)
But, v2 - v1
∴ a = 0/(t2 - t1) or a = 0
(ii) v/t graph for uniformly accelerated motion:
Motion Class 9 Notes Science Chapter 7In uniformly accelerated motion, there will be an equal increase in velocity in equal intervals of time throughout the motion of the body.
(iii) v/t graph for non-uniformly accelerated motion:
Motion Class 9 Notes Science Chapter 7a2 ≠ a1
(iv) v/t graph for uniformly decelerated motion:
Motion Class 9 Notes Science Chapter 7a1' = a2'
(v) v/t graph for non-uniformly decelerated motion:
Motion Class 9 Notes Science Chapter 7

Note: In v/t graph, the area enclosed between any two time intervals, t2 - t1, will represent the total displacement by that body.

Motion Class 9 Notes Science Chapter 7
Total distance travelled by body between t2 and t1
= Area of ∆ABC + Area of rectangle ACDB = ½ × (v2 – v1)×(t2 - t1) + v1× (t2 - t1)

Example: From the information given in the s/t graph, which of the following body ‘A’ or ‘B’ will be faster?
Motion Class 9 Notes Science Chapter 7Sol. vA > vB

Question for Chapter Notes: Motion
Try yourself:
Which of the following statements is true about acceleration?
View Solution

Equations of Motion by Graphical Method

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1. First Equation: v = u + at

Final velocity = Initial velocity + Acceleration × Time
Graphical Derivation
Suppose a body has initial velocity ‘u’ (i.e., velocity at time t = 0 sec.) at point ‘A’ and this velocity changes to ‘v’ at point ‘B’ in ‘t’ secs. i.e., final velocity will be ‘v’.
Motion Class 9 Notes Science Chapter 7

For such a body there will be an acceleration. a = Change in velocity/Change in Time
⇒ a = (OB - OA)/(OC-0) = (v-u)/(t-0)
⇒ a = (v-u)/t
⇒ v = u + at

2. Second Equation: s = ut + ½ at2

Distance travelled by object = Area of OABC (trapezium)
= Area of OADC (rectangle) + Area of ∆ABD
= OA × AD + ½ × AD × BD
= u × t + ½ × t × (v – u)
= ut + ½ × t × at  
⇒ s = ut + ½ at2  (∵a = (v-u)/t)

3. Third Equation: v2 = u2 + 2as

s = Area of trapezium OABC
Motion Class 9 Notes Science Chapter 7

Example 1: A car starting from rest moves with a uniform acceleration of 0.1 ms-2 for 4 mins. Find the speed and distance travelled.
Sol: u = 0 ms-1 (∵ car is at rest), a = 0.1 ms-2, t = 4 × 60 = 240 sec.
v = ?
From, v = u + at
v = 0 + (0.1 × 240) = 24 ms-1

Example 2: The brakes applied to a car produces a deceleration of 6 ms -2 in the opposite direction to the motion. If a car requires 2 sec. to stop after the application of brakes, calculate the distance travelled by the car during this time.
Sol: Deceleration, a = − 6 ms-2; Time, t = 2 sec.
Distance, s =?
Final velocity, v = 0 ms-1 (∵ car comes to rest)
Now, v = u + at
⇒ u = v – at = 0 – (-6×2) = 12 ms-1
s = ut + ½ at= 12 × 2 + ½ (-6 × 22) = 24 – 12 = 12 m

Question for Chapter Notes: Motion
Try yourself:A car is moving with an initial velocity of 20 m/s. If it accelerates at a rate of 5 m/s? for 4 seconds, what is its final velocity?
View Solution

Uniform Circular Motion

  • If a body is moving in a circular path with uniform speed, then it is said to be executing the uniform circular motion.
  • In such a motion the speed may be the same throughout the motion but its velocity (which is tangential) is different at each and every point of its motion. Thus, uniform circular motion is an accelerated motion.Direction at different points while executing circular motion
    Direction at different points while executing circular motion

Motion Class 9 Notes Science Chapter 7

The document Motion Class 9 Notes Science Chapter 7 is a part of the Class 9 Course Science Class 9.
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FAQs on Motion Class 9 Notes Science Chapter 7

1. What is the difference between distance and displacement?
Ans.Distance is a scalar quantity that refers to the total length of the path traveled by an object, regardless of direction. Displacement, on the other hand, is a vector quantity that refers to the shortest straight line distance from the initial position to the final position of the object, along with the direction.
2. How do you calculate average speed?
Ans.Average speed is calculated by dividing the total distance traveled by the total time taken. The formula is Average Speed = Total Distance / Total Time. It provides an overall measure of how fast an object is moving during a journey.
3. What is retardation or deceleration?
Ans.Retardation, also known as deceleration, is a type of acceleration that occurs when an object slows down. It is defined as a negative acceleration, meaning that the velocity of the object decreases over time. This can happen when an external force acts against the direction of motion.
4. How can motion be represented graphically?
Ans.Motion can be represented graphically using distance-time graphs and velocity-time graphs. In a distance-time graph, the slope represents speed, while in a velocity-time graph, the slope signifies acceleration. The area under the velocity-time graph indicates the distance traveled.
5. What are the equations of motion, and how can they be derived using graphs?
Ans.The equations of motion describe the relationship between displacement, velocity, acceleration, and time. They can be derived graphically by analyzing the slope and area under the curves in distance-time and velocity-time graphs. These equations help in predicting an object's future position and velocity based on its current state.
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