Quick recap for CBSE class 9 Students

1. Chapter-8
Motion
By
N.Aarthi, B.E, M.Tech
SASTRA UNIVERSITY

2. Motion

3. Reference point and reference frame
• To describe the position of an object we need a
reference point or origin. An object may seem to
be moving to one observer and stationary to
another.
• Example: A passenger inside a bus sees the other
passengers be at rest, whereas an observer
outside the bus sees the passengers are in
motion.
• In order to make observations easy, a convention
or a common reference point or frame is needed.
All objects must be in the same reference frame.

4. Distance and Displacement
• The magnitude of the length covered by a
moving object is called distance. It has no
direction.
• Displacement is the shortest distance
between two points or the distance between
the starting and final positions with respect to
time. It has magnitude as well direction.
• Displacement can be zero, but distance
cannot.

5. Distance and Displacement

6. What is Distance?
• Distance is the total movement of an object
without any regard to direction.
• We can define distance as to how much
ground an object has covered despite its
starting or ending point.

7. Distance here will be = 4m + 3m + 5m = 12 m
Distance Formula
Δd=d1+d2

8. How is Displacement defined?
• Displacement is defined as the change in
position of an object. It is a vector
quantity and has a direction and magnitude.
• It is represented as an arrow that points from
the starting position to the final position.
• For example- If an object moves from A
position to B, then the object’s position
changes. This change in position of an object
is known as Displacement.

10. Displacement = Δx=xf−x0
xf = Final Position
x0 = Initial Position
Δx = Displacement

11. Examples of Distance and Displacement
Question 1.
John travels 250 miles to North but then back-tracks to South for
105 miles to pick up a friend. What is John’s total displacement?
Answer: John’s starting position Xi= 0.
Her final position Xf is the distance travelled N minus the
distance South.
Calculating displacement, i.e.D.
D = ΔX = (Xf – Xi)
D = (250 mi N – 105 mi S) – 0
D = 145 mi N

12. Question 2. An object moves along the grid through points A, B,
C, D, E, and F as shown below. The side of square tiles
measures 0.5 km.
a) Calculate the distance covered by the moving object.
b) Find the magnitude of the displacement of the object.

13. Solution:
a) The distance covered by the moving object is
calculated as follows:
AB + BC + CD + DE + EF
3 + 1 + 1.5 + 0.5 + 0.5 = 6.5 km
The distance covered by the moving object is 6.5
km.

14. b) The initial point is A and the final point is F,
hence the magnitude of the displacement is
equal to the distance AF which is calculated by
applying Pythagoras’s theorem to the triangle
AHF as shown in the figure below

15. Applying the Pythagorean formula, we get
AF2=AH2+HF2
Substituting the formula, we get
AF2=(0.5×4)2+(0.5×3)2=6.25
AF=square root(6.25km)=2.5km
The magnitude of displacement is 2.5 km.

16. Distance vs Displacement

17. Magnitude
• Magnitude is the size or extent of a physical
quantity. In physics, we have scalar and vector
quantities.
• Scalar quantities are only expressed as
magnitude. E.g: time, distance, mass,
temperature, area, volume
• Vector quantities are expressed in magnitude
as well as the direction of the object. E.g:
Velocity, displacement, weight, momentum,
force, acceleration, etc.

19. Time, Average Speed and Velocity
Time and speed
Time is the duration of an event that is
expressed in seconds. Most physical phenomena
occur with respect to time. It is a scalar quantity.
Speed is the rate of change of distance. If a body
covers a certain distance in a certain amount of
time, its speed is given by
Speed = Distance/Time
Average speed = Total distance travelled / Total
time taken

21. Uniform motion and non-uniform motion
• When an object covers equal distances in
equal intervals of time it is in uniform motion.
• When an object covers unequal distances in
equal intervals of time it is said to be in non-
uniform motion.

23. Uniform Motion
Definition: This type of motion is defined as the
motion of an object in which the object travels
in a straight line and its velocity remains
constant along that line as it covers equal
distances in equal intervals of time, irrespective
of the duration of the time.

25. Example of Uniform Motion:
• If the speed of a car is 10 m/s, it means that
the car covers 10 meters in one second. The
speed is constant in every second.
• Movement of blades of a ceiling fan.

26. Non Uniform Motion:
Definition: This type of motion is defined as the
motion of an object in which the object travels with
varied speed and it does not cover same distance in
equal time intervals, irrespective of the time interval
duration.

27. If a body is involved in rectilinear motion, and if the
motion is not consistent, then the acceleration of the
body must be non-zero.

29. What is Rectilinear Motion?
• When we require only one co-ordinate axis
along with time to describe the motion of a
particle it is said to be in linear motion or
rectilinear motion.
• Some examples of linear motion are a parade
of soldiers, a train moving along a straight line.

31. Comparison table Between these two
types of motions:

32. Uniform Circular Motion:
• This type of motion is seen if an object is
travelling at a constant speed around a fixed
axis or center point, and the object travels
around a curved path and it maintains a
constant radial distance from the center point
at any given time and it moves in a tangent to
the curved path.

35. Non Uniform Circular Motion
This type of motion is inconsistent because if an
object is travelling at a variable angular speed
around a fixed axis or center point, then the
motion of that object is said to be inconsistent
because the object covers a curved path and it
will have some variable radial acceleration due
to which its velocity would change every second.

36. Velocity
The Rate of change of displacement is velocity. It
is a vector quantity. Here the direction of motion
is specified.
Velocity = Displacement/Time

38. Speed: Speed is a scalar quantity which means it
has no direction. It denotes how fast an object is
moving. If the speed of the particle is high it
means the particle is moving fast and if it is
low, it means the particle is moving slow.

41. Velocity:
• Velocity is a vector quantity which means it has
both magnitude and direction.
• It denotes the rate at which the object is moving
or changing position.
• The direction of the velocity vector is easy to find.
Its direction is same as the direction of the
moving object.
• Even if the object is slowing down, and the
magnitude of velocity is decreasing, its direction
would still be same as the direction in which the
object is moving

45. Average Speed
The average speed of a body in a certain time
interval is the distance covered by the body in
that time interval divided by time. So if a
particle covers a certain distance s in a
time t1 to t2, then the average speed of the
body is:

46. 1.) In travelling from Pune to Nagpur , Rahul drove his
bike for 2 hours at 60 kmph and 3 hours at 70 kmph.
Sol 1) We know that, Distance = Speed × Time
So, in 2 hours, distance covered = 2 × 60 = 120 km
in the next 3 hours, distance covered = 3 × 70 = 210 km
Total distance covered = 120 + 210 = 330 km
Total time = 2 + 3 = 5 hrs

47. Average Velocity
• The average velocity of a body in a certain
time interval is given as the displacement of
the body in that time interval divided by time.
So if a particle covers a certain displacement

48. Calculate the average velocity at a particular time
interval of a person if he moves 7 m in 4 sec and 18 m
in 6 sec along x-axis?
Sol)
Initial distance traveled by the person, xi = 7 m,
Final distance traveled, xf = 18 m,
Initial time interval ti = 4 sec,
Final time interval tf = 6 sec,

49. Acceleration

50. Uniform Acceleration

51. Acceleration
• The rate of change of velocity is called
acceleration. It is a vector quantity. In non-
uniform motion, velocity varies with time, i.e.,
change in velocity is not 0. It is denoted by “a”
• Acceleration = Change in Velocity / Time

55. Motion Visualised
Distance-Time graph
• A distance-time graph shows how far an
object has travelled in a given time. It is a
simple line graph that denotes distance versus
time findings on the graph.
• Distance is plotted on the Y-axis.
• Time is plotted on the X-axis.

57. • We can use distance time graph to determine
the speed of an object.
• Consider small part AB of graph.
• Draw parallel line to x-axis from point A.
• Draw another line parallel to y-axis from point
B.
• These 2 lines meet at C to form a triangle ABC.
• AC denotes time interval t2-t1.
• BC denotes distance s2-s1

58. • We can see that object travels from A to B, it
covers distance s2-s1 in time t2-t1
• Speed v of object can be represented as

60. Example:
• For better understanding, let us consider an
example of uniform motion. A bus driver
drives at a constant speed which is indicated
by the speedometer and the driver measures
the time taken by the bus for every kilometre.
The driver notices that the bus travels 1
kilometre in every 2 minutes.

62. • By this table, he had a clear idea about the
speed which is: ½ × 60 = 30 km/hr.
• The graph is a straight line and the motion of
the bus is also uniform. Also, from the graph,
we can find the speed of the bus at any
instant of time. The initial and final position of
the car can be found as the following:
• Speed = (Final Position-Initial position)/Time

64. Conclusion:
The following things can be concluded now:
• If the distance-time graph is a straight line then
the motion is uniform.
• If the distance-time graph of anybody is given, its
speed can be calculated using the slope of the
graph.
• The slope of the straight-line graph is the same
irrespective of the interval which is chosen. This
implies that the speed of an object under uniform
motion remains constant.

65. Velocity-Time Graph
• Velocity-Time graphs show the change in
velocity with respect to time.
• Slope gives acceleration
• The area under the curve gives displacement
• Line parallel to x-axis implies constant
velocity-

66. OA = constant acceleration, AB = constant
velocity, BC = constant retardation

68. • Figure shows velocity time graph of car
moving with uniform velocity 40 km/hr
• Velocity of 40km/hr is represented by height
AC or BD.
• Time (t2-t1) is represented by length AB
• So distance s moved by the car in time t2-t1
can be expressed as

70. • A person sitting next to driver records its
velocity after every 5 seconds by noting the
reading of speedometer of car.

71. • Velocity changes by equal amounts in equal intervals
of time. So we get straight line graph.
• We can determine distance moved by car using
velocity-time graph.
• Distance travelled is given by area ABCD.
• Since magnitude of velocity of car is changing due to
acceleration, the distance s travelled by car will be
given by area ABCDE under velocity-time graph.

72. Case 1: Velocity-time graphs with
constant velocity (zero acceleration)
When the velocity is constant, the velocity time
graph, with Y-axis denoting velocity and X-axis
denoting time, will be like:

73. Case 2: Velocity-time graphs with constant
acceleration
• When the acceleration is constant (positive),
and the initial velocity of the particle is zero,
the velocity of the particle will increase
linearly as predicted by the equation:
• v = u + at
• Since u = 0
• v = at

74. As shown in the figure, the velocity of the
particle will increase linearly with respect to
time. The slope of the graph will give the
magnitude of acceleration.

75. Case 3: Velocity-time graphs with increasing
acceleration
When the acceleration is increasing with time, the
velocity-time graph will be a curve as predicted
from the equation:
v = u + at
Since u = 0
v= at
Since acceleration is a function of time, the velocity-
time graph will be a curve.
Note: Since the acceleration is continuously
increasing with time, the magnitude of the slope
will also continuously increase with time.

76. Velocity-time graphs with increasing
acceleration

77. Equations of Motion
The motion of an object moving at uniform
acceleration can be described with the help of
three equations, namely
• v = u + at (8.5)
• s = ut + (1/2)at2 (8.6)
• 2as =v2 – u2 (8.7)

78. Derivation of velocity-time relation by graphical
method

79. • Initial velocity of object is u(at point A).
• Velocity increases to v(at point B)
• Initial velocity is given by OA.
• Final velocity is given by BC.
• Time interval t is given by OC.
• BD=BC-CD(represents change in velocity in
time interval t)
• Draw AD parallel to OC.

80. • From graph we observe that
BC=BD+DC=BD+OA
Substituting BC=v and OA=u
We get v=BD+u
Or BD=v-u (8.8)
Acceleration of object is given by

81. Derivation of position-time relation by graphical
method

82. • Consider object travelled a distance s in time t
under uniform acceleration.
• Distance travelled by object is obtained by
area enclosed within OABC under velocity
time graph AB
• Thus distance s travelled by object is given by
s=area OABC
= area of rectangle OADC + area of triangle ABD

84. Derivation of position-velocity relation by graphical
method

86. Uniform circular motion
• If an object moves in a circular path with
uniform speed, its motion is called uniform
circular motion.
• Velocity is changing as direction keeps
changing.
• Acceleration is constant

87. Following are the examples of uniform circular
motion:
• Motion of artificial satellites around the earth is
an example of uniform circular motion. The
gravitational force from the earth makes the
satellites stay in the circular orbit around the
earth.
• The motion of electrons around its nucleus.
• The motion of blades of the windmills.
• The tip of second’s hand of a watch with circular
dial shows uniform circular motion.

94. Extra Questions
1. An object has moved through a distance. Can
it have zero displacement? If yes, support
your answer with an example.
Solution Yes, an object moving a certain
distance can have zero total displacement.
Displacement refers to the shortest distance
between the initial and the final positions of the
object. Even if an object moves through a
considerable distance, if it eventually comes
back to its initial position, the corresponding
displacement of the object would be zero.

95. 2. A farmer moves along the boundary of a
square field of side 10m in 40 s. What will be the
magnitude of displacement of the farmer at the
end of 2 minutes 20 seconds from his initial
position?
Solution
Given that the farmer covers the entire
boundary of the square field in 40 seconds,
the total distance traveled by the farmer in 40
seconds is 4*(10) = 40 meters.

99. 3. Which of the following is true for displacement?
(a) It cannot be zero. (b) Its magnitude is greater
than the distance travelled by the object.
Solution
Neither of the statements are true.
(a) is false because the displacement of an object
which travels a certain distance and comes back
to its initial position is zero.
(b) Statement (b) is false because the displacement
of an object can be equal to, but never greater
than the distance traveled.

101. 5. Under what condition(s) is the magnitude of
average velocity of an object equal to its average
speed?
Solution
Since average speed is the total distance
traveled in a time frame and velocity is the total
displacement in the time frame, the magnitude
of average velocity and average speed will be
the same when the total distance traveled is
equal to the displacement.

102. 6. What does the odometer of an automobile
measure?
Solution
The odometer measures the total distance
traveled by the automobile.
7. What does the path of an object look like
when it is in uniform motion?
Solution
The path of an object in uniform motion is a
straight line.

103. 8. During an experiment, a signal from a spaceship
reached the ground station in five minutes.
What was the distance of the spaceship from the ground
station?
The signal travels at the speed of light, that is, 3 × 108
m/s.
Solution
Given that the signal travels in a straight line, the distance
between the spaceship and the ground station is equal to
the total distance traveled by the signal.
5 minutes = 5*60 seconds = 300 seconds.
Speed of the signal = 3 × 108 m/s.
Therefore, total distance = (3 × 108 m/s) * 300s
= 9*1010 meters.

104. 9. When will you say a body is in (i) uniform
acceleration? (ii) non-uniform acceleration?
Solution
Uniform Acceleration: In this type of acceleration,
the body moves along a straight line and its velocity
increases/decreases at a uniform rate (it changes at
a constant rate in any constant time interval).
Non-Uniform Acceleration: In this type of
acceleration, the body moves along a straight line
and its velocity increases/decreases at a rate that is
not uniform (it changes at a different rate for a
given constant time interval).

105. 10. A bus decreases its speed from 80 km h–1 to 60 km
h–1 in 5 s. Find the acceleration of the bus.
Solution

129. Thank You
For queries contact
aarthisam2003@yahoo.co.in