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Chapter 7: MOTION
Motion is perceived when an object’s position changes over time, but it can also be inferred indirectly. E.g., moving dust or swaying leaves indicate air movement.
Sunrise, sunset, and changing seasons are caused by the Earth's motion.
We don't directly perceive Earth's motion due to its steady and consistent
movement.
Motion is relative. i.e., an object may seem
moving to one person and stationary to another. E.g., passengers in a moving
bus see trees moving backward, while a person on the roadside sees the bus
moving. But a passenger
inside the bus sees fellow passengers are at rest.
DESCRIBING
MOTION
To describe the position of an object, a reference point called the origin must be specified.
MOTION
ALONG A STRAIGHT LINE
It is the simplest type of motion.
Consider an object moving from point O (reference point) along a straight path through points C, B, and A, then back to C through B.
The total path length (distance) covered by the object is 95 km
(OA + AC = 60 km + 35 km).
Distance
requires only magnitude (numerical value), not direction.
Displacement is the
shortest distance (straightline distance) from the initial to final position of an
object.
The magnitude of displacement may differ from the distance. E.g., from O
to A and back to B, the distance is 85 km (60 km + 25
km), while the displacement is 35
km.
The magnitude of displacement can be zero if the object returns to its
starting point, but the distance covered would be the total path length. i.e., OA + AO = 60 km + 60 km = 120 km.
Distance & displacement are distinct physical quantities used to describe the motion and final position of an object relative to its initial position.
In automobiles, odometer shows the distance travelled. But displacement is determined using a road map or geographic tools.
UNIFORM
MOTION AND NONUNIFORM MOTION
If an object covers equal distances in equal intervals of time, it is called uniform motion. Here, the time interval should be small. E.g., moving of an object along a straight line 5 m in each second.
If an object covers unequal distances in equal intervals of time, it is called nonuniform motion. E.g., a car moving on a crowded street, a person jogging in a park.
Table showing uniform and nonuniform motion of two different
objects A and B.
Time 
Distance travelled by object A in m (Uniform motion) 
Distance travelled by object B in m (Nonuniform motion) 
9:30 am 
10 
12 
9:45 am 
20 
19 
10:00 am 
30 
23 
10:15 am 
40 
35 
10:30 am 
50 
37 
10:45 am 
60 
41 
11:00 am 
70 
44 
MEASURING
THE RATE OF MOTION
Speed is a measure of how fast an object moves. It is the distance travelled
by an object in a given unit time.
SI unit of
speed is metre per second (m s^{–1} or m/s). Other units are centimetre
per second (cm s^{–1}) and kilometre per hour (km h^{–1}). E.g., In cricket, bowling speed may be 140 km/hr. Traffic sign indicates speed limit like
50 km/hr.
To specify
the speed, only its magnitude is required (no direction).
The speed
of an object is often not constant due to nonuniform motion. So, the rate of motion
is described as average speed.
If an
object travels a distance s in time t, its speed v is,
E.g., If a
car travels a distance of 100 km in 2 h, Its average speed is 50 km h^{–1}.
The car might not have travelled at 50 km h^{–1} all the time.
Problem: An object travels 16 m in 4 s and
then another 16 m in 2 s. What is the average speed of the object?
Solution:
Total distance travelled= 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
SPEED
WITH DIRECTION

The speed
of an object moving in a definite direction is called velocity. Here,
speed and direction are specified.

Velocity
of an object can be uniform or variable. It can change by altering the object's
speed, direction, or both.

When an
object moves along a straight line at variable speed, its rate of motion is expressed as average velocity.

If the
velocity changing at a uniform rate, average velocity (v_{av}) is
the arithmetic mean of initial velocity (u) and final velocity (v)
for a given period of time.
Speed and velocity have
the same units (m s^{–1} or m/s).
The sound of thunder
takes some time to reach after seeing lightning because light travels faster
than sound.
To
measure the distance of lightning:
· Use a
stopwatch to record the time between seeing
lightning and hearing thunder.
· Multiply
the time by speed of sound in air (346 m/s).
E.g., if the time interval is 5 seconds:
Distance= 346 m/s × 5 s= 1730 m
Problem: The odometer of a car reads 2000
km at the start of a trip and 2400 km at the end of the trip. If the trip took
8 h, calculate average speed of the car in km h^{–1} and ms^{–1}.
Solution:
Distance covered by the car, s = 2400– 2000 = 400 km
Time elapsed, t = 8 h
Average speed of the car is,
Problem: Usha swims in a 90 m long pool. She covers 180 m in one minute by
swimming from one end to the other and back along the same straight path.
Find the average speed and average velocity of Usha.
Solution:
Total distance covered by Usha in 1 min is 180 m.
Displacement of Usha in 1 min = 0 m
RATE OF CHANGE OF VELOCITY
During
uniform motion along a straight line, the velocity remains constant. Here,
change in velocity of the object for any time interval is zero.
In nonuniform
motion, velocity varies with time. So, change in velocity during any time
interval is not zero.
The change
in velocity of an object is described by acceleration. It is the rate of
change in the velocity per unit time.
If the
velocity of an object changes from an initial value u to the final value
v in time t, the acceleration a is,
This kind
of motion is known as accelerated motion.
The SI
unit of acceleration is ms^{–2}.
Acceleration
is positive if it is in the direction of the velocity and negative if it is
opposite to the direction of the velocity.
If an
object travels in a straight line and its velocity changes by equal amounts in
equal intervals of time, the acceleration is uniform (uniformly accelerated motion).
E.g., motion of a freely falling body.
If an
object’s velocity changes at a nonuniform rate, the acceleration is
nonuniform. E.g., if a car travelling along a straight road increases its
speed by unequal amounts in equal time intervals, it is in nonuniform
acceleration.
Activity Answer:
·
Acceleration
in the direction of motion: E.g., Accelerating car.
·
Acceleration
against the direction of motion: E.g., Car
braking.
·
Uniform
acceleration: E.g., Freefalling object in a
vacuum.
·
Nonuniform
acceleration: E.g., Car accelerating in
traffic.
Example: Starting from a stationary position,
Rahul paddles his bicycle to attain a velocity of 6 m s^{–1} in 30 s.
Then he applies brakes such that the velocity of the bicycle comes down to 4 m
s^{1} in the next 5 s. Calculate the acceleration of the bicycle in
both the cases.
Solution:
In the first case:
initial velocity, u
= 0;
final velocity, v
= 6 m s^{–1};
time, t = 30 s.
In the second case:
initial velocity, u =
6 m s^{–1};
final velocity, v =
4 m s^{–1};
time, t = 5 s.
The acceleration of the
bicycle in the first case is 0.2 ms^{–2} and in the second case, it is
–0.4 ms^{–2}.
GRAPHICAL
REPRESENTATION OF MOTION
DISTANCE–TIME
GRAPHS
The change
in the position of an object with time can be represented on the distancetime
graph (time on the xaxis and distance on the yaxis).
They are used under conditions where objects move
with uniform speed, nonuniform speed, remain at rest
etc.
When an object travels equal distances in equal
intervals of time, it moves with uniform speed. i.e., distance is directly proportional
to time. Thus, for uniform speed, a graph of distance travelled against time is
a straight line.
Distancetime
graph can be used to determine the speed of an object. Consider a
segment AB. Draw a horizontal line from point A and a vertical line from point
B. These lines intersect at point C, forming a triangle ABC. The time interval
is represented by AC (t_{2} – t_{1}) and the distance by
BC (s_{2} – s_{1}). As the object moves from point A to
B, it covers a distance (s_{2} – s_{1}) in the time
interval (t_{2} – t_{1}). Hence, the speed, v is
Distancetime
graph can also be used to plot for accelerated motion. Table shows the distance
travelled by a car in a time interval of two seconds.
Distance travelled by a car at
regular time intervals 

Time in seconds 
Distance in metres 
0 
0 
2 
1 
4 
4 
6 
9 
8 
16 
10 
25 
12 
36 
The graph
shows nonlinear variation of the distance travelled by the car with time (nonuniform
speed).
VELOCITYTIME
GRAPHS

The change in velocity over time for an object
moving in a straight line is represented by a velocitytime graph. Here,
time is plotted on xaxis and velocity on yaxis.
If the object moves with uniform velocity, the
graph will be a straight line parallel to the xaxis.
The
product of velocity and time give displacement of an object moving with uniform
velocity. The area enclosed by velocitytime graph and the time axis will be equal
to the magnitude of the displacement.
To know
the distance moved by the car between time t_{1} & t_{2},
draw perpendiculars from the points corresponding to t_{1} and t_{2}
on the graph. The velocity of 40 km h^{–1} is represented by the height
AC or BD and the time (t_{2} – t_{1}) is represented by the
length AB.
So, the
distance s moved by the car in time (t_{2} – t_{1}) is
s = AC × CD
= [(40 km h^{–1})
× (t_{2} – t_{1}) h]
= 40 (t_{2}–
t_{1}) km
= area of the rectangle
ABDC.
Velocitytime
graph is used to study uniformly accelerated motion. E.g., Consider a
car is moving along a straight road. A passenger records the car's velocity
every 5 seconds by observing the speedometer.
Time (s) 
Velocity of the car 

(m s^{–1}) 
(km h^{–1}) 

0 
0 
0 
5 
2.5 
9 
10 
5.0 
18 
15 
7.5 
27 
20 
10.0 
36 
25 
12.5 
45 
30 
15.0 
54 
The
velocitytime graph for the car's motion illustrates that velocity changes by
equal amounts in equal time intervals. Hence, for uniformly accelerated motion,
the velocitytime graph is a straight line.
For
uniform velocity, the distance is represented by the area ABCD under the graph.
However, for accelerated motion, the distance s is given by the area
ABCDE.
i.e., s = area ABCDE
= area of the rectangle ABCD + area of the triangle ADE

For
nonuniform acceleration, the velocitytime graph takes various shapes as shown
below:
Fig. (a)
shows the velocity of the object is decreasing with time (negative acceleration).
Fig. (b)
shows the nonuniform variation of velocity of the object with time (irregular
acceleration).
Activity
•
The times
of arrival and departure of a train at three stations A, B and C and the
distance of stations B and C from station A are given below:
Station 
Distance from A (km) 
Time of arrival (hours) 
Time of departure (hours) 
A 
0 
08:00 
08:15 
B 
120 
11:15 
11:30 
C 
180 
13:00 
13:15 
•
Plot and
interpret the distancetime graph for the train assuming that its motion
between any two stations is uniform.
Answer:
Activity
•
Feroz and
his sister Sania go to school on their bicycles. Both of them start at the same
time from their home but take different times to reach the school although they
follow the same route. The table shows the distance travelled by them in
different times.
Time 
Distance travelled by Feroz (km) 
Distance travelled by Sania (km) 
8:00 am 
0 
0 
8:05 am 
1.0 
0.8 
8:10 am 
1.9 
1.6 
8:15 am 
2.8 
2.3 
8:20 am 
3.6 
3.0 
8:25 am 
– 
3.6 
•
Plot the
distancetime graph for their motions on the same scale and interpret.
Answer:
EQUATIONS
OF MOTION
When an
object moves in a straight line with uniform acceleration, its velocity,
acceleration, and distance can be related by
equations of motion. The key equations are:
v = u + at 
Velocitytime relation 
s = ut + ½ at^{2} 
Positiontime relation 
2 as = v^{2} – u^{2} 
Positionvelocity relation 
Here, u
is the initial velocity, v is the final velocity, a is the
acceleration, t is the time, and s is the distance.
Example: A train starting from rest attains
a velocity of 72 km h^{–1} in 5 minutes. Assuming that the acceleration
is uniform, find (i) the acceleration and (ii) the distance travelled by the
train for attaining this velocity.
Solution:
u = 0; v = 72 km h^{–1} = 20 m s^{1} and t
= 5 minutes = 300 s.
(i) v = u + at
(ii) 2as = v^{2} – u^{2} = v^{2} – 0
Example: A car accelerates uniformly from 18 km h^{–1 }to 36 km h^{–1}
in 5 s. Calculate (i) the acceleration and (ii) the distance covered by the car
in that time.
Solution:
u = 18 km h^{–1} = 5 m s^{–1}
v = 36 km h^{–1} = 10 m s^{–1}
t = 5 s.
(i) v = u + at
(ii) Distance covered, s
= ut + ½ at^{2}
= 5 ms^{–1} × 5 s + ½ × 1 ms^{–2} × (5 s)^{2}
= 25 m + 12.5 m = 37.5 m
Example: The brakes applied to a car produce
an acceleration of 6 ms^{2} in the opposite direction to the motion.
If the car takes 2 s to stop after the application of brakes, calculate the
distance it travels during this time.
Solution:
a = – 6 m s^{–2}; t
= 2 s and v = 0 m s^{–1}.
v = u + at
0 = u + (– 6 ms^{–2})
× 2 s
u = 12 m s^{–1}.
s = ut + ½ at^{2}
= (12 ms^{–1}) ×
(2 s) + ½ x (–6 m s^{–2}) (2 s)^{2}
= 24 m – 12 m = 12 m
Thus, the
car moves 12 m before it stops after the application of brakes. It shows the
importance of maintaining a safe following distance to account for braking
distance, ensuring safety on the road.
Uniform
Circular Motion
In some
cases, an object does not change its magnitude of velocity but only its
direction of motion.
Consider an
athlete runs along a closed path of rectangular track ABCD at a uniform speed
on the straight parts AB, BC, CD, and DA. To complete one round of the track, he
needs to change direction at each corner. Thus, the direction of motion is
changed 4 times.
As the
number of sides increases indefinitely, the track’s shape becomes a circle, and
the length of each side decreases. If the athlete moves with constant speed
along this circular path, the only change in velocity results from the
continuous change in direction. Hence, motion along a circular path is an
example of accelerated motion.
The circumference
of a circle of radius r is 2Ï€r. If the athlete takes t seconds
to go once around the circular path of radius r, the speed v is
When an object moves in a circular path with
uniform speed, it is called uniform circular motion.
Activity
•
Take a
piece of thread and tie a small stone at one end. Move the stone to describe a
circular path with constant speed by holding the thread at the other end.
Release the
stone at different points of the circular path. It consistently moves tangentially
to the circle at the point of release. This shows that direction of motion
changes continuously while moving along circular path.
When an
athlete throws a hammer or a discus, it is held in a circular motion by
rotating his body. Upon release, the hammer or discus continues moving in the
direction it was traveling at the moment of release.
Other
examples of uniform circular motion: The moon
orbiting Earth, satellites in circular orbits, and a cyclist on a circular
track at constant speed.
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