TOPIC: MECHANICS AND PROPERTIES OF MATTER
General Objective: The learner should be able to use knowledge of motion and its equations to
understand relationship between force, energy and motion.
SUB-TOPIC: Linear motion
SPECIFIC OBJECTIVES
The earner should be able to;
• Define speed and average speed.
• Calculate speed and average speed.
• Define displacement, velocity and acceleration.
• Define uniform velocity and uniform acceleration.
• Draw and interpret velocity graphs for linear motion.
• Use equations of motion to solve numerical problems.
• Use ticker-timer t, find velocity and acceleration.
• Define acceleration due to gravity, g.
• Describe a simple experiment to determine, g.
LINEAR MOTION
This involves study of motion of a body in a straight line.
Distance
This is the total length of path travelled by a body.
Or
Is the measurement along the exact path followed by a moving body.
The SI unit of distance is metre (s)
Displacement
This is the distance moved in a specified direction.
If student followed the path from Labs, she would cover a distance of 120m. However, if she
moved in a straight line, she would cover a displacement of 70 m.
If the student moved from Kivebulaya house to the Laboratory and back to the house along the
path, she would have covered a total distance of 240 m, but her resultant displacement would be 0
m.
So, displacement is a vector because it is described by both magnitude and direction. On the other
hand, distance is a scalar.
Speed
Speed is the rate of change of distance with time.
Speed =
distance
time taken
S.3 MOTION NOTES 2020 2
Its SI unit is metre per second (ms
−1
).
It is a scalar since it is specified by magnitude only.
Uniform speed
A body is said to move with uniform speed if it covers equal distances in unit time intervals.
However, quite often a body moving between two points does so with varying speeds. Such a body
is said to move with non-uniform speed. In such a case the speed between the two points is
called average speed.
Average speed =
Total distance
total time taken
Examples
1. What is the speed of a racing car in metres per second if the car covers 360km in 2 hours?
(50ms−1
)
2. A car is moving along a straight road ABC as below maintains an average speed of 90kmh−1
between points A and B and 36kmh−1 between points B and C.
Calculate the:
(a) total time taken in seconds by the car between points A and C. (300s)
Average speed =
total distance
total time
total time between A and B =
total distance between AB
average spedd
=
1.5
90
hours =
1.5
90
× 3600 = 60 s
total time between B and C =
total distance between BC
average speed
=
2.4
36
hours =
2.4
36
× 3600 = 240 s
Total time between A and C = 60 + 240 = 300 s.
(b) average speed in metres per second of the car between points (13ms−1
)
Average speed =
total distance
total time
=
(1.5+2.4)×1000 m
300 s
=
39
3
= 13 ms−1
.
Velocity
This is the rate of change of displacement with time.
Velocity =
distance moved in a particular direction (displacement)
time taken
S.3 MOTION NOTES 2020 3
The SI unit is metres per second (ms
−1
).
It is a vector, since it is specified by both magnitude and direction.
In some cases, the velocity of a moving body keeps on changing. In such cases, it is better to
consider the average velocity of the body.
Average velocity =
total displacement
time taken
A particle is said to move with uniform velocity if its displacement changes by equal amounts in
equal time intervals.
When the velocity in a particular direction is constant, the velocity is referred to as uniform
velocity.
Uniform velocity is the constant rate of change of displacement with time.
For example, the table below shows the displacement of a girl and the corresponding time taken.
Displacement (m) 0 4 8 12
Time taken (s) 0 2 4 6
A displacement-time graph of the girl’s motion can be plotted as below:
A straight line graph is obtained.
The slope of the graph is the constant velocity at which the girl is moving.
Slope, v =
change in displacement
change in velocity
= velocity.
Slope, v =
(16−4) m
(8−2) s
=
12
6
= 2 ms
−1
.
The velocity after every two seconds is 2ms
−1
, hence its velocity is uniform.
The girls motion can also be represented on a velocity – time graph as below:
Displacement (m) 0 4 8 12
Time taken (s) 0 2 4 6
Velocity (ms
−1
) 0 2 2 2
S.3 MOTION NOTES 2020 4
A straight line parallel to the time axis is obtained and this confirms that the girl is moving at a
constant velocity and zero acceleration.
The total displacement covered by the girl in the 8 seconds can obtained as the area under the
velocity –time graph as follows:
Displacement = area under velocity time graph
Displacement = velocity × time = 2 × 8 = 16 m.
Example
A car travelled from town A to town B 200km east of A in 3hours. The car changed direction and
travelled a distance of 150km due north from town B to town C in 2 hours as shown below.
Calculate the average;
(a) speed for the whole journey. (70 kmh
−1
)
(b) velocity for the whole journey. (50 kmh
−1
from A to C.)
Acceleration
A body is said to be accelerating when its velocity changes.
Definition
This is the rate of change of velocity with time.
Acceleration =
change in velocity of body
time taken
Its SI unit is metres per square second (ms
−2
).
If the acceleration of a body is 4ms
−2
, it means that its velocity is increasing by 4ms
−1 every
second.
NOTE
Acceleration can be positive or negative. If the acceleration is increasing then it is taken to be
positive and if it is decreasing (decelerating or retarding), it is taken to be negative.
A body moving with uniform velocity has zero acceleration since there is no change in
velocity.
When the rate of change of velocity with time is constant, the acceleration is referred as uniform
acceleration.
Consider a body moving with velocity, v, in time, t, as shown in the table below.
velocity (ms
−1
) 0 5 10 15 20
Time taken (s) 0 2 4 6 8
The information can be plotted on a velocity-time graph as below
S.3 MOTION NOTES 2020 5
A straight line graph is obtained. The slope of the graph is a constant value obtained as below:
slope =
change in velocity
change in time
= acceleration, a
slope =
20 − 5
8 − 2
=
15
6
= 2.5 ms
−2 = acceleration, a
i.e. the velocity increases by 5ms
−1
for every 2 seconds. Thus, the body is said to be accelerating
uniformly at 2.5 ms
−2
The displacement of the body in the 8 seconds is found as the area under the velocity-time
graph as follows.
Displacement, s =
1
2
× velocity × time =
1
2
× 8 × 20 = 80 m.
Example:
The table below represents the velocity of a vehicle after a given time.
Velocity (ms-1) 0 3 6 9 12 15 15 15 15 15
Time taken (s) 0 1 2 3 4 5 6 7 8 9
(a) Plot a velocity – time graph representing the motion of the vehicle.
(b) Find the slope of the graph in the first 5 seconds of the vehicle’s motion.
(c) Use the graph to describe the motion of the vehicle in the 9 seconds.
(d) Use the graph to determine the total displacement of the vehicle in the 9 seconds of its
motion.
Solution.
(a)
(b) slope =
change in velocty
change in time
=
(15−0) ms
−1
(5−0) s
=
15
5
= 3 ms
−2
.
(c) The vehicle starts from rest accelerates uniformly at 3 ms
−2 until it attains a
velocity of 15 ms-1 in the first 5 seconds. It then moves with this velocity constantly for the
next 4 seconds.
(d) Displacement = area under the velocity time graph
Displacement, s = area of the trapezium =
1
2
× 15 × (9 + 4)
Displacement, s = 97.5 m
S.3 MOTION NOTES 2020 6
Example
A car accelerates from rest to a velocity of 20ms
−1
in 5s. thereafter it decelerates to rest in 8s.
Calculate the acceleration of the car.
(a) in the first 5s, (4ms
−2
)
Acceleration =
change in velocity of body
time taken
Acceleration =
final velocity − initial velocity
time taken
(rest means velocity is zero)
Acceleration =
20−0
5
= 4ms
−2
(b) in the next 8s. (-2.5ms
−2
)
Acceleration =
final velocity − initial velocity
time taken =
0 − 20
8
−2.5ms
−2 Or deceleration = 2.5ms
−2or retardation = 2.5ms
−2
MOTION –TIME GRAPHS AND THEIR INTERPRETATION.
The motion of objects can be represented graphically. During the objects’ motion, the
displacement and velocity of the body usually changes with time. Consideration will be given to:
1. Displacement-time graphs.
2. Velocity-time graphs.
and the information we derive from them.
1. Displacement –Time Graphs
(a) Moving with constant velocity/uniform velocity/steady velocity
(b) Accelerating uniformly (non-uniform velocity)
This graph is an example of a stone that drops from rest, the displacement covered in each second
is not equal but rather increasing.
(c) A body at rest.
When a body is at rest (stationary), its position remains the same even as time passes by.
Time (s)
Displacement (m)
Time (s)
Displacement (m)
S.3 MOTION NOTES 2020 7
(i) The displacement-time graph represents a body that is at rest at its original position.
(ii) The displacement-time graph represents a body that is at rest 5 m after or away from its
original position.
(iii) The displacement-time graph represents a body that is at rest and 5 m before its original
position.
Note: (i) The gradient of a displacement-time graph gives the velocity of the particle.
(ii) When the body is stationary the gradient of the graph is zero and hence
the velocity is zero.
A body may be moving to the left or right away from the reference point, O.
The displacement to the right of the reference point is considered to be positive, while that to the
left is negative.
Suppose the body is moving to the left of its reference point, O, graphs are as shown below.
(i) Displacement-time graph (ii) Velocity-time graph
The slope of the displacement-time graph is change in displacement
change in time
=
−20
10
= −2ms−1
, the negative sign
means the body is moving in the opposite direction.
2. Velocity – Time Graphs
Here, the acceleration is zero Since the velocity is the same i.e. does not increase or decrease.
Time (s)