MOTION IN A STRAIGHT LINE  
Motion:  
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Motion is the process of continuously changing in position of an object from  
one place to another.  
Types of Motion  
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Linear motion  
➢
Circular motion  
Linear Motion  
➢
Is a motion of an object in a straight line.  
Circular Motion  
➢
➢
Is a motion of an object around the circle.  
Example of circular motion: Rotation of the earth and motion of electrons  
around the nucleus  
Terms used to describe Motion  
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•
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Distance and displacement  
Speed and velocity  
Acceleration and Retardation  
Distance and displacement  
Distance  
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•
•
•
I
s the length of path taken by an object in motion.  
Distance is represented by letter  
The distance is a scalar quantity.  
The SI unit of distance is Metre (m). Other units are Centimeter (cm), and  
Kilometer (km)  
s
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The figure below shows the distance between two points(AB)  
Displacement  
o
o
o
I
s the distance moved by an object in a specific direction  
.
It is a vector quantity.  
The SI unit of displacement is metre (m). Other units are Centimeter (cm),  
and Kilometer (km)  
o
The diagram below shows the distance in a particular direction between the  
two pints.  
Speed and Velocity  
Speed  
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•
•
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Is the distance moved per unit time.  
OR Is the rate of change of distance.  
The speed is represented by letter v.  
It is a scalar quantity.  
풅풊풔풕풂풏풄풆  
풔
Speed (v)  
=
=
풕
푻풊풎풆 풕풂풌풆풏  
•
The SI unit of speed is Meter per Second (m/s). Other unit used is kilometer  
per hour (km/h)  
Velocity  
➢
➢
•
I
s the rate of change of displacement  
It is a vector quantity.  
풅풊풔풑풍풂풄풆풎풆풏풕  
풔
Velocity (v)  
=
=
푻풊풎풆 풕풂풌풆풏  
풕
➢
The SI unit of velocity is Meter per Second (m/s). Other units used is kilometer  
per hour (km/h)  
NB: 10 m/s = 36 km/h  
NB:  
➢
➢
Initial velocity is the velocity of the body at the starting point of observation.  
Final velocity is the velocity of the body at the ending point of observation.  
풖 + 풗  
➢
Average velocity is the mean of initial and final velocities.  
→
Va =  
ퟐ
➢
OR Average velocity is the ratio of the total displacement to the total time.  
풔
Average velocity,  
Va  
=
풕
➢
➢
Uniform velocity is the type of velocity in which the rate of change of  
displacement with time is constant  
.
Instantaneous velocity is the velocity of the body at any instant.  
Example  
1. An object travelled 20 m to the right in 4 s and then 12 m to the left in 3s, for its  
total motion. What was its average speed & its average velocity.  
Data given  
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Total distance traveled, s = 20 m + 12 m = 32 m  
Total time, t = 4s + 3 s = 7 s  
ퟑퟐ  
풕풐풕풂풍 풅풊풔풕풂풏풄풆  
=
= ퟒ. ퟓퟕ 풎/풔  
푨풗풆. 푺풑풆풆풅 =  
풕풐풕풂풍 풕풊풎풆  
ퟕ
•
When the object is moving to right its displacement is positive and when to the left  
its displacement is negative  
Total displacement (s) = 20 + -12 m = 8 m  
풕풐풕풂풍 풅풊풔풑풍풂풄풆풎풆풏풕  
풕풐풕풂풍 풕풊풎풆  
ퟖ
푨풗풆. 풗풆풍풐풄풊풕풚 =  
=
ퟕ = ퟏ. ퟏퟒ 풎/풔  
Individual Task – 1  
1. A ball is dropped from a height of 20m above the ground. It hits the ground in  
2s and bounces back up to a height of 12.7m in 1.6s .What are its average  
velocity  
(ANS: 2.03 m/s)  
2. A 100m runner finishes the race in 10s. What is her average speed?  
(ANS: 10 m/s)  
3. A body covers a distance of 480 m in 6sec. Calculate its speed.(ANS: 80 m/s)  
Acceleration and Retardation  
Acceleration  
➢
Is the rate of change of velocity  
.
➢
It is denoted by small letter “a”  
풇풊풏풂풍 풗풆풍풐풄풊풕풚−풊풏풊풕풊풂풍 풗풆풍풐풄풊풕풚  
풗−풖  
풕
풂풄풄풆풍풆풓풂풕풊풐풏 =  
=
풕풊풎풆  
풗 − 풖  
풕
∴ 풂 =  
➢
➢
Its SI unit is metre per second square (m/s2)  
Uniform acceleration is the type of acceleration in which the rate of  
change of velocity is CONSTANT  
.
Retardation (Deceleration)  
➢
➢
➢
Is the rate of decreasing of velocity  
.
It is referred as negative acceleration.  
Uniform retardation  
i
s the one in which the rate of decreasing of velocity does  
not change  
NB  
:
✓
✓
When a body starts moving from rest its initial velocity become zero, u = 0m/s2  
When a body is brought to rest by the application of brakes its final velocity, v = 0m/s2  
✓
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When a velocity of a moving object increases its acceleration become positive  
When the velocity of a moving object decreases its acceleration become negative  
✓
When a body is moving with a uniform velocity its acceleration becomes zero,  
a = 0 m/s2  
Example  
1. An object is moving at 15 m/s to the right after 8 sec later it is moving at 5 m/s  
to the left, what was the acceleration of the object?  
Solution  
Initial velocity, u = +15 m/s  
Final velocity, v = -5 m/s  
Time taken, t = 8s  
Acceleration, a =?  
풗 – 풖  
풕
−ퟓ – ퟏퟓ  
ퟖ
= −ퟐ. ퟓ 풎/풔 2  
∴ 풂 =  
=
= − ퟐퟎ  
ퟖ
Individual Task – 2  
1. A car brakes and slows down from 20 m/s to 5 m/s in 3 sec. find its  
acceleration  
(ANS: a = -5 m/s2 )  
2. Starting from rest, a sports car accelerate to a velocity of 96 km/h in 16 sec.  
find its acceleration (ANS: a = 1.67 m/s2  
)
3. A car travels at 10 m/s and increase its velocity to 30 m/s in 10 sec. find  
acceleration of the car(ANS: a = 2 m/s2)  
4. A car travels at 45 m/s and decreases its velocity uniformly to 20 m/s in 5 sec.  
find acceleration  
(
ANS: a = -5 m/s2)  
5. A car with a velocity of 90km/h under uniform retardation and brought to rest  
after 10s. Calculate its acceleration (ANS: a = -2.5m/s2)  
Position – time graphs  
Displacement, velocity and acceleration can be represented on a graph.  
Distance (displacement) time graphs  
✓
Displacement time graph is the graph which shows the displacement (y-axis)  
versus time (x-axis).  
Velocity Time Graph  
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I
s the graph which shows the velocity (in y-axis) versus time (in x-axis).  
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Consider a body starts moving from rest and accelerates uniformly to a  
velocity, v after time, t1. It then moves with this velocity for time, t2 and then  
comes to stop after another time, t3.  
The above information can be represented on the velocity time graph as shown.  
Deduction from velocity time graph  
a) The Area under the velocity time graph = Total distance travelled by a body  
b) The slope of the velocity time graph represents acceleration  
Example  
1. A car travel with uniform velocity of 30m/s for 5 second and then comes to rest  
10 second with uniform deceleration.  
i) Draw a velocity-time graph of the motion.  
ii) Find the total distance travelled.  
iii) Find the average velocity.  
Solution:  
i)  
velocity-time graph of the motion  
ii. Total distance travelled, s =?  
S = (30 x 5) + (30 x 10)/2 = 150 + 300/2 = 150 + 150 = 300 m  
iii. From velocity time graph  
Total distance, s = 300 m  
Total time taken, t = 15 s  
ퟑퟎퟎ  
ퟏퟓ  
풔
∴
Average velocity  
=
=
= ퟐퟎ 풎/풔  
풕
Equations of uniformly accelerated Motion  
(Equations of Linear Motion)  
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First equation of motion  
Second equation of motion  
Third equation of motion  
Consider a body moving with a constant acceleration  
풂
from an initial velocity  
풖, to  
a final velocity  
풗
.The body covers a displacement ,풔, after sometime  
풕.  
❖
Derivation of first equation  
Now, find the acceleration of the body  
풗−풖  
∆풗  
From: a =  
=
, 풎풂풌풆 풗 푡ℎ푒 푠푢푏푗푒푐푡  
풕
풕
푎푡 = 푣 − 푢 → 푣 = 푢 + 푎푡  
∴ 푇ℎ푒 1푠푡 푒푞푢푎푡푖표푛 푖푠 푔푖푣푒푛 푏푦  
풗 = 풖 + 풂풕  
❖
Derivation of the second equation  
Find the average velocity of the body  
풗+풖  
풔
푽 =  
=
풃풖풕 풗 = 풖 + 풂푡  
풂
ퟐ
풕
풖+풂풕+풖  
ퟐ풖+풂풕  
풔
풕
푽풂 =  
=
=
ퟐ
풕
풔 -------------------------  
ퟐ풖+풂풕  
=
multiply by t each side  
ퟐ
풕
ퟐ
ퟐ풖풕+풂풕  
ퟏ 풂풕ퟐ  
ퟐ
풔 =  
= 풖풕 +  
ퟐ
ퟏ
ퟐ
풂풕ퟐ  
∴ 풕풉풆 ퟐ풏풅 풆풒풖풂풕풊풐풏 풊풔 품풊풗풆풏 풃풚 풔 = 풖풕 +  
❖
Derivation of third equation  
Consider the 1st equation:  
풗 = 풖 + 풂풕  
Then square the equation in each side: (v)2 = (u+at)2  
This gives: v2 = u2 + 2uat + a2t2 = u2 + 2a(풖풕 +  
ퟏ 풂풕ퟐ)  
ퟐ
ퟏ 풂풕ퟐ  
풔 = 풖풕 +  
But:  
ퟐ
Therefore풗ퟐ = 풖ퟐ + ퟐ풂풔  
∴ 풕풉풆 ퟑ풓풅 풊풔 풆풒풖풂풕풊풐풏 풊풔 품풊풗풆풏 풃풚  
풗ퟐ = 풖ퟐ + ퟐ풂풔  
Example  
1. A body moving with a velocity of 30m/s is accelerated uniformly to a velocity of  
50m/s in 5s.Calculate the acceleration and the distance traveled by the body.  
Data given  
Initial velocity (u) = 30m/s,  
Final velocity (v) = 50m/s, Time (t) = 5s  
Acceleration =?  
풗−풖  
ퟓퟎ−ퟑퟎ  
ퟓ
• 풂 =  
=
= ퟒ 풎풔−ퟐ  
풕
Distance traveled =?  
ퟏ
풖풕 + ퟏ 풂풕ퟐ = ퟑퟎ 풙 ퟓ +  
•
풙 ퟒ 풙 ퟓ 풙 ퟓ = ퟐퟎퟎ 풎  
S
=
[
Motion under Gravity  
❖
All bodies on the earth will always fall down towards the earth’s surface when  
released from a point. What makes these bodies falling downwards is the  
acceleration of free falling body called acceleration due to gravity which is 9.8 or  
10 N/kg.  
❖
❖
Acceleration of free falling body is denoted by ‘g’. Light bodies like feathers, paper  
etc are observed to fall down more slowly than iron balls. This is because light  
bodies are very much affected by air resistance.  
There are two important characteristics of free fall  
(a) Free falling objects do not encounter air resistance  
(b) All free falling objects on earth accelerate downwards at a rate of 9.8  
m/s2 (often approximated as 10 m/s2)  
•
For a body moving downwards the following formulae are applied  
(Here 풂 = 품 풂풏풅 풔 = 풉)  
풗 = 풖 + 품풕  
ퟏ
2푛푑 푒푞푢푎푡푖표푛 푖푠 푔푖푣푒푛 푏푦  
풉 = 풖풕 + 품풕ퟐ  
ퟐ
3푟푑 푖푠 푒푞푢푎푡푖표푛 푖푠 푔푖푣푒푛 푏푦 풗ퟐ = 풖ퟐ + ퟐ품풉  
•
When the body moves upwards ,the formulae will change to:  
(Here 풂 = −품 풂풏풅 풔 = 풉)  
풗 = 풖 − 품풕  
2푛푑 푒푞푢푎푡푖표푛 푖푠 푔푖푣푒푛 푏푦  
풉 = 풖풕 − ퟏퟐ 품풕ퟐ  
3푟푑 푖푠 푒푞푢푎푡푖표푛 푖푠 푔푖푣푒푛 푏푦 풗ퟐ = 풖ퟐ − ퟐ품풉  
Example  
A stone is thrown vertically upward from the ground with a velocity of 30 m/s. find  
a) Maximum height reached  
b) Time taken for maximum height  
c) Time taken for reach ground again  
d) The velocity reached half-way to the maximum height  
SOLUTION  
a) Maximum height reached, s = H =?  
Data given:  
Initial velocity, u = 30 m/s  
Final velocity, v = 0 m/s  
Acceleration, a = -g = -10 m/s2  
From: third equation of motion  
풗ퟐ = 풖ퟐ − ퟐ품풉 ----------------------- make h the subject  
ퟐ
ퟐ
ퟐ
ퟐ
풖 −풗  
ퟑퟎ −ퟎ  
ퟗퟎퟎ  
ퟐퟎ  
풉 =  
=
=
= ퟒퟓ 풎  
ퟐ품  
ퟐ풙ퟏퟎ  
b) Time taken for maximum height, t =?  
From: first equation of motion  
풗 = 풖 − 품풕 ----------------- make t the subject  
풖−풗  
품
ퟑퟎ−ퟎ  
ퟏퟎ  
ퟑퟎ  
ퟏퟎ  
풕 =  
=
=
= ퟑ 풔  
c) Time taken for reach ground again,(T = 2t) ?  
T = 2t = 2 x 3 = 6 sec  
d) Velocity reached half-way to the maximum height, v =?  
When stone is halfway to maximum height, the height attained is  
푯
ퟒퟓ  
풉 =  
=
= ퟐퟐ. ퟓ 풎  
ퟐ
ퟐ
Data given:  
u = 30 m/s and a = -10 m/s2  
From: third equation of motion  
v2 = u2 – 2gh  
풗 = √풖ퟐ − ퟐ품풉 = √ퟑퟎퟐ − ퟐ풙ퟏퟎ풙ퟐퟐ. ퟓ = ퟗퟎퟎ − ퟒퟓퟎ = ퟒퟓퟎ  
√
√
∴ 풗 = ퟐퟏ. ퟐ 풎  
Simple Pendulum  
•
Simple pendulum is a small heavy body suspended by a light inextensible string  
from a fixed support  
•
When pendulum bob swings it reached maximum displacement called  
Amplitude and the angle between string and vertical axis is called Angular  
Amplitude  
θ = Angular Amplitude  
A = Amplitude  
•
When the length of string change while the mass of pendulum bob is constant,  
the period is always constant and that constant time is given by  
푳
푻 = ퟐ흅  
√
품
Where:  
T = Period to complete oscillation  
L = Length of string  
g = Acceleration due to gravity  
Application of gravitational force  
•
Used to launch satellites and space vehicle into space requires overcoming the  
gravitational attraction forces for take off  
•
•
•
Used to keep satellite rotating on the earth’s orbit  
It causes everybody to be attracted towards the earth’s surface  
It is used to calculate the time taken by object to reach to ground for all objects  
near the earth’s surface. E.g. army aircraft when firing bombs, parachutist move  
under free fall,(a = g)