INTRODUCTORY PHYSICS
Semester 1.
Lecturer :
email :
Catherine Walsh
Room
cwalsh@wit.ie
326
Lectures : 3 hours per week
Tutorial : 1 hour per week
Practical : 2 hours per week
SYLLABUS
1.
INTRODUCTION. S.I. units, prefixes, scientific notation. Vectors and
scalars.
2.
MECHANICS. Motion in one direction, Newton’s laws of motion. Work, Energy
and Power. Uniform circular motion. Conservation laws, mechanical energy, total
energy momentum.
3.
PROPERTIES OF MATTER. Structure of matter. Pressure. Elasticity, viscosity,
surface tension.
4.
WAVE MOTION AND VIBRATIONS. Wave classification and properties.
Sound waves, characteristics of sound. Doppler effect.
For this module no previous knowledge of physics is assumed.
This module will be delivered by a combination of lectures, tutorials and practical
work.
Lectures will be used to present new topics and their related concepts. Problem
solving will be used in lectures and tutorials to analyse physical situations and find
numerical solutions.
There is also an integrated practical programme designed to run in parallel with the
lectures and the develop experimental skills.
Lectures. Your attendance is expected at all lectures. You are responsible for all
material covered in lectures as well as any problem assignments given.
Lectures will cover ideas, concepts, and sample problems. The lectures will facilitate
the acquisition of factual knowledge the application of the knowledge by problem
solving.
You will be required to have a calculator at physics lectures.
Tutorials. The class will be divided into smaller groups for a one hour tutorial every
week.
The purpose of the tutorial is to facilitate the learning process by providing more
individual attention to science students to help them learn and understand physics
concepts and problem-solving skills.
Tutorials offer the time for feedback and discussion. Again you will require a
calculator.
Practicals : Each student will have a two hour laboratory session every week. You
will be required to have two note books for practical work.
1.
A direct log book, this note book should be used as a work book during each
practical session for recording data etc. This note book should also be signed by
your lab supervisor at the each of each practical.
2.
A report book for the formal writing of the practical you have carried out. Your
lab supervisor will make arrangement for the collection of these report books.
You are required to attend ALL practicals.
If you should miss more than one session per term you should contact your lab
supervisor immediately.
The final mark for this subject will be derived from a combination of assignment,
laboratory mark and final exam.
Final Exam
Continuous Assessment
50%
50%
Physics is a fundamental science concerned with the interactions of matter and
energy, time and space.
Physics is concerned with how things behave and in discovering the general principles
which explain natural phenomena.
This course is designed to impart an understanding of the basic physics principles to
enable a student to solve a variety of problems and to develop laboratory skills.
MEASUREMENT.
All measurement involves a number and a unit. Scientists use the Systeme
International ( S. I. ) units which was recommended by the general conference on
Weights and Measures in 1960. In physics there are six basic units
Quantity
Symbol
Unit
Symbol
Length
l
metre
m
Mass
m
kilogram
kg
Time
t
second
s
Electric
Current
I
Ampere
A
Temperature
T
Kelvin
K
Luminous
Intensity
I
Candela
cd
Definitions.
Metre : 1 metre = 1,650,763.73 wavelengths of
Light emitted by a Krypton-86 Atom.
Kilogram : 1 kilogram = mass of international
Standard platinum-iridium cylinder kept at the
International bureau of weights And measures in Serves, France.
Second : 1 second = 9,192,631,770 periods of vibration of the radiation emitted by a
caesium-133 atom.
The other three units we will define as we meet them later in the course.
DERIVED UNITS.
Derived units can be constructed in terms of the independent base units.
The derived units can be written in terms of the basic units and are derived from the
definition of the quantity.
Example
Area
Area rectangle  length x width
Area square  a 2  a.a
base.height b.h
Area triangle 

2
2
Area circle   .r 2   .r.r
In each case length, width, radius , base etc the basic S.I. unit is meter (m)

Therefore the S.I. unit of area is m2
Quantity
Symbol
Unit
Symbol
m2
Area
A
metre squared
Volume
V
metre cubed
m3
Velocity
v
metre per second
m.s-1
m.s-2
Acceleration
a
metre per second
squared
Density
d
kilogram per
metre cubed
kg.m-3
There are other frequently used quantities which are given individually named units.
In each case the unit name honour famous scientists for their work in the fields which
use these units.
Quantity
Symbol
Unit
Symbol
Force
F
Newton
N
Pressure
P
Pascal
Pa
Energy,Work
E,W
Joule
J
Power
P
Watt
W
Each of these units can be written in terms of the base units
e.g.
Newton = kg m s-2
Pascal = kg m-1 s-2
Question 1. Given the following equations derive the S.I. units for each of the following
quantities
distance travelled
time taken
change in velocity
acceleration 
time taken
velocity 
Density 

mass
Volume
Question 2. Given that
Force  mass  accelerati on
Work  force  distance
work done
Power =
time taken
Determine the unit of Force, Work and Power in terms of the basic S.I. units.
SCIENTIFIC NOTATION.
In Physics we use scientific notation to express the numerical value for a physical
quantity. Using the POWERS OF TEN
100 = 1
101 = 10
102 = 101 x 101 = 10 x 10 = 100
103 = 101 x 101 x 101 = 10 x 10 x 10 = 1000
RULE 1. WHEN MULTIPLYING ADD THE INDICES
10m x 10n = 10 m + n
100 = 1
10-1 = 1 / 10 = 100 / 101
10-2 = 1 / 100 = 100 / 102
10-3 = 1 / 1000 = 100 / 103
RULE 2: WHEN DIVIDING SUBTRACT THE INDICES
10a / 10b = 10 ( a – b )
RULE 3 : WHEN USING SCIENTIFIC NOTATION PUT ONE DIGIT
BEFORE THE DECIMAL POINT AND ALL OTHERS AFTER
e.g.
190 m = 1.9 x 102 m
Rule 4
For every place the decimal point moves to the right SUBTRACT one from the power
of 10
Rule 5
For every place the decimal point moves to the left ADD one from the power of 10
Rule 5
10n
= 1 x 10n
USE OF CALCULATOR
When entering a number in scientific notation into a calculator it is important that you
follow the correct procedure
Pressing the exponential EXP button on the calculator is the same as
Multiplied by 10 to the power of 10
Examples
103
is entered into a calculator as
1 EXP
3
10-3 is entered as
1 EXP
+/- 3
5.3 x 10-7
5
.
3
EXP
+/-
7
NOTE 1: You do not have to enter the 10 into the calculator the EXP button takes
care of that
NOTE 2: 5.3 x 10-7 will be displayed on the calculator as
5.3 –07
Always remember to write this down as
x 10-7
5.3
as 5.3 –07 written down on a sheet of paper means 5.3 to the power of –7 NOT
5.3 by ten to the power of -7
always remember when transferring this back down to a sheet of paper it is
5.3 x 10-7
Question 1. Calculate each of the following values
5 x 106 (2x 103) (3x 103)
_________________________
______5 x 10 4
(4 x 106) (5x 10-3)
___________________
(8 x 10 -4 )(5x 103)
Question 2. Write the following numbers in scientific notation.
1. 1001
6. 0.13592
2. 53
7. -0.0038
3. 6,926,300,000
8. 0.00000013
4. -392
Question 3. Calculate each of the following values.
1. 102 x 105 2. 1015 x 1039
3.1016 x 10-13
4. 10-8 x 102
5. 10-8 x 10-5
5. 0.00361
For quantities very much larger or smaller than the standard unit multiples or submultiples are used
Multiple
Prefix
Symbol
109
giga-
G
106
mega-
M
103
kilo-
k
10-2
centi-
c
10-3
milli-
m
10-6
micro-
10-9
nano-
n
10-12
pico-
p
1 cm
=
1 x 10-2 m
1 mm
= 1 x 10-3 m
1 km
=
1 ms
= 1 x 10-3 s
1 x 103 m
-6
s
Question 1. Use prefixes to express the following
(i) 0.000001 m (ii) 3 x 10-9 m
(iii) 4560000 m (iv)9 x 109 N
Question 2. Express the following quantities using standard notation
(i)
(iv) 589 nm (v)1.4 MW
AREA CONVERSIONS.
S.I. unit of area is m2 however area is often given in either cm2 or mm2
units. We need to relate these to m2.
1 cm2 = 1 cm x 1 cm
= 1 x 10-2 m x 1 x 10-2 m
= 1 x 1 x 10-2 x 10-2 x m x m
1 cm2 = 1 x 10-4 m 2
SIMILARLY
1 mm2
=
1 mm x 1 mm
= 1 x 10-3 m x 1 x 10-3 m
= 1 x 1 x 10-3 x 10-3 x m x m
= 1 x 10-6 m2
VOLUME CONVERSIONS
1 cm3 = 1 cm x 1 cm x 1 cm
= 1 x 10-2 m x 1 x 10-2 m x 1 x 10-2m
= 1 x 1 x 1 x 10-2 x 10-2 x 10-2 x m.m.m
= 1 x 10-6 m3
1 mm3 = 1 mm x 1 mm x 1 mm
= 1 x 10-3 m x 1x 10-3m x 1 x 10-3m
= 1 x 1 x 1 x 10-3 x 10-3 x 10-3 x m.m.m.
1 mm3 = 1 x 10-9 m3
Question 1. Convert each of the following into correct S.I. units
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
an area of 500 cm2
an area of 1600 mm2
the area of a circle of diameter 5cm
the area of a square of length 155mm
a volume of 6578 cm3
a volume of 442 mm3
the volume of a sphere of radius 23cm
the volume of a cylinder of diameter 7cm and height 15.54cm.
Orders of Magnitude
The following tables give some feeling of the "typical" numbers that are associated
with each measured quantity. You need not worry about the details of each number just the orders of magnitude.
Object or Distance
Size( m )
Universe
1026
Milky Way Galaxy
1021
Nearest Star
1016
Solar System
1012
The Sun
109
The Earth
106
A Mountain
103
Humans
100
A Cell
10-5
An Atom
10-9
The Nucleus
10-14
Duration/Age
Time
(s)
Age of The Solar System
1017
Last Ice Age
1012
Human Life Time
109
A Day
104
A Lecture
103
A Moment/A Second
100
Question 1.
(a)The speed of light is 3 x 10 8 meters/second. If the sun is 1.5x1011 meters from earth, how
many seconds does it take light to reach the earth. Express your answer in scientific notation.
(b)If the earth is a sphere of Radius = 6.3×106 m.
and Mass = 5.9742 × 10 24
kilogram. Calculate an approximate value for the density of the earth in kg/m 3 (Note:
this method will only yield an approximate value for the density of the earth).
Question 2. Determine the conversion factor from
(i)
kmhr-1 to ms-1
(ii)
mileshr-1 to ms-1
Where 5 miles = 8 km
Question 3 An airplane travels at
750 km/h
(i)
How long does it take to travel 100km
(ii)
How long does it take to fly from Waterford to London
Question 4 Estimate each of the following
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
(ix)
the distance from B07 to B18
the distance from Waterford to Dublin
the floor area of B18
the volume of B18
the number of litres of water a human drinks in a lifetime
the volume of your body
cross-sectional area of a 5 cent piece
how long it would take one person to mow a football pitch using a
mower which has a speed of 1 km/h and a 0.5m width
how many books can be shelved in a college library with 3500 m2 of
floor space. Assume 6 shelves high having books on both sides with
aisles 1.5 m wide.
PHYSICAL QUANTITIES.
Physical quantities can be divided into two categories
Those which have a direction and
Those which do not.
SCALAR.
A quantity which is completely specified by magnitude only is defined as a scalar
Magnitude - Numerical value and Associated Unit
e.g mass, time, length,
speed
VECTOR.
A quantity which is specified by two values magnitude and direction is called a
vector
e.g.
displacement, velocity, force
Vector- Numerical value and unit and direction
Typical the difference is shown by the terms speed and velocity
Speed is a scalar while velocity is a vector
Speed
10 m . s-1
SCALAR
Velocity 10 m . s-1 due East
VECTOR
VECTOR ALGEBRA.
The Mathematics of scalar quantities is the ordinary algebra with which we are
very familiar.
However when dealing with vectors we must conform to the rules of vector
algebra.
Vectors are represented by straight lines where
(i)
The magnitude of the vector is proportional to the length of the line
(ii)
The direction of the vector is given by the direction of the line shown
using an arrow on the line.
e.g.
10 m s-1 due east represented by a horizontal line of length 10 cm
m s-1 due west may be represented by a horizontal line of length 10 cm in
the opposite direction
10
5 m s –1 due east may be represented by a horizontal line of length 5 cm
The negative of a vector is that vector with its direction reversed
+5N
-5N
When we write about vector quantities to distinguish them from scalars, the longstanding custom in Physics is to draw a small arrow over the symbol that represents a
vector
e.g.
s
v
representing displacement and velocity
ADDITION OF VECTORS
In geometry
1. A parallelogram is a quadrilateral with two sets of parallel sides. The opposite sides
of a parallelogram are of equal length, and the opposite angles of a parallelogram are
congruent(equal)
A
B
A
B
2. A triangle has three sides of lengths a, b and c and angles A, B and C. Notice that
the side of length a is opposite the angle A, and similarly for b and c
A + B + C = 1800
3.
If a diagonal is drawn through a parallelogram two triangles are formed.
A
B
A
B
The total angle therefore is 3600
2 ( A + B ) = 3600
A + B = 1800
Parallelogram Law
If two vectors are represented both in magnitude and direction by the adjacent sides
of a parallelogram, their resultant will be represented by the diagonal of the
parallelogram drawn from the point.
The point being a common starting point for both vectors.
ALSO
If two vectors A and B are added to give a resultant vector C then by the reverse
process the vector C may be represented by the sum of two components A and B.
IF
A
+
B
=
C
THEN
C
=
A
+
B
C is also a vector quantity i.e. it has both a magnitude and a direction.
This process is called RESOLVING VECTORS and we will return to this later.
Example 1. Add two forces where
F1 = 2 N south
and
F2 = 4 N east.
To solve this we can use two methods
Method 1. By Construction using a scaled diagram
Decide on a scale
1. Construct a vector diagram to scale, so that the two vectors form two adjacent
sides of a parallelogram
2. Complete the parallelogram, using a protractor and ruler.
3. Draw in the diagonal of the parallelogram. Measure the length of the diagonal of
the parallelogram.
4. The magnitude of the resultant is calculated by measuring the length of the
diagonal of the parallelogram and converting to real units using the scaling value.
5. The direction of the diagonal represents the direction of the resulting vector.
Method 2. By calculation using the COSINE and SINE RULES.
1.Draw a vector diagram to represent the vectors to be added (You do not need to use an
accurate scale). The two vectors need to form two adjacent sides of a parallelogram

2. Complete the parallelogram. Draw in the diagonal of the parallelogram. The diagonal
divides the parallelogram into two triangles.
B
c
B
C
A
a
b
3. Usually we then look at the triangle formed by the two vectors and the resultant
vector and use the Cosine rule to evaluate the magnitude of the resultant vector
c
B
C
A
Cosine Rule
b
a
Often the triangles occurring in real problems are not right-angled, in which case we
can use the "sine rule" and the "cosine rule" to help
Cosine Rule For any triangle
c  a  b  2abCosC
2
2
2
Angles in a triangle A + B + C =180
Therefore C = 180 – (A+B)
Recall A + B =  (the angle between the two vectors a and b)
Therefore
C = 180 -
The equation can be written as
c 2  a 2  b 2  2abCos(180   )
c
a 2  b 2  2abCos(180   )
4. Using the same triangle use the sine rule to calculate the direction of the vector
a
b
c


SinA SinB SinC
b
c

SinB SinC
SinB 
bSinC
c
The equation can be rewritten as

bsin C 
1bSin(180   ) 
B  Sin 
 Sin 

 c 


c
1

Example 1. Calculating Vector Forces
A force 1 of magnitude 5 kN is acting in a direction 80o from a force 2 of magnitude
8 kN. Assume that the first force acts along the +x direction.
By construction.
1. Decide on a scale
1 kN = 1 cm
2.
Draw the vector diagram.
Draw vector 1 using appropriate scale and in the direction of its action
From the starting point of vector 1 draw vector 2 from the same point using the same
scale in the direction of its action
3.
Complete the parallelogram by using vector 1 and 2 as sides of the
parallelogram include the diagonal.
4.
Measure the length of the diagonal of the parallelogram and convert to real
units using the scale.
5.
Measure the angle between the diagonal (resulting vector ) and the +x axis.
This is also the direction of the resulting vector.
The resulting force can be calculated as
FR = [ (3(kN))2 + (8(kN))2 - 2 5(kN) 8(kN) cos(180o - (80o)) ] 1/2
= 9 kN
The angle between and the resulting vector and the +axis can be calculated as
B = sin-1[ 3(kN) sin(180o - (80o)) / 9(kN) ]
= 19.1o
PROBLEM SHEET.
Question 1. Forces of 8 N and 10 N act at the origin and are inclined at 60o to each
other. Assuming that the 8 N force acts along the + x direction calculate the
resultant force using both methods.
Question 2. Two forces each of magnitude 8 N act at a point O and are inclined at
95o to each other. Calculate the resultant.
Question 3. Find the magnitude and direction of the resultant for a force of 5 N due
east and a force of 8 N 45o North of east.
SUBTRACTION OF VECTORS.
The vector - A means a vector equal in magnitude but opposite in direction to the
vector + A.
A
-
B
=
A
+
(-B )
B
RESULTANT
ALSO
A -
A
= 0
A
+
AND IF
C
=
THEN
- C is called the EQUILIBRANT
Example : A is a force of magnitude 5 N acting along the + x axis, B is a force of
magnitude 8 N acting along the + y axis. Calculate
(i)
Their resultant
(ii)
Their equilibrant and
(iii)
A
-
B
RESOLVING VECTORS.
We can also use the parallelogram law to convert a single vector into two
components.
When this is done the vector is said to be resolved into its components.
The vectors are resolved into two chosen directions at right angles to each other.
The resolved part of the vector, in a particular direction, tells us the effect of the
vector in that direction.
The direction of the components are usually taken along an x axis and y axis i.e.
horizontal and vertical directions.
The magnitudes of the components can be found by using a scaled diagram or by
calculation.
By construction.
Select a scale and accurately draw the vector to scale in the indicated direction.
Include the x and y axis.
To find the component of a vector in a particular direction drop a perpendicular
from the vector to the direction of the components i.e. to the x axis and to the y
axis.
Measure the length of each of the components and use the scale to determine the
magnitude of the components in real units.
NOTE : A vector has NO component at right angles to itself.
COMPONENTS OF VECTORS.
A component of a vector in a given direction e.g
Fx
Fy
F1
F2
Tells us the effect of the vector in a particular direction.
The directions of the components do not necessarily have to be in the x and y
directions but they do have to be at 90o to each other.
e.g. Suppose C is a VECTOR where
Cx is the component of C in the x direction
and
Cy is its component in the y direction
To find Cx and Cy by construction.
Draw a vector diagram to represent C, show x and y axis
Cy
C
x
Drop a perpendicular from C to the x axis Then the vector from the origin to the
point along the x axis represents the x component Cx
Drop a perpendicular from C to the y axis for Cy

By calculation.
In a right angle triangle
adj
Cos 
hyp
CX
Cos 
C
Cross multiply
CX  C.Cos
opp
Sin 
hyp
CY
Sin 
C
CY  C.Sin
Where C = Magnitude of the vector and

Question 1 . A force of 100 N makes an angle of 40o to the + x axis calculate its x
and y components.
Question 2. A force acting at an angle of 30o to the + x direction has a y component
of 55 N calculate (i) the value of the force and (ii) the x component.
Kinematics describes motions of objects as a function of time but does not consider the
causes of the motion.
The study of the causes of motion is called dynamics.
The branch of physics which deals with the kinematics and dynamics of macroscopic
( large scale ) objects is called MECHANICS.
Physical Quantities.
To fully specify a physical quantity we will use the following procedure, using
acceleration as an example.
1.
Think about what you already know about the quantity. What concept
or physical principle is associated with the quantity.
e.g acceleration has to do with
 motion,
 moving ,
 speeding up
 increasing speed, decreasing speed,
 changing velocity
 Force ( what causes the acceleration?)
2.
Define the physical quantity using one or more sentences.
WORDS
Example acceleration is defined as the rate of change of velocity
3.
Derive a mathematical expression from the definition. Simplify the expression using
appropriate symbols
MATHEMATICAL EQUATION
e.g
change in velocit y
acceleration 
time taken
a
4.
vu
t
Determine the basic S.I. units for the quantity from the mathematical
expression. Rename the basic units if appropriate.
(ms 1  ms 1 )
S .I .Units 
 ms  2
s
5.
Decide whether the quantity is a vector or a scalar quantity. This will
determine whether you need to use vector or scalar mathematics to
calculate a value for the quantity.
VECTOR QUANTITY
A vector quantity is defined by both a magnitude (the numerical value ) and a
direction
A scalar is defined totally by the magnitude.
DISPLACEMENT :When a body moves from one location to another it undergoes a
DISPLACEMENT ( Distance with an associated direction ). A change from one
position x1 to another position x2 is called a displacement
2
-x1
SPEED : Defined as the rate of change of distance.
Where Rate means per unit time
CONSTANT SPEED: Constant speed is when a body travels equal distances in
equal periods of time.
AVERAGE SPEED : Average speed is defined as the total distance travelled during
a particular time divided by that time interval.
INSTANTANEOUS SPEED. This is the speed of an object at a given instant.
Average speed 
s
total distance
 T
total time taken tT
S .I .Units
meter
m
  ms 1
second
s
[ SCALAR ]
VELOCITY : Velocity is defined as the rate of change of displacement.
OR
Speed in a given direction.
OR
Distance travelled in unit time in a given direction.
S.I. Units ms-1
[VECTOR]
AVERAGE VELOCITY : is the ratio of the displacement
1
- x2
1
S. I. Units ms-1
- t2
[VECTOR]
INSTANTANEOUS VELOCITY:
This is the rate at which the objects position x is changing with time at any given
instant. The instantaneous velocity is obtained from the average velocity my making
Therefore
Instantaneous velocity
x
lim
t 0 t
S. I. Units
ms-1
dx

dt
VECTOR
Example 1. A train travels a distance of 270km in 2 hours 35 minutes due East.
Calculate (i) total distance travelled (ii) total time taken (iii) average speed and
(iv) average velocity
ACCELERATION : is defined as the rate of change of velocity.
AVERAGE ACCELERATI ON 
Change in velocit y
time taken
v v
v
Average accelerati on  2 1 
t 2  t1
t
ms -1
S.I. Units 
 ms  2
s
VECTOR
Instantaneous acceleration is the derivative of the velocity with respect to time.
dv d 2 x
acceleration  a 

dt dt 2
The acceleration of an object at any instant is the rate at which its velocity is changing
at that instant.
Example. A car is uniformly accelerating from rest to a velocity of 30 m s-1 due east
in 15 seconds. Calculate its acceleration.
Acceleration due to gravity
An object falling freely under the gravitational attraction of the earth is moving with
constant acceleration of
9.81 m s-2 in the vertical downward direction.
Acceleration due to gravity is denoted by the symbol g and it is positive in the
downward vertical direction.
g = 9.81 m s-2
Vertically down
Deceleration : Negative acceleration .
If an object is thrown vertically upward it slows down i.e the object experiences a
deceleration.
Acceleration = - g =
- 9.81 m s-2
Since the direction is opposite the sign is opposite i.e negative
Example 1 : Calculate the velocity with which an object hits the ground if it is
dropping for 2.26 s.
Example 2. How long will it take an object to reach its maximum height if it is
thrown vertically upward with an initial velocity of 16ms-1
PROBLEM SHEET
Question 1. A car travels along a straight road in time t as shown
s ( m)
0
t ( s)
0
v ( ms-1)
0
8
2
32
72
4
8
128
6
16
Plot a graph of distance s versus time t
Plot a graph of velocity v versus time t and use it to find
(i)
The total distance travelled and
(ii)
The acceleration of the car.
8
24
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PROBLEM SOLVING GUIDELINES.
1.
Read the entire question carefully from start to finish and then return to the
start of the question.
2.
Write down and name all the quantities given in the question, one at a time, as
they are given
3.
Check for S.I. Units and correct into S.I. where necessary, then assign the
appropriate label or symbol
4.
Find a mathematical relationship between the known and unknown quantities
5.
Solve the equation by substituting values both numerical and units
6.
Consider whether or not the answer is reasonable.
EQUATIONS OF MOTION.
For all equations we will use the following symbols
u = initial velocity in m s-1
v = final velocity after time t in m s-1
s = distance travelled in time t in a given
direction i.e displacement in metres
t = time taken in seconds.
Note :
u, v, s, a are all VECTOR quantities and measured in S.I. units
Usually all vectors in a given application are in the same direction.
To use the equations of motion the value of the acceleration, a , must be constant for a
given application.
If an application has an acceleration value that changes it must be divided into a
number of stages which has a constant acceleration value for each stage, and
therefore each stage can be solved using equations of motion.
acceleration = Rate of change of velocity
From definition
change in velocity
acceleration  a 
time taken
vu
a
t
Cross multiply
at = v - u
rewritten as
v = u + at

First equation of motion
From definition
average velocity =
total distance travelled
time taken
uv s

2
t
Cross multiply
s=
u + v t
2
From equation 1 we have
v = u + at
Therefore
s=
u + u + at  t
2
Giving
s = ut + 1 2 at 2

Second equation of motion
From equation 1
v  u  at
Square both sides
v 2  u 2  2.u.a.t  a 2 t 2
Rearranging
v 2  u 2  2.a.(ut  1 at 2 )
2
From equation 2
ut  12 at 2  s
Giving
v 2  u 2  2.a.s Third equation of motion

PROBLEM SHEET EQUATIONS OF MOTION
Question 1. An object starts from rest and moves with constant acceleration of 8m.s-2
along a straight line. Calculate (i) the velocity after 5 seconds (ii) distance travelled
in 5 seconds and (iii) average speed for 5 seconds.
Question 2. A truck travelling east increases speed uniformly from 15 km hr-1 to 60
km.hr-1 in 20 seconds. Calculate (i)acceleration (ii)total distance travelled
(iii)average speed(iv) average velocity
Question 3. A bus moving at a velocity of 20 m.s-1 west decelerates at a rate of 3ms-2.
How far does it travel before stopping.
Question 4. An object starts from rest and moves at constant acceleration of 2 m.s-2
for 10 seconds, it then travels at constant speed for a further 1 minute before
decelerating to rest in a distance of 200 m.
Calculate (i) total distance travelled
(ii) total time taken and (iii) average speed for the journey.
Question 5. A stone is thrown vertically upward and reaches a height of 20 metres.
Calculate
(i)
with what speed was it thrown up
(ii)
how long would it take for the stone to drop back down to ground from 20
m.
Question 6. A stone is thrown vertically upward with a speed of 30 m.s-1 from the top
of a tower 80m high. Calculate
(i) The total distance travelled by the stone before it hits the ground and
(ii)
the velocity with which it hits the ground.
Question 7. A cyclist starts from rest and accelerates uniformly at a rate of 1 m.s-2 for
8 s, then continues his journey at constant velocity for 1 minute before decelerating to
rest at double the acceleration rate. Calculate (i) the total distance travelled (ii) the
total time taken (iii) the average speed and (iv) plot a velocity versus time graph to
and use it to calculate the total distance travelled.