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Linear Motion Explained With Worked Examples
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DOI: 10.13140/2.1.1581.4729
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LINEAR MOTION
EXPLAINED WITH
WORKED EXAMPLES
BY
S H E F I U S . Z A K A R I YA H
PREFACE
After a successful dissemination of the previous books1, which are available online, in
your hands is another book for potential scientists and engineers. This current work –
Linear Motion Explained with Worked Examples – offers 100 worked examples. There are
two main parts in this book; one gives a broad explanation of the topic and the other
presents worked examples. The questions used in this work are similar to those in
physics, mathematics and / or engineering textbooks designed for A-level, college and
university students. Advanced learners, particularly those returning to study after a
break from the academic environment, will also find this helpful. Additionally, it could
be used as a reference guide by teachers, tutors, and other teaching staffs during classes
and for assessment (quizzes, home works and examinations).
Finally, many thanks to my colleagues who have offered suggestions and comments,
especially Khadijah Olaniyan (Loughborough University, UK), Shazamin Shahrani
(University of Sussex, UK) and Dr. Abdul Lateef Balogun (Saudi).
Pertinent suggestions, feedback and queries are highly welcome and can be directed to
the author at the address below.
Coming soon in this series are:

Worked Examples on Circuit Theorems

Worked Examples on Calculus

Worked Examples on Partial Fractions

Worked Examples on Balancing Chemical Equations
© Shefiu S. Zakariyah 2014
Email: shefiuz@theiet.org | S.Zakariyah@derby.ac.uk
1
These (and future publications) are available at https://independent.academia.edu/ShefiuZakariyah/ or
http://www.researchgate.net/profile/Shefiu_Zakariyah .
i
Disclaimer
The author has exerted all effort to ensure an accurate presentation of questions and
their associated solutions in this book. The author does not assume and hereby
disclaims any liability to any party for any loss, damage, or disruption caused by errors
or omissions, either accidently or otherwise in the course of preparing this book.
ii
CONTENTS
PREFACE ................................................................................................................................................................. I
DISCLAIMER ........................................................................................................................................................... II
CONTENTS ............................................................................................................................................................ III
FUNDAMENTALS OF LINEAR MOTION .................................................................................................................... 1
WORKED EXAMPLES ............................................................................................................................................ 11
SECTION 1.
SECTION 2.
SECTION 3.
SECTION 4.
SECTION 5.
SECTION 6.
SECTION 7.
EQUATIONS OF MOTION ..............................................................................................................11
SPEED AND DISTANCE ..................................................................................................................14
LINEAR MOTION - HORIZONTAL (BASIC - INTERMEDIATE) ........................................................19
LINEAR MOTION - HORIZONTAL (INTERMEDIATE - ADVANCED) ..............................................26
GRAPHICAL SOLUTION OF ONE-DIMENSIONAL MOTION ..........................................................41
FREE FALL MOTION (BASIC) .........................................................................................................57
FREE FALL MOTION (ADVANCED) ...............................................................................................63
BIBLIOGRAPHY AND FURTHER READING ............................................................................................ 75
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FUNDAMENTALS OF LINEAR MOTION
1. Introduction
Motion of objects – living and non-living matter - is one of the key branches of physics.
It finds applications in numerous fields such as engineering, medicine, geology, and
sport science among others. Whether you drive (or you are driven), walk, jog or fly in
the air, you are exercising some form of motion. When in this state, habitually one is
interested in how much longer, when, and how quickly one can get to his / her
destination. Having answers to these and similar questions would be very useful to
individuals, and this is exactly the discourse of this book.
In this current work, we will be considering motion in one dimension called linear
motion. In other words, attention is only given to vertical, horizontal and any other
forms of straight line motion, such as motion on an inclined surface. Furthermore,
motion will be analysed without making reference to its causes, i.e. forces. This is to say
that this book deals with kinematics; dynamics on the other hand will be dealt with in
the nearest future – keep fingers crossed. One more thing that I should add here is that,
for this topic we will keep our discussion concise and focus more on the examples.
However, further information on the topic can be obtained from the reference list
provided at the end of this book.
2. Variables (or Quantities)
Let us take a little time to review the variables or terms commonly used in this topic. By
a variable, we mean ‘what varies’ of course, but more specifically it refers to physical
quantities that we measure. So what are the variables commonly used in this subject?
Here we go:
(a) Distance: is a change in position relative to a reference (or zero) point. It is a scalar2
quantity, measured in metre3 (m) and as such, it can only be positive.
(b) Displacement: is a change in position relative to a reference (or zero) point in a
particular direction. It is a vector4 quantity and also measured in metre (m). Various
letters are used to represent both displacement and distance, but the most
2
It is a physical quantity that has (or is described with) a magnitude only.
This is the fundamental SI unit for distance, but other units can also be used or found in use.
4
It is a physical quantity that has (or is described with) both a magnitude and a direction. As a result, it can be
positive or negative in value depending on the chosen direction of reference. A positive sign is generally omitted
but a negative value is indicated unless if this is substituted with a word or term that indicates such.
3
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commonly used ones are , and . Displacement, being a vector, can be positive or
negative.
(c) Average Speed: is the rate of change of distance. It is a scalar quantity, measured in
metre per second (m/s or ms-1). Since speed is likely to change over the course of
motion, it is often useful to give the average speed, which can be obtained using:
When we refer to the speed at a given point in time, we are talking about what is
‘technically’ called instantaneous speed (or simply speed). This is the speed recorded
by a car’s speedometer. It is mathematically given as
where
, and
(d) Average Velocity: is the rate of change of displacement and is also measured in
metre per second (m/s or ms-1). Unlike speed, it is a vector quantity, which is
expressed as
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Velocity at a specified position or a particular point in time is called instantaneous
velocity (or simply velocity). This is obtained by reducing the change in time, in
principle, to zero and is expressed mathematically as
or simply
where
,
, and
Alternatively, we can define velocity as the rate of change of distance in a particular
direction or simply speed in a particular direction. Note the phrase ‘in a particular’
which shows the distinction between them on the basis of being either a ‘scalar’ or a
‘vector’. In the same vein, we can say that speed is a velocity without any direction
associated to it (either in words or writing). It is therefore not surprising that they are
misunderstood as synonyms and sometimes used interchangeably especially in
conversation.
When the symbols and are used together, refers to the initial velocity and the
final velocity.
or
can also be used to represent the initial velocity or you may find
other symbols being used. Furthermore, it is possible to find that the symbol is used
for speed, but this is more appropriately ‘reserved’ for distance.
At this point we need to clear a misconception that could occur (and this is indeed
found among students). In data analysis (or statistics) for example, average or mean is
computed by diving the sum of all the values with the number of the items as
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This is not generally the case in linear motion. For instance, if a journey is undertaken at
three different velocities denoted as ,
, and
the average velocity denoted as is
not equal to the sum of the three velocities divided by three. In other words,
The average value in linear motion can only be computed similarly to the average value
in statics if the time spent at each of the velocity is equal. That is to say
where
are the times spent while moving at
In general, the average velocity is found by
respectively.
(e) Average Acceleration: is the rate of change of velocity and is measured in metre per
second squared (m/s2 or ms-2). It is also a vector quantity and can be evaluated
using
Acceleration occurs due to a change in the:
(i)
(ii)
(iii)
magnitude of the velocity only,
direction of the velocity only, or
magnitude and direction of the velocity.
In other words, the velocity can remain constant while a body accelerates (possibly due
to a change in direction). A typical example is a body moving in a circular path at a
constant velocity. Since the direction of the motion keeps changing at every particular
point, the body is said to accelerate although there is no change in the value of the
velocity.
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Instantaneous acceleration (or simply acceleration) is the acceleration at a specified
position or a particular point in time. It is obtained by reducing the change in time, in
principle, to zero as
or simply
where
,
, and
In general, the letter is used for acceleration, however, an object undergoing vertical
motion in the vicinity of the earth experiences a uniform acceleration, irrespective of the
characteristics (shape, mass, or density) of the object. This is termed acceleration due to
gravity (or free-fall acceleration); it is denoted by the letter and has a value of
(correct to two significant) at sea level. This value slightly varies as one moves
from the equator to either the South or North pole and also on the elevation (or altitude)
of the object from the Earth’s surface. This free fall acceleration acts as though it is
pulling a body towards the centre of the Earth.
For simplicity, it is customary to use
when carrying out ‘quick’
calculations. In this book, we will be using both
and
.
Sometimes large accelerations are expressed as multiples (or in ‘unit’ ) of . For
instance, an acceleration of
can be written as
because
. Note also that the equations of motion are valid only for free falling objects
near the Earth’s surface provided that the effect of air is negligible. For this a positive
value of is used for a body falling towards the surface of the earth (or downwards)
and a negative value of i.e.
is used for a body moving away from the
earth’s surface (or upwards).
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When acceleration is negative, it is called deceleration. It is worth noting that when the
velocity decreases it does not necessarily mean the body is decelerating, it rather
implies that acceleration is in an opposite direction from the chosen reference axis.
However, the word retardation can be correctly used to mean that a body slows down
or its speed decreases. To erase any confusion, we can say that
Case 1.
Case 2.
If the velocity increases, then the sign of both velocity and acceleration
must be the same (positive or negative).
If the velocity decreases, then the sign of velocity and acceleration must be
opposite, i.e. if one is positive the other must be negative.
Note that when an object is thrown vertically upwards, its velocity decreases until it
reaches zero at its maximum height; therefore, and based on what was stated above,
velocity and acceleration must have different signs. Thus velocity is positive and
acceleration (due to gravity) is negative. Similarly, if a particle falls freely, it gains
velocity so both velocity and acceleration must have the same sign, a positive sign.
Time is another important variable that is inevitably used in this topic and it is one of
the fundamental units; as such I believe we do not require a ‘formal’ definition for this
or do we? Table 1 below gives a summary of what has been presented on these six
variables and these will be our ‘tools’ for analysis.
Table 1: Variables used in Linear Motion
Quantity
1 distance
Scalar /
Unit
Vector
metre ( )
scalar
2 displacement metre ( )
vector
3 speed
metre per second (
)
scalar
4 velocity
metre per second (
)
vector
5
Calculus5
∫ ( )
These are instantaneous values (velocity and acceleration) given as differentials and / or integrals.
6
∫ ( )
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5 acceleration
6 time
metre per second squared
(
)
seconds
(
)
vector
scalar
Before we leave this section, it will be useful to mention a term which frequently occurs
in motion or in mechanics as a whole. The term is ‘particle’, and it refers to a point-like
matter which has a relatively negligible size and mass. One final quick note on average
and instantaneous values is that when acceleration or velocity remains constant its
average and instantaneous values are both equal. Hopefully this is clear, or you may
want to reflect on this.
3. Equations of (Constant Acceleration) Motion
The equations presented so far can be used to analyse simple motion problems.
However, a complex problem particularly when the object is accelerating will require
that an advance equation is used. These equations called equations of motion are
summarised in the table below. Note that these equations are valid when the
acceleration is constant (or can be considered, by approximation, to be constant)
otherwise these equations cannot be used. Furthermore, equation 5 in Table 2 is not
distinctive because it can be derived by combining equations 1 and 2 in the same Table
2 as demonstrated below.
Table 2: Equations of Motion
Equations (general) Equations (gravity)
1
2
3
7
Variables
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(
)
(
)
5
From equation (1),
Substitute the above in equation (2) of Table 2, we will have
(
)
This is the reason why some textbooks only list the first four equations. Anyway, note
the following subtle differences between equations 2 and 5 from Table 2:
(i)
(ii)
Equation 2 has an initial velocity while equation 5 has a final velocity.
Equation 2 has a positive sign between its two terms while equation 5 has a
negative sign in the same place.
Furthermore, if we carefully look at the above table of equations, we will discover that
each equation has four variables. To use an equation, we need to carry out two quick
checks, namely:
Check 1. First and foremost, it must have the variable we are solving for.
Check 2. It must have the other three variables in the chosen equation, either given
in the problem or obtained from another calculation.
Once these are satisfied, we then need to substitute the known variables in order to
solve for the unknown quantity. In a situation where the ‘check 2’ above is not
completely satisfied, perhaps we have two known variables, it is likely that we would
need to combine two equations from Table 2 in order to arrive at an answer.
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On a final note, although it is worthwhile to know how certain equations are derived,
but we will not be showing the ‘how’ for equations of motion. If you are interested in
this kindly do refer to the reference / further reading list or any other source of your
choice.
4. Graphs of Linear Motion
Graphs are used for various purposes in science and engineering, and for onedimensional motion there is no exception. In particular, complex motion problems,
which could prove difficult with the equations of motion, can easily be solved
graphically. For this we will be using a two-axis graph where only two variables are
involved. In each of the graph, time is by default the horizontal axis (or
). Since
we have six physical quantities, it implies that there are five different types of graphs,
namely:
(i)
(ii)
(iii)
(iv)
(v)
Distance-time graph,
Displacement-time graph,
Speed-time graph,
Velocity-time graph, and
Acceleration-time graph.
In each case, the first named quantity is plotted on the vertical axis (or
). For
example, in a distance-time graph, distance and time are plotted on the vertical and
horizontal axes respectively. Furthermore, while it is general to consider graphs (i) & (ii)
and (iii) & (iv) as separate, the pairs are however identical. Hence, in practice we have
three different graphs, i.e. (i) distance-time or displacement-time graph, (ii) speed-time
graph or velocity-time graph, and (iii) acceleration-time graph.
For a uniform motion, the paths under the graphs are straight lines (vertical, horizontal
or slanting) but for a non-uniform motion, the graphs can be of any shapes such as
parabola, exponential, etc. Our focus will be on the former (straight line graphs).
Whatever the graphs, two things are usually of interest for analysis. These are the slope
and under the area on the graph. Each of the two (slope and area) will result into (or be
equivalent to) one of the aforementioned six variables (apart from time) depending on
the type of graph as shown in Table 3.
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Table 3: Slope and Area in Linear Motion Graphs
Graph
Slope (Gradient)
Area
speed
1 Distance-time (
)
meaningless
velocity
2 Displacement-time (
3 Speed-time (
)
meaningless
)
meaningless
distance
acceleration
4 Velocity-time (
5 Acceleration-time (
)
displacement
)
meaningless
velocity
END OF FUNDAMENTALS OF LINEAR MOTION
AND
BEGINNING OF WORKED EXAMPLES
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WORKED EXAMPLES
Section 1.
Equations of Motion
[3] Given , ,
in a problem, decide which
equation to use to find .
INTRODUCTION
In this first section of the Worked Examples, we
will begin by getting familiar with the equations
of motion. This is achieved by giving certain
variables and deciding on which equation is to be
used to determine an unknown variable. In
addition, we will put into test our ability to
substitute values into an equation in order to find
unknown variable(s). Do not worry as this will be
basic; but if you are confident with this already,
feel free to move to any section of your choice.
[1] Given , ,
Solution
The equation that connects the four variables and
which should be used to solve this problem is
What Next 1.
Substitute the values.
What Next 2.
Simplify the expression on the
right hand side.
in a problem, decide which
What Next 3.
equation to use to find .
[4] Given , ,
Write the final answer.
in a problem, decide which
equation to use to find .
Solution
The equation that connects the four variables and
Solution
which should be used to solve this problem is
The equation that connects the four variables and
which should be used to solve this problem is
What Next 1.
Substitute the values.
What Next 2.
Simplify the expression on the
What Next 3.
[2] Given , ,
(
)
right hand side.
What Next 1.
Substitute the values.
Write the final answer.
What Next 2.
Simplify the expression on the
right hand side.
in a problem, decide which
What Next 3.
equation to use to find .
[5] Given , ,
Solution
Write the final answer.
in a problem, decide which
equation to use to find .
The equation that connects the four variables and
Solution
which should be used to solve this problem is
The equation that connects the four variables and
What Next 1.
Substitute the values.
What Next 2.
Simplify the expression on the
which should be used to solve this problem is
right hand side.
What Next 3.
What Next 1.
Write the final answer.
11
Substitute the values.
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Simplify the expression on the
What Next 1.
right hand side.
What Next 3.
[6] Given , ,
Make
the subject of the
formula.
Write the final answer.
in a problem, decide which
equation to use to find .
What Next 2.
Substitute the values.
What Next 3.
Simplify the expression on the
right hand side.
Solution
What Next 4.
The equation that connects the four variables and
[9] Given , , in a problem, decide which
which should be used to solve this problem is
What Next 1.
Make
Write the final answer.
equation to use to find .
the subject of the
Solution
formula.
The equation that connects the four variables and
What Next 2.
Substitute the values.
which should be used to solve this problem is
What Next 3.
Simplify the expression on the
right hand side.
What Next 4.
[7] Given , ,
What Next 1.
Write the final answer.
Make
the subject of the
formula.
in a problem, decide which
equation to use to find .
What Next 2.
Substitute the values.
What Next 3.
Simplify the expression on the
right hand side.
Solution
What Next 4.
The equation that connects the four variables and
Write the final answer.
which should be used to solve this problem is
NOTE
In questions 7, 8 and 9, making the unknown
What Next 1.
Make
variable the subject of the formula before
the subject of the
substituting the known values is optional.
formula.
What Next 2.
Substitute the values.
What Next 3.
Simplify the expression on the
[10] Find
[8] Given , ,
,
and
.
right hand side.
What Next 4.
when
Write the final answer.
Solution
in a problem, decide which
Step 1. Decide on the formula.
equation to use to find .
We need to use
()
Solution
Step 2. Solve for
The equation that connects the four variables and
.
Substituting the given values in equation (i), we
which should be used to solve this problem is
will have
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are dealing with vector quantities. It could be that
the motion occurs in two directions for example.
[11] Find
when
,
Let find the value of the initial velocity ‘
and
to see
what is going on.
.
Solution
Step 1. Decide on the formula.
We need to use
(
Step 2. Solve for
)
()
This suggests that the initial and final velocities
.
are equal but opposite, which might imply that
Substituting the given values in equation (i), we
equal distance is covered in two opposing
will have
directions with a net value of zero.
[13] Find
[12] Find
when
,
,
,
.
and
.
Solution
Step 1. Decide on the formula.
Solution
We need to use
Step 1. Decide on the formula.
()
We need to use
Step 2. Re-arrange the formula.
()
Step 2. Solve for
when
Since we are solving for the acceleration (a), we
.
need to make
Substituting the given values in equation (i), we
the subject of the above equation
as
have
implies
therefore
( )
NOTE
Step 3. Solve for
One may wonder why the distance, s, is zero
.
Substituting the given values in equation (ii), we
when the velocity is not. This is possible since we
will have
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Step 4. The average speed for the whole
journey.
Section 2.
Speed and Distance
.
INTRODUCTION
By now we should have developed a degree of
confidence with the variables used in linear
motion and how to choose a suitable equation for
obtaining the unknown quantity. In this section, it
is time we engage in, at least for now, basic
problems involving speed and distance. Let us get
on with this right away.
/
.
.
/
/
[15] Mu’aadh lives in Jeddah and wants to travel
to Makkah to perform a lesser pilgrimage. If
the distance between his residence in Jeddah
and Makkah is
[14] A plane flies from London Heathrow Airport
and the maximum safe
to Dubai International Airport, a distance of
driving speed is
approximately
time he can stop to rest if he must get to
of
at an average speed
Makkah within
. The return trip was made at an
average speed of
. Find the average
, what is the longest
?
Solution
speed for the whole journey.
Step 1. Write out the given values.
Solution
Step 1. Write out the given values.
Step 2. Calculate the shortest possible time for
the journey.
If
is the short time possible to use for the trip,
Step 2. Trip time from London to Dubai.
then
Step 3. Trip time from Dubai to London.
Therefore, the longest time to stop for rest is
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This means that Mu’adh can wait (to rest) on his
and
way to Makkah for up to 39 minutes (maximum).
[16] A university driver drove from Windhoek to
Ongwediva campus for
average speed of
at an
. If a third of the
distance was driven at
, calculate
Thus, from (i)
the distance between these two campuses and
the average speed for the remaining part of
the journey.
Solution
From (iii),
Step 1. Write out the given values.
Step 2. Calculate the distance between the
campuses.
NOTE
Distance between Windhoek and Ongwediva
Alternatively,
campus is the same as the distance, , travelled.
This is given by
that is
as before.
Step 3. Calculate the average speed for the
Similarly, from (iv),
remaining part of the journey.
Let
and
be the average speeds of a third and
the remaining part (i.e. two-third) of the journey
respectively. Similarly, let us denote the time
taken for these with
and
Hence, the average speed for the rest of the
respectively. Thus,
journey can be found from the equation (ii) as,
()
( )
and
( )
( )
But
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[17] In a learner training exercise, David has to
drive a distance of
. For the first
he drives at a constant speed of
. At what constant speed must he
drive the remaining distance if his average
From (v)
speed for the whole distance is to be
?
Solution
Step 1. Write out the given values.
From (ii)
From (ii)
Step 2. Calculate the average speed for the rest
of the journey.
Let
and
be the speeds of the first and second
[18] Yusuf a.k.a. Joseph plans to visit a friend at
part of the journey respectively. Similarly, let us
Loughborough University in Loughborough
denote the time taken for the first and second part
of the journey with
and
town. He intends to drive from Birmingham
respectively. It
to Loughborough via the motorways; he
therefore follows that
covers distances along the motorways at
()
, single carriage-ways at
( )
and those in built-up areas (of towns and
and
cities) at
( )
. Find his average speed for
each of the following:
( )
(i)
( )
within Birmingham city (built-up
area) and then
(ii)
From (iv)
on the motorways.
on single carriage-way and then
in Loughborough town (built-up
area).
Solution
Step 1. Write out the given values.
From (i)
We are given the following
16
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(
)
(
Thus,
)
(
)
Also,
Now let
(
)
(
)
(
)
(
Hence,
)
(
)
(
)
Step 2. Calculate the average speed for
within Birmingham and
on the
. /
.
/
motorways.
.
/
Thus,
(
)
[19] ETS North bound train from Kualar Lumpur
(KL Sentral) to Ipoh (South-west of KL) is
Also,
scheduled to take
(including the time spent waiting at stations)
to complete its
Therefore,
journey. It stops at 10
stations on the way with 1 minute waiting
time (for passengers to alight and get on
board) at each of the stations.
. /
.
/
.
(i) Calculate the average speed of the train.
/
(ii) What would be the average speed if the
stop at each station was increased to
(
)
?
Step 3. Calculate the average speed for
single carriage ways and
on
Solution
in
Step 1. Write out the given values.
Loughborough.
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(
Step 2. Calculate the average speed.
)
[20] From Abuja, a motorist travels
southwards heading to Port Harcourt
But the total time
. Due to an emergency, he makes a
U- turn and travels
northwards at
to a nearby service station.
Determine the difference between the average
speed and the average velocity on this
Thus,
journey assuming he travels on a straight
road.
. /
Solution
(
Step 1. Write out the given values.
)
Step 3. Calculate the average speed when the
time spent to stop at each station is
increased to 9
.
Step 2. Calculate the average speed.
The actual time spent on the journey, excluding
If
and
are the time spent for the journey
the waiting times at the 10 stations, in the
towards the north and south respectively, then
previous case is:
(
)
Also,
Also, the time spent at the stations in the current
case is:
(
)
Therefore, the average speed,
Thus,
.
/
. /
18
Shefiu S. Zakariyah
shefiuz@theiet.org
from a velocity of
(
of
)
to
at a rate
. How far does she travel while
accelerating?
Step 3. Calculate the average velocity.
Since velocity is a vector quantity we need to
Solution
specify a positive direction. So let us take the
Step 1. Write out the given values.
north as our positive direction, then the average
velocity
NOTE
You see here that we have taken
(
initial velocity and
)
as the final velocity
although the question did not specify. This is
Step 4. Calculate the difference between the
because the word ‘initial’ means the first in the
average speed and the average velocity
series, so
on the trip.
If
as the
is the first velocity in the part
we are considering.
represents the difference between the average
speed and the average velocity then
Furthermore, it should be added that the final
velocity in one part of a journey could represent
(
)
the initial velocity in the succeeding part of the
same journey. We will come across this later in
(
Section 3.
)
this book.
Linear Motion - Horizontal
Step 2. Calculate the distance travelled while
(Basic - Intermediate)
accelerating.
Using
INTRODUCTION
In section 1, we have had a look at how to choose
and substitute variables into the equations of
motion. In the current section it is time to present
some basic uniform acceleration problems, which
will require the use of one of the equations of
motion previously introduced. Note that from
now on, we will need to be able to determine
variable quantities from expressions or problems
so do get ready.
Re-arrange the formula to make the subject
Substitute the values to determine
[21] During her daily exercise, Sarah accelerates
19
Shefiu S. Zakariyah
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Step 1. Write out the given values.
[22] The safe take-off velocity of a particular
passenger plane is set at
. Find the
minimum acceleration that the airplane needs
to move on a
runway.
Step 2. Calculate the acceleration.
Solution
Using
Step 1. Write out the given values.
Make
the subject
Substitute the values to have
Step 2. Calculate the acceleration necessary to
achieve this final velocity.
Using
Because
is negative, it implies that the
acceleration is in the opposite direction to the
Re-arrange the formula to make
the subject
direction of velocity, which is westerly in this
case. Hence the acceleration is
First we need to convert the velocity from
to
NOTE
as
Alternative ways of saying this includes:
(1) Acceleration of
(2) Deceleration of
(3) Acceleration of
Substitute the values in the above equation in
order to determine
[24] A tractor is initially at rest and accelerates at
for
. Find its velocity after this
time.
[23] A coach is travelling eastwards at
After
its velocity is
.
Solution
in the same
Step 1. Write out the given values.
direction. What is the magnitude and
direction of its acceleration?
Solution
20
Shefiu S. Zakariyah
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The collision between the ball and the racket
Step 2. Calculate the velocity.
takes place for
. What is the average
acceleration of the ball during this collision?
Solution
Step 1. Write out the given values.
NOTE
The word ‘rest’ is typically used to mean that the
velocity is zero. As a result, we assign
.
This is a general principle and you will find this
Step 2. Calculate the average acceleration.
frequently used in linear motion.
Average acceleration
is
[25] A motorist leaving a village to a town, drives
on a straight road at an average velocity of
. If the motorist enters the town with
a velocity of
If we take the direction of rebound as positive
, what was his velocity
then
. Now substitute to have
(
at the village, assuming that acceleration was
)
constant during this journey?
Solution
[27] Sultan, a motorcyclist, is initially moving with
Step 1. Write out the given values.
a velocity of
. He then accelerates
uniformly at a rate of
Calculate his final velocity.
Step 2. Calculate the initial velocity.
(
Solution
Step 1. Write out the given values.
)
Thus
(
)
Step 2. Calculate the velocity.
[26] During a table tennis game, the Ping-Pong
ball hits Tanyaluk’s racket with a speed of
and rebounds with a speed of
.
21
for
.
Shefiu S. Zakariyah
shefiuz@theiet.org
[28] In a traffic accident involving a car and a
maintaining a constant speed. How far has it
truck, the car's velocity changed from
to
in
covered during the acceleration period?
.
Solution
(i) What is the acceleration of the car?
Step 1. Write out the given values.
(ii) Express the acceleration of the car as
multiple acceleration due to gravity, , to
the nearest whole number.
Solution
Step 2. Calculate the distance.
Step 1. Write out the given values.
(
)(
)
Step 2. Calculate the driver’s acceleration.
Using
[30] Two different motorists are travelling at
and
Substitute the values
on a motorway, both
driving within the regulated limits. If they are
both required to stop, how much further apart
would they be after coming to rest (assuming
that both retard at equal rates)?
Solution
Step 3. Calculate the multiples of
of the
Step 1. Write out the given values.
driver’s acceleration.
If the multiples of the acceleration is denoted by
⁄
then
⁄
| |
|
|
|
|
Step 2. Choose the equation to calculate the
distance travelled after braking.
For this problem, we need to use
[29] A passenger bus starts from rest at a bus stop
Re-arrange the formula to make the subject
on a straight road and moves with a uniform
acceleration of
for
before
22
Shefiu S. Zakariyah
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Solution
Case 1. Distance for braking from
Step 1. Write out the given values.
.
Substitute the given values in the above equation
to determine
Step 2. Calculate the distance.
(
Case 2. Distance for braking from
.
)
Substitute the values, we will have
Substitute the given values in the above equation
(
to determine
)
[32] A car starting from rest at a traffic light
reaches a speed of
.
.
/
.
/
and
) and the
distance travelled.
instead of
Find the ratio of the two distances
. Find the
acceleration of the car (in
Step 3. Calculate how much further to travel for
braking from
in
Solution
as
Step 1. Write out the given values.
Step 2. Calculate the acceleration.
or
The above shows that the distance travelled by
Therefore
one motorist is 4 times the other. Hence, we can
say that the distance between the two motorists
would be 3 times the distance travelled by the
motorist braking from
to rest.
[31] An athlete accelerates uniformly from
to
in
Step 3. Calculate the distance travelled.
. Find the distance covered
(
by him / her during this time.
23
)
Shefiu S. Zakariyah
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If we take the direction of rebound as positive
Substitute the values in the above equation to
then
(
have
)
The original velocity is equal to
which implies
that the ratio of the original velocity to the
NOTE
acceleration is
Alternatively, we can say
[34] A car with an initial velocity of
accelerates uniformly at
reaches a velocity of
as before.
,
until it
. Calculate the
time taken and the distance travelled during
this acceleration.
[33] Somdee kicks a ball and hits an opponent
player directly on the chest. Consequently, the
Solution
direction of the ball is completely reversed
Step 1. Write out the given values.
with its velocity being halved. If the ballopponent contact lasts for
, what is the
ratio of the original velocity to the
acceleration of the ball?
Solution
Step 2. Calculate the time.
Step 1. Write out the given values.
Using
Substitute above values to have
Step 2. Calculate the average acceleration.
Average acceleration
is
Step 3. Calculate the distance travelled.
We can use
24
Shefiu S. Zakariyah
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Therefore
Substituting the values, we will have
√
( )( )
Substitute the values, we will have
√
√
NOTE
Step 3. Convert the speed from m/s to km/h.
Alternatively, we can use
This implies that
Yes it is safe, but it will be better to stop
accelerating at this stage.
Substitute the values, we will have
[36] In a
race, Yuyan covers three quarters
of the track in
. If he is initially at rest and
maintains a constant acceleration, what is his
velocity when he crosses the finish line?
Leave the answer correct to 2 significant
as before.
figures.
Solution
[35] On a motorway, a vehicle is driven from rest
with an acceleration of
velocity after a 5
Step 1. Write out the given values.
. Find its
drive. Is it safe to
(
continue to accelerate if the speed limit is
)
for this type of vehicle?
Solution
Step 2. Calculate the velocity.
Step 1. Write out the given values.
Step 2. Calculate the velocity.
25
(
)
(
)
Shefiu S. Zakariyah
Section 4.
shefiuz@theiet.org
Linear Motion - Horizontal
(i) the speed after
,
(ii) the distance travelled in
(Intermediate - Advanced)
.
Solution
INTRODUCTION
Unlike previous sections, in this section we will be
solving more challenging problems. Are you
ready? Then, let us get started.
Step 1. Write out the given values.
[37] A tram starts from rest and moves with a
uniform acceleration of
for
. Determine the value of , if it
covers
during this journey.
Step 2. Calculate the velocity after
.
Step 3. Calculate the distance after
.
Solution
Step 1. Write out the given values.
( )( )
Step 2. Calculate the distance.
Using
NOTE
Alternatively, we can say
Substituting above values, we will have
(
(
(
(
)
(
)
)
)
)
This implies that
as before.
[39] In an electronics factory, a conveyor belt is
[38] In a speed controlled zone, a car is initially
travelling at a speed of
to accelerate uniformly at
used to move circuit boards. Initially at rest, a
before it begins
board moves from the production stage to the
. Calculate:
assembly stage with acceleration of
26
.
Shefiu S. Zakariyah
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Find the velocity with which the board
reaches the assembly section if the length of
the belt between the two stages is
(
.
)
Step 3. Calculate the time.
Using
Solution
Step 1. Write out the given values.
Substituting the values, we will have
Step 2. Calculate the velocity.
Substitute the values, we will have
.
/
.
/
(
[41] A motor vehicle is uniformly retarded from a
√
velocity
in
and finally come to a
complete halt after
[40] The speed of a particle increases from
to
)
. Determine:
(i) the rate at which it slowed down, and
. Calculate the rate of change
(ii) the distance covered during this period.
and the time taken for this increment.
Solution
Solution
Step 1. Write out the given values.
Step 1. Write out the given values.
Step 2. Calculate the acceleration.
Step 2. Change the velocity from
Using
Substituting the values, we will have
Step 3. Calculate the retardation.
Substituting the values, we will have
27
to
.
Shefiu S. Zakariyah
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Let
be the time it takes to go from A to B and
the time taken to travel between B and C.
Therefore,
Step 4. Calculate the distance covered.
Substituting the values, we will have
(
Tip: Since the velocity is constant for the whole
)
journey, we can find velocity first and then use it
to find the distance between A and B.
Step 3. Calculate the velocity.
But the time, , is
NOTE
Alternatively, we can use
(
Hence
Substituting the values, we will have
(
)
)( )
Step 4. Find the
.
as before.
[42] A passenger bus moves with a constant
[43] A van skids to a halt from an initial speed of
velocity along a straight road having three
consecutive bus stops, ,
to move from
from
to
to . If
covering a distance of
and . It takes
and
to move
, find
.
acceleration of the van (assumed constant)
and the time it takes to stop.
Solution
Solution
Step 1. Write out the given values.
Step 1. Sketch the journey.
A
B
. Find the
C
Step 2. Write out the given values.
28
Shefiu S. Zakariyah
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Step 2. Calculate the acceleration.
Substitute the values, we will have
Therefore
Step 3. Calculate the time taken to stop.
Note that for this part of the journey, we have
because Hafsah comes to rest.
Step 3. Calculate the time taken.
Therefore,
Substitute the values in above equation to have
Substituting the values, we will have
[44] Hafsah is running at
up to
in
. She then speeds
.
[45] Alcohol is one of the factors that affect human
(i) Determine her acceleration.
(ii) If she thereafter slows down at
reaction times (RT). Initially travelling at
,
, how much farther would a drunk
how long does she take to come to a final
driver travel before he/she reacts to an
stop?
Emergency Road Sign when compared to a
‘normal’ driver given that the former level of
Solution
alcohol would increase his / her RT to
Step 1. Write out the given values.
Take
?
as the ‘normal’ RT.
Solution
Step 1. Write out the given values.
Step 2. Calculate the acceleration.
Step 2. Convert the initial velocity from
.
29
to
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NOTE
Alternatively, we can say
Step 3. Calculate the distance covered by the
(
normal driver before hitting the brake.
)
(
)
Let the distance covered by the ‘normal’ driver’s
car be
, thus
where
is the reaction time for the ‘normal’
as before.
driver, therefore
[46] Asmaa’ is cycling her bike initially at 1.5 m/s
before she decides to accelerate at
Step 4. Calculate the distance covered by the
.
What will be the time taken to cover
drunk driver before hitting the brake.
a
on
straight road leading to her school.
Let the distance covered by the drunk driver’s car
be
Solution
, thus
Step 1. Write out the given values.
Where
is the reaction time for the drunk
driver, therefore
Step 2. Calculate the time.
Step 5. Calculate how much farther the drunk
Using
driver would travel.
Let ‘how much farther’ distance travelled by the
drunk driver’s car be
, thus
Substitute above values to have
(
Divide through by 1.5,
Now multiply through by 5,
(
Either
30
)(
)
)
Shefiu S. Zakariyah
shefiuz@theiet.org
or
Step 4. Calculate the velocity.
Since time cannot be negative, it follows that the
only solution (the time taken to cover
[48] A taxi driver moving at a velocity of 10 m/s
distance) is
realised that he had 35 seconds to get to his
[47] The front of a multi-coach train
destination which is 800 m away. He therefore
long
(approx.) passes a signal at a level crossing
accelerated at
with a speed of
journey. Did he succeed in getting to his
. If the rear of the
train passes the signal
for the rest of the
destination on time?
later, determine:
(i) the acceleration of the train, and
Solution
(ii) the speed at which the rear of the train
Step 1. Write out the given values.
passes the signal.
Solution
Step 1. Write out the given values.
Step 2. Calculate the time.
Substituting the values in above equation, we will
Step 2. Convert
to
have
.
( )
Step 3. Calculate the acceleration.
Multiply through by 2,
Using
(
Either
( )( )
or
31
)(
)
Shefiu S. Zakariyah
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Step 3. Find the smallest non-zero time when (a)
the velocity is zero, and (b) the object is
Since time cannot be negative, it follows that the
at the origin.
only solution is
 The smallest non-zero time when the object is
at the origin can be found at
Yes, he succeeded in getting to his destination on
(i) as
time, i.e. 15 seconds earlier.
[49] A particle moves along a straight line with
acceleration
(
. It starts its motion at
the origin with velocity
)
Therefore, either
.
(i) Write down equations for its position and
velocity at time
in equation
or
.
(ii) Find the smallest non-zero time when
(a) the object is at the origin,
(b) the velocity is zero.
 The smallest non-zero time when the velocity
(iii) Sketch the position-time, velocity-time
is zero can be found at
and speed-time graphs for the interval
in equation (ii)
as
.
Solution
Step 1. Write out the given values.
Step 4. Sketch the position-time, velocity-time
and speed-time graphs for
.
Step 2. Write down equations for its position
and velocity at time
position-time graph
seconds.
30
is
(
25
position (m)
 Its position at time
)
()
 Its velocity at time
20
15
10
5
is
0
0
(
)
( )
32
2
4
6
time (s)
8
10
12
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the velocity at
velocity-time graph
.
15
Solution
velocity (m/s)
10
Step 1. Write out the given values.
5
Step 2. Calculate the position of the particle at
0
0
2
4
6
8
10
12
specified times.
-5
 At
, the position of the particle is
-10
( )
-15
( )
( )
time (s)
 At
, the position of the particle is
speed-time graph
( )
12
( )
( )
speed (m/s)
10
8
 At
, the position of the particle is
6
( )
4
( )
( )
2
0
0
2
4
6
8
10
12
 At
tine (s)
, the position of the particle is
( )
[50] A particle moves such that its position
( )
( )
metres at time seconds is given by the
expression
.
 At
(i) Determine the position of the particle
when times
, , ,
,
and .
, the position of the particle is
( )
( )
( )
(ii) Construct a table showing the position of
the particle at these times.
(iii) Draw a position- time graph.
 At
, the position of the particle is
(iv) State the times when the particle is at the
origin and describe the direction in which
( )
it is moving at those times.
(v) Using the graph in (iv) or otherwise, find
33
( )
( )
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Step 3. Construct a table.
Taking the values of
The table showing the positions of the particle at
find the gradient
the interval
and at two points, we can
(
is shown below.
( )
0
1
2
3
4
5
( )
0
-12
-32
-42
-24
40
)
38.3
Hence, the velocity at
Step 4. Sketch a graph of the position against
is
correct to 2
significant figures.
time.
NOTE
Graph of position against time
Alternatively, we can find the velocity using
50
40
differentiation as
position, x (m)
30
20
Point 1
10
( )
0
-10
0
1
2
3
4
5
6
(
-20
)
-30
-40
-50
time (s)
The above is a velocity as a function of time. To
Point 2
find the velocity at
Step 5. State the times when the particle is at the
, we simply substitute for
in the expression as
origin and describe the direction in
which it is moving at those times.
( )
( )
( )
From the above graph, the particle is at the origin
(where the curve crosses the x-axis) at

, and it is decreasing or going in the
as before.
negative direction.

, and it is increasing or going in the
[51] An object moving with a uniform acceleration
positive direction.
Step 6. Find the velocity at
covers distances
seconds.
and
in the first two
equal and consecutive intervals of time .
Since the graph is not a straight line, we need to
Express
make a tangent to the curve at t = 4 as shown
in terms of ,
and .
above. The velocity at this time is equal to the
Solution
slope of the tangent at that point.
The initial velocity, , of the second stage of the
From the graph, the slope of the tangent is
motion is equal to the final velocity of the first
stage. Using
34
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and
Step 2. Calculate the time it takes Faisal to cover
Thus for
the distance of
distance we have
.
()
Similarly for
distance
( )
This is the time it takes Faisal to reach the finish
because
line. We can use this to know what Edward has
covered in the same time interval.
Subtracting equation (i) from (ii), we have
(
)
(
Step 3. Calculate the distance covered by Edward
in
)
.
[52] Towards the end of a 400m race, Faisal and
Edward are leading and are both running at
. While Faisal is
line Edward is
from the finish
Therefore, Edward did not succeed because when
from the finish line.
Faisal finished, Edward has 13.6 m distance to
Realising this and to beat Faisal, Edward
cover since he was 100 m away from the finish
decides to accelerate uniformly at
line.
until the end of the race while Faisal keeps on
NOTE
the same constant speed. Does Edward
Alternatively, we can find the time taken by
succeed in beating Faisal?
Edward to reach the finish line as follows
Solution
Step 1. Write out the given values.
 Faisal
Using quadratic equation formula
 Edward
35
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and
√
and
represent the distances of T1
and T2 respectively from the International
Airport. Assuming the motion is considered
we have
( )(
( )
√
to be on a straight path,
)
(i) Write down an expression for
√
of the
train at time .
(ii) Write down an expression for
either
of the
train at time
√
(iii) When and where do the trains meet?
Solution
or
Step 1. Write out the given values.
√
Since time cannot be negative, it implies that the
time taken by Edward to reach the finish line is
13.59 seconds. This indicates that he reached the
finish line
(
)
Step 2. Calculate the acceleration of T1 and T2.
later than Faisal. Therefore, he did not succeed as
The acceleration of T1 and T2 can be found using
previously determined.
 For T1
[53] Between two terminals E1 and D1 for
We have
international and domestic flights
respectively, trains are used to transfer
Since
passengers. If a train T1 starts from terminal
E1 and accelerates uniformly for
until it reaches a maximum speed of
it implies that
or
.
At the same time, a train T2 starts from D1
and accelerates uniformly for
until
it reaches the same maximum speed of
. The two trains then maintain the
maximum speed of
for
 For T2
after
Similarly we will have
leaving their respective terminals. The
distance between the two terminals is
Again, since
36
it implies that
Shefiu S. Zakariyah
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or
Step 5. Determine when the two trains meet.
The two trains meet when their respective
distances from terminal E1 is equal. In other
words
Step 3. Write an expression for
time
.
The distance
which implies that
of train T1 from the International
Airport E1 after
Where
Thus
for T1 after
is
and
are the distances covered
The two trains meet 59 seconds after leaving their
during acceleration and constant velocity
origins, i.e. terminals E1 and D1.
respectively. Thus
Step 6. Determine where the two trains meet.
[
]
, (
)-
The place where the trains meet can be found by
substituting
which implies that
(
[
,
-
,
)( ) ]
,
(
in either
or
(
{
)-
as
)
or
Step 4. Write an expression for
time
)
.
The distance
of train T2 from the International
Airport E1 after
is
The two trains meet at
(
Where
(
{
for T2 after
and
)
[54] When John rows his boat, the two oars are
are the distances covered
both in water for
during acceleration and constant velocity
water for
respectively. Thus
]
,
(
)-}
. This pattern is repeated for a
(i) Find the change in speed that takes place
in water if the boat accelerates at a
which implies that
( )( ) ]
{[
*
and then both out of the
race lasting 1 minute.
{[
*,
from terminal E1.
-
,
,
(
constant rate of
)-}
.
(ii) Find the change in speed that takes place
-+
out of water if the boat decelerates at a
+
37
Shefiu S. Zakariyah
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constant rate of
.
The positive value indicates that the total change
(iii) Calculate the change in the boat’s speed
for each
in speed for each 7 s cycle is increase.
cycle.
Step 4. Calculate the finishing speed.
(iv) What is John’s speed as he crosses the
The speed with which John would finish his race
finishing line, if he starts the race from
is given by
rest?
Since he starts from the rest and takes
Solution
to
complete then
Step 1. Find the change in speed that takes place
in water.
But
If the acceleration in the water is
then
(
)
Thus
(
)
[55] As soon as the traffic light changed to green,
Step 2. Find the change in speed that takes place
Abdullah accelerates uniformly with his
out of water.
motorcycle at
If the deceleration out of the water is
then
for
. He then
maintains a steady velocity for a two-third of
his
journey.
(i) How far does he travel from the traffic
light junction until he reached a
maximum velocity?
(ii) What maximum velocity does he reach?
(iii) How long in total does he take to come to
Step 3. Calculate the change in the boat’s speed
rest if he then slows down at
for each 7 s cycle.
The change in the boat’s speed for each
cycle is
Solution
the algebraic sum of the changes in and out of the
Step 1. Write out the given values.
water, which is
(
)
(
)
(
(
)
)
38
?
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shefiuz@theiet.org
Step 2. Calculate the distance travelled during
the acceleration.
The time spent on retardation can be found from
We need to use the formula
where
Substituting the values, we will have
(
)
(
)( )
(
)
Step 3. Calculate the final velocity reached.
We need to use
( )
(
)
Substitute the values to have
√
Step 4. Calculate the total time taken to come to
a stop.
Thus
Note that the total time spent is equal to the sum
of the time spent in accelerating, the time spent at
the constant velocity and the time spent on
retardation. Let these be respectively represented
as
,
and
If the total time spent is denoted
as , then
NOTE
This can be also solved using graphical method.
But,
This method is dealt with in section 5.
We know that
[56] ,
and
are three Non-Stop stations for a
High Speed train connecting two major cities.
Assuming ,
and
are a on a straight
railway track and station
Where is the distance covered during a constant
velocity, which is
, i.e. two-third of
between stations
.
and . If the speeds with
which the train passes
Substituting the values, we will have
and
39
is mid-way
and
are
respectively, what is the speed
Shefiu S. Zakariyah
shefiuz@theiet.org
with which the train passes station .
further
.
The motion of the car is modeled by taking
Solution
the acceleration to be constant with a value of
Step 1. Sketch the journey.
A
.
B
(i) By considering the part of the journey
C
from the T-junction to the tunnel, show
that
Step 2. Calculate the speed at point .
Let the distance between
and
.
(ii) Find another second equation involving
be . It follows
and .
that
(iii) By elimination method or otherwise,
()
solve for
Also, let the speed with which the train passes
and
in the two equations.
be . This is equal to the final velocity of the
Solution
journey between
Step 1. Find an expression for the journey
and
of the journey between
and the initial velocity
and . Using
between the junction and the tunnel in
terms of
and
.
we can therefore write:
The distance between the T-junction and the
 For
tunnel can be calculated using
journey
( )
 For
Substituting the values, we will have
journey
(
)
( )
Now equate the right-hand sides of equations (ii)
Divide through by 12
and (iii)
()
Step 2. Find another expression involving
√ (
’.
)
The distance from the junction to the end of the
tunnel can be calculated using
[57] From a speed of
at a T-junction, Barrak
accelerates his car along a straight highway
For this case,
road. He observes that he reaches a tunnel
from the junction
crosses the tunnel, which is
and
and that he
long, in a
40
Shefiu S. Zakariyah
shefiuz@theiet.org
between the two bus stops.
Now substituting the values, we will have
Solution
)
Step 1. Draw the graph.
Vel. / (m/s)
(
Divide through by 18
( )
B
Step 3. Find the values of
and
E
.
To find , subtract equation (i) from (ii),
C
A
From equation (i),
D
20
60
Step 2. Calculate the final velocity
( )
Section 5.
F
70
time / (sec)
’.
The slope of AB is equal to the acceleration, i.e.
Graphical Solution of OneStep 3. Calculate the total distance travelled by
Dimensional Motion
the train.
INTRODUCTION
In the previous sections, we have attempted to
solve linear motion problems using one of the
equations of motions (and employing suitable
methods). Sometimes, questions are easily dealt
with using a graphical method and this is the goal
of this section.
The total distance travelled during the journey is
equal to the area of the trapezium ABEF
[58] A passenger bus starts from a stop and travels
[59] A roller skater, Hawwah, starts off with an
with a uniform acceleration of
(
initial velocity of
for
before she gets to a
sloppy surface where she gains speed at a
until it reached a maximum velocity. It
travels with this maximum velocity for
constant rate. If after
another
moving at
, the brake is then applied so that
she is already
,
(i) Find an expression of her speed and
a uniform retardation is obtained and then
come to a halt at the next bus stop after
)
distance seconds after she started
.
gaining speed.
Sketch the velocity-time graph of this motion
(ii) Given that the length of the slope is
and use this to calculate the total distance
41
,
Shefiu S. Zakariyah
shefiuz@theiet.org
(
what is her speed at the bottom of the
)
slope?
( )
Solution
Step 1. Draw the graph.
Vel. / (m/s)
NOTE
Alternatively, the distance (s) travelled after
seconds is
E
C
(
)
B
A
D
F
Step 4. Find the speed at the bottom of the slope.
time / (sec)
The time taken to reach the bottom of the slope
can be found using equation (iii) as
Step 2. Find an expression for her speed as a
function of time
The speed
.
Multiply by 4 and re-arrange to have,
after the push is
Using quadratic equation formula
The acceleration
BC, i.e.
is equal to the slope of the line
√
( )(
√
)
( )
Therefore,
This implies that
()
Step 3. Fid an expression for the distance
travelled as a function of time
.
or
The distance (s) travelled after t seconds is equal
to the area of the trapezium ABEF
(
)
Therefore, the only valid value is
The final speed at the bottom of the slope can be
( )
found using equation (i) as
Substitute equation (i) in equation (ii), we have
42
.
Shefiu S. Zakariyah
shefiuz@theiet.org
(
)
Similarly, the slope of EF is equal to the
deceleration, i.e.
[60] A train is travelling along a straight path
between two stations
and . Initially at
station A, the train accelerates uniformly from
rest to a speed of
speed for
and maintains this
Therefore, the journey time
. It then decelerates
uniformly until it comes to rest at station . If
the acceleration and deceleration are
and
(
respectively, find the total journey
)
Step 3. Calculate the distance travelled during
time and the total distance between the two
the journey.
stations. Give the answer correct to 3
The distance travelled during the journey is
equal to the area of the trapezium ABEF
significant figures.
Solution
(
)
Step 1. Draw the graph.
Note that the time has been converted from
(
minutes to seconds.
)
[61] Yaasir is driving his car behind a coach at a
Vel. / (m/s)
velocity of
while the coach is moving
at a constant velocity of
is
B
. When the car
behind the front of the coach, Yaasir
E
accelerates uniformly at
. The car
continues at the same acceleration until it
reaches a velocity of
A
C
D
, which he
maintains until passing the front of the coach.
F
Calculate the distance the car travels while
time / (sec)
accelerating correct to 2 significant figures.
Step 2. Calculate the total journey time.
The total journey time, , is given by
Solution
Step 1. Draw the graph.
The slope of AB is equal to the acceleration, i.e.
43
Shefiu S. Zakariyah
shefiuz@theiet.org
Vel. / (m/s)
distance.
(i) Find the time that the driver takes to react
to the situation.
D
(ii) Using, a velocity-time graph, calculate the
Car
deceleration once the car starts braking.
C
B
E
(iii) What is the stopping distance for another
Coach
vehicle travelling at
if the reaction
time and the deceleration are the same as
before?
F
A
Solution
time /(sec)
Step 2. Calculate the time
Step 1. Draw the graph.
spent in
Vel. / (m/s)
accelerating.
The acceleration, , is equal to the slope of the line
CD, thus
C
B
Therefore,
A
Case 2
Case 1
D
E
F
G
H
time /(sec)
Step 3. Calculate the distance the car travels
Step 2. Calculate the time taken to react.
while accelerating.
The time, , taken to react is
The distance travels while accelerating, , is equal
to the area of the trapezium ACDF
(
[62] A car is travelling at
)
Step 3. Calculate the deceleration.
when the driver
The deceleration, a, is equal to the slope of the line
EG, thus
has to perform an emergency stop due to a
sudden shock. The car has travelled a distance
of
()
before the driver decided on what to
do and applied a brake. The car then comes to
But the distance covered after applying the brake
is equal to the area of the triangle EFG, thus
rest with constant deceleration in a further
44
Shefiu S. Zakariyah
shefiuz@theiet.org
(
(
before it catches up ?
)
)
Solution
Therefore,
Vel. / (m/s)
Step 1. Draw the graph.
( )
Substitute equation (ii) in equation (i), we have
Step 4. Calculate the stopping distance.
Car A
From equation (ii),
time / (sec)
The deceleration of the case 2 is equal to the slope
of the line DH. Since the deceleration is the same
in both cases then the slope of the line DH is also
equal to
. Therefore,
Step 2. Calculate the time.
Let
Using the figure above,
(
The stopping distance, , is equal to the area of the
trapezium ACDH
(
[63] In a convoy, Car
and passed a
waiting to join the convoy.
Five seconds later, car
at
)
is travelling with a
uniform velocity of
stationary car
Make
. How far will
starts and accelerates
have to travel
45
the subject
)
Shefiu S. Zakariyah
shefiuz@theiet.org
Since the two cars must have covered the same
distance, it implies that
, therefore
(
Make
the subject
(
)
)
(
Simplify the equation and solve for
)(
)
Since the two cars must have covered the same
distance, it implies that
, therefore
(
)(
(
)
)
Simplify the equation and solve for
Therefore, either
or
(
)(
)
Therefore, either
Substitute the values of in either
or
to
determine how far will have to travel before it
catches up . Note that cannot be negative so
use
only.
(
)
or
Substitute the values of in either or
to
determine how far will car have to travel before
it catches up car ,
NOTE
Vel. / (m/s)
Alternatively, using figure below
as before.
or
Car A
It is important to note that
cannot be
a valid answer since after 5 seconds, car A must
have travelled a distance of 100 m, i.e.
, a distance car B needs, at a minimum, to
travel in order to catch up with car A.
time / (sec)
(
)
[64] A motorist drives past a police patrol team
with an over speed-limit of
continues with that speed for
46
, and
,
Shefiu S. Zakariyah
shefiuz@theiet.org
when he noticed that he was being chased. He
(
)
therefore applied the brake and is brought to
rest in a further
On the other hand,
the patrol team started from rest
Step 3. Calculate the distance between the two
later, speed up and attained a speed of
in
vehicles after stopping.
. They directed the motorist to stop
Distance covered by the motorist equals the area
when they are abreast each other. As a result,
of the trapezium ABEH
the team also applied a brake and are brought
(
to rest under uniform retardation in
another
)
. Assuming uniform
acceleration and retardation in both cases,
Distance covered by the patrol team is equal to
using a suitable graph,
the area of the triangle DCG
(i) find the value of ;
(
)
(ii) the distance between the two vehicles
after both have finally stopped.
Therefore, the distance between the motorist’s car
and the patrol team car after both have finally
Solution
stopped is
Step 1. Draw the graph.
Vel. / (m/s)
(
)
This shows that the patrol team car will be
in front of the motorist’s car.
[65] Three junctions ,
C
and
are on a straight
road. A motorcyclist passes junction
speed of
from where he accelerates at
a constant rate of
B
until he gets to
junction . He therefore pulls up with a
E
constant retardation of
D
F
with a
G
Step 2. Calculate the value of
complete halt at junction . Given that the
H
A
and comes to a
time / (sec)
total distance between junctions
and
is
, find:
.
(i) the speed of the motorcycle at junction ,
When the two are abreast, they must have
travelled the same distance. Thus, the area of
triangle DCF is equal to the area of rectangle
ABEF.
and
(ii) the distance from junction
47
to junction .
Shefiu S. Zakariyah
shefiuz@theiet.org
Solution
(
)
(
)
( )
Step 1. Draw the graph.
Vel. / (m/s)
Multiply through by 6
(
)
(
)
( )
Junction Y
C
This implies that
B
Junction X
A
√
E
D
Junction Z
time /(sec)
Step 2. Calculate the value of
Step 3. Calculate the distance from
.
If the distance between junction and is
denoted by
, then
is equal to the area of
the trapezium ABCD given by
The velocity, , is the speed at which the
motorcycle passes junction .
If the distance covered by the motorcycle from
junction to is denoted by
, then
is the
sum of the areas of the trapezium ABCD and
triangle CDE. Thus,
(
[
)]
[
to .
(
)
Using equation (ii),
]
)(
(
()
)(
(
Acceleration is the slope of the line BC given by
( )(
[66] A train takes
( )
)
)
)
to travel from
Stansted Airport to Cambridge. The train
accelerates from rest at a rate of
Also, the deceleration is the slope of the line CE
given by
for
. It then travels at a constant speed before
it is finally brought to rest in
with a
constant deceleration. The motion is modelled
as a linear motion on a straight railway track.
( )
Substitute for
and
(i) Sketch a velocity-time graph for the
journey.
in equation (i), thus
(ii) Find the steady speed, the deceleration
48
Shefiu S. Zakariyah
shefiuz@theiet.org
and the distance from Stansted Airport to
Cambridge station.
(
Solution
)
Step 4. Calculate the distance from Stansted
Step 1. Draw the graph.
Airport to Cambridge.
Note: The time in the graph is in minutes, but this
will be converted to seconds during the
calculations where appropriate.
Let the distance covered by the train be , which is
equal to the area of the trapezium ABEF.
Vel. / (m/s)
,(
)
-
E
B
-NOTE
A
C
D
F
Alternatively, let the distance covered by the
train be , which is equal to the sum of the
distance covered in the different stages of the
journey. If ,
and are the distances
covered in stages 1, 2 and 3 respectively, then
time / (min)
Step 2. Calculate the steady speed.
Let the steady speed be , therefore
where
which implies that
or
Step 3. Calculate the deceleration.
The deceleration is the slope of the line EF,
therefore
(
)
Therefore,
which implies that
49
Shefiu S. Zakariyah
shefiuz@theiet.org
Step 3. Calculate the distance covered during
the uniform speed.
as before
Let the distance covered during the uniform
speed be denoted as
. This is equal to the area
of the rectangle BCDE. Therefore,
[67] A motorcycle starting from rest reaches a
maximum speed of
in
with this speed for another
; it continues
Step 4. Calculate the distance covered while
before it is
finally brought to rest in another
decelerating.
. Using a
Let the distance covered during the deceleration
be denoted as
. This is equal to the area of the
triangle DEF. Therefore,
graphical method, determine the distance
covered by the car when
(i) it was accelerating
(ii) it was moving with uniform speed
(iii) it was decelerating.
Step 5. Calculate the distance travelled by the
Hence or otherwise, calculate the total
car.
distance travelled by the car.
Let the total distance travelled by the car be
denoted as , which is equal to the total area of
the shape. Therefore,
Solution
Vel. / (m/s)
Step 1. Draw the graph.
E
B
NOTE
Alternatively, the total distance travelled, , is
equal to the area of the trapezium ABEF.
Therefore,
C
D
F
A
(
time / (sec)
)
Step 2. Calculate the distance covered while
accelerating.
as before.
Let the distance covered during the acceleration
be denoted as
. This is equal to the area of the
triangle ABC. Therefore,
[68] From an initial velocity of
,a
professional cyclist accelerates uniformly
50
Shefiu S. Zakariyah
shefiuz@theiet.org
until he attains a maximum velocity of
.
Hence,
He maintains this velocity for some time
before decelerating uniformly to rest. The
total time taken for the journey is
total distance travelled is
[ (
and the
. If the time
(
[
spent accelerating is two-thirds of that at
,
, (
) ]
-
,
)-
))]
,
-
constant velocity, calculate the deceleration.
This implies that
Solution
Vel. / (m/s)
Step 1. Draw the graph.
Thus, from equation (i) deceleration is
C
( )
F
[69] A train accelerates uniformly from rest at a
B
station
D
E
to a maximum speed of
.
The constant maximum speed is maintained
G
A
for a period of time and the train then
time / (sec)
Step 2. Calculate the deceleration.
decelerates uniformly until it comes to a stop
The deceleration, , is the slope of line FG. This is
given as
at station . The distance between the two
railway stations is
takes
and the journey
. If the magnitude of the
acceleration is half that of deceleration, by
using graphical method, determine:
()
(i) the acceleration, in metre per second, and
The total distance covered is equal to the area of
ABCFG in the above figure. Let this area be
denoted as and ,
and
to represent the
area under acceleration, constant velocity and
deceleration respectively. Therefore,
(ii) the time, in minutes, during which the
train travels at its maximum speed.
Solution
Step 1. Draw the graph.
( )
where
51
Shefiu S. Zakariyah
shefiuz@theiet.org
Vel. / (m/s)
( )
The distance between the two stations is equal to
the area of the trapezium ABEF. Hence,
B
E
,
(
)-
substitute equation (iii), thus
C
D
(
F
)
(
A
time / (sec)
)
Let us do some conversions as follows
 Speed from
to
This implies that
 Time from minutes to seconds
Thus, from equation (ii) acceleration is
Step 2. Calculate the acceleration.
.
The acceleration, , is the slope of line AB. This is
given as
/
Step 3. Calculate the time, in minutes, during
which the train travels at its maximum
speed.
()
From the graph, the time for constant speed is
Because the magnitude of the deceleration is
double that of acceleration implies that
(
( )
)
Therefore, time in minutes is
Equating (i) to (ii), we have
(
)
[70] On her way to Kuala Lumpur International
Airport, Katrina was initially driving at
Therefore,
52
Shefiu S. Zakariyah
shefiuz@theiet.org
. On getting to one end of the SMART
From the graph, the acceleration of the car is equal
to the slope of the line BC, thus
(Stormwater Management and Road Tunnel),
she began to increase her speed such that she
gained
every
. She continued like
this until reaching a maximum speed of
where is the time the car reached its maximum
speed. Therefore,
which she maintained. Assuming the
tunnel is modeled as a straight-line path,
using a speed-time graph or otherwise,
(i) Find the time taken to reach the
Step 3. Calculate the time taken to drive through
maximum speed.
the tunnel.
(ii) If the tunnel is approximately 10 km long,
Let the distance covered after be , which is
equal to the sum of the distance covered in the
two different stages of the journey. If
and are
the distances covered in stages 1 and 2
respectively, then
find the time, in minutes correct to 2
significant figures, taken to drive through
the tunnel.
(iii) Write down expressions for the speed of
the car as a function of time in seconds.
where
Solution
Step 1. Draw the graph.
Vel. / (m/s)
(
(
C
F
But
B
Therefore,
Stage 2
Stage 1
D
E
A
time / (sec)
Step 2. Calculate the time taken the car to reach
its maximum speed of
If the car gains
the acceleration is
every
.
, it implies that
(
53
)
)
)
Shefiu S. Zakariyah
shefiuz@theiet.org
Solution
Step 4. Determine expressions for the speed of
Step 1. Draw the graph.
the car as a function of time in seconds.
 Stage 1 :
Vel. / (m/s)
There are two stages of the motion, each has a
different expression for the speed.
or
C
At this stage, the speed (v) as a function of time (t)
is
D
B
 Stage 2:
Stage 1
or
Stage 2
Stage 3
Stage 4
E
A
time / (sec)
At this stage, the speed (v) as a function of time (t)
is
Step 2. Calculate the speed
at point B.
The slope of AB is equal to the acceleration of the
car between A and B, thus
[71] A motorist is initially at rest at a point
on a
which implies that
straight road between two small villages. He
moves with a constant acceleration of
for
where he attained a velocity of
.
or
He thereafter changed to a higher gear and
moves with a uniform acceleration
until reaching a maximum speed limit of
in a further
at point
maintained this speed for
Step 3. Calculate acceleration of the car when
and
travelling from B to C.
to reach point
The slope of BC is equal to the acceleration of the
car between B and C, therefore
. As he comes close to a traffic light at , the
brake is applied and the car gradually comes
to rest in
. Draw a velocity-time graph for
the motion, and find:
(i) the speed
But
of the car at
(ii) the acceleration a of the car when
travelling from
to ,
or
(iii) the retardation of the car when
travelling from
to , and
(iv) the total distance from
to .
54
, thus
Shefiu S. Zakariyah
shefiuz@theiet.org
complete stop in
. Assuming the motion is
Step 4. Calculate the distance between A and E.
modeled to be a straight line, find the distance
Let the distance covered by the car be , which is
equal to the sum of the distance covered in the
different stages of the journey. If , ,
and
are the distances covered in stages 1, 2, 3 and 4
respectively as shown in the graph, then
covered if the whole journey takes
.
Solution
Vel./ (m/s)
Step 1. Draw the graph.
where
B
E
F
Stage 2
Stage 1
C
Stage 3
Stage 4
D
G
H
A
(
)
time / (sec)
Step 2. Calculate the total distance covered.
Let the total distance covered be , which is equal
to the sum of the distance covered in the different
stages of the journey. If , , and are the
distances covered in stages 1, 2, 3 and 4
respectively, then
where
Hence,
(
[72] From a petrol station, Benjamin increases his
speed at a constant rate to
)
in the first
of his journey. On seeing a temporary
speed limit on the road, he then decreases
uniformly to
in a further
(
,
remaining constant for some times before he
decelerates uniformly from this speed to a
55
)
Shefiu S. Zakariyah
shefiuz@theiet.org
Step 2. Calculate the distance the coach has
travelled from the garage.
Let the distance covered by the train be , which is
equal to the sum of the distance covered in the
different stages of the journey. If , , ,
and are the distances covered in stages 1, 2, 3, 4
and 5 respectively, then
Thus,
Note: remember to convert the time to seconds by
multiplying by 60.
[73] A coach driver leaves a garage and accelerates
at a constant rate for
. He then
maintains a constant speed of
for
before he begins to slow down
where
uniformly as he gets close to a set of signals.
After
the coach is travelling at
(
but the signal changes to green on his
)
approach. He therefore increases speed
uniformly for
speed of
until reaching a
(
. A road signal then orders
)
the coach to stop, which he obeys by slowing
down uniformly for
. Using a graph
motion, calculate the distance, kilometres
(
)
(
)
(
)
(
)
correct to 2 significant figures, covered by the
driver.
Solution
Vel. / (m/s)
Step 1. Draw the graph.
(
I
B
Therefore,
E
F
Stage 1
Stage 2
C
Stage 3
D
G
Stage 4
Stage 5
H
J
A
time / (min.)
56
)
Shefiu S. Zakariyah
Section 6.
shefiuz@theiet.org
[75] A light weight stone is thrown vertically
Free fall Motion (Basic)
upwards from the ground and hits the
INTRODUCTION
So far we have had a look at linear motion in an
horizontal direction; that is to say motion that is
not under the influence of the earth’s gravity. In
this section, we will be spending time to analyse
motion in the vertical direction, which is
constantly under the effect of acceleration due to
gravity. Unless otherwise is stated in the question,
we will take the value of to be
.
ground after
. Calculate the maximum
height reached by the stone during this
journey.
Solution
Step 1. Write out the given values.
[74] A metal coin is thrown straight upwards with
an initial velocity of
. Calculate the
distance covered (from the point of
projection) and the velocity after
NOTE
. Take
There are certain tips to know in order to be able
.
to solve this question more comfortably, namely:
(i) The given time is the time taken to travel up
Solution
from and down to the ground; it thus follows
Step 1. Write out the given values.
that the time taken to reach the maximum
height is half of this, i.e.
.
(ii) At maximum height the ball comes to rest. In
other words, final velocity for the upward
motion is zero.
Step 2. Calculate the distance.
(iii) Height is the same as the distance in the
equation of motion.
We will use these tips frequently in this and
Substituting the values, we will have
(
subsequent questions.
)( )
Step 2. Calculate the height.
Step 3. Calculate the velocity.
Substituting the values, we will have
Substitute the values, we will have
( )
(
(
57
)
)
Shefiu S. Zakariyah
shefiuz@theiet.org
[76] An object is thrown straight upwards with an
initial velocity of
. How long does it
take to reach the greatest height?
Step 3. Calculate the time taken to reach the
Solution
greatest height.
Step 1. Write out the given values.
This implies that ,
Step 2. Calculate the time.
At the greatest height the ball is at rest. This
implies that the final velocity is zero, i.e.
[78] An apple fruit falls freely from a tree at a
This implies that ,
height of
. How long does it take to
reach the ground? Disregard the air resistance
and leave the answer correct to 2 significant
figures.
Solution
[77] A helium-filled balloon is thrown straight
upwards with an initial velocity of
Step 1. Write out the given values.
.
Assuming the air resistance is negligible,
calculate the greatest height reached and the
time taken to reach it. Take
.
Step 2. Calculate the time.
Solution
Step 1. Write out the given values.
Because
Make
, we will have
the subject of the formula
Step 2. Calculate the greatest height reached.
Using
it implies that ,
√
(
)
58
Shefiu S. Zakariyah
shefiuz@theiet.org
Substituting the values, we will have
√
Tip: Note that the initial velocity of the package is
the same as the velocity of the helicopter.
√
However, since the package is moving in the
opposite direction, we will assign a negative sign
[79] A book falls from a bookshelf which is at a
height of
to it.
from the floor. Determine its
Step 2. Calculate the height from which the
velocity just before striking the floor.
package was thrown.
If is the height above the ground from where the
Solution
package was thrown then
Step 1. Write out the given values.
Substituting the values, we will have
(
)(
)
Step 2. Calculate the time.
At the initial height the ball is at rest. This implies
that the initial velocity is zero.
Because
[81] An object falls off from the edge of a blade of
, we will have
an off-shore wind turbine. If the tower
is
This implies that
, high find the velocity with which the
object hits the surface of the water below it.
√
Leave the final answer in surd form.
Substitute the values, we will have
√
Solution
Step 1. Write out the given values.
[80] A package is dropped from a helicopter
moving upwards at
. If it takes
before the package reaches the ground, how
high above the ground, correct to 2 s. f., was
Step 2. Calculate the velocity.
the package when it was released?
Solution
This implies that ,
Step 1. Write out the given values.
(
59
)(
)
Shefiu S. Zakariyah
shefiuz@theiet.org
√
√
√
[82] An object falls freely from a height through a
distance of
√
. Calculate the velocity
attained by the object, taking
[84] A 500 g parcel is dropped from a height of
.
, from a plane which is moving upwards
Solution
with a velocity of
Step 1. Write out the given values.
(i) the initial velocity of the parcel,
. Determine:
(ii) the time taken for the parcel to reach the
ground? Disregard air resistance and take
the value of
Step 2. Calculate the velocity.
to be
.
Solution
Step 1. Write out the given values.
This implies that ,
(
)(
)
√
Step 2. Calculate the initial velocity.
[83] A fountain is designed such that water can be
Provided the air resistance is negligible, the initial
projected vertically upwards to a height of
velocity of the parcel is the same as that of the
. Determine the speed at which the water
balloon. In other words,
must be leaving the fountain nozzle, leaving
the answer in surd form.
Step 3. Calculate the time.
Solution
Substituting the values, we will have
Step 1. Write out the given values.
(
Step 2. Calculate the initial velocity.
When the water reaches a height of
, the
(
velocity is zero, i.e.
Either
This implies that ,
(
)( )
60
)(
)
)
Shefiu S. Zakariyah
shefiuz@theiet.org
or
Step 2. Calculate the height.
Since
Since time cannot be negative, it follows that the
, it implies that ,
(
only solution is
)(
)
[85] A metallic ring is released from rest from the
top of a cliff
high. Find the time it takes
[87] A child, standing on the floor of a balcony of a
the ring to reach the ground.
multi-storey building, throws a ball straight
upwards with an initial velocity of
Solution
the floor is
Step 1. Write out the given values.
. If
high from the ground, find:
(i) the maximum height above the ground
reached by the ball, and
(ii) the velocity with which the ball strikes the
ground.
Step 2. Calculate the velocity.
Solution
Step 1. Write out the given values.
This implies that ,
(
)( )
√
[86] An object at rest is subjected to a free fall from
the top of a building. How high is the
building if it takes
the ground? Take
for the object to reach
to be
.
Solution
Step 1. Write out the given values.
61
Shefiu S. Zakariyah
shefiuz@theiet.org
Step 2. Calculate the velocity.
Step 1. Write out the given values.
Let
Step 2. Calculate the time.
We need to use the equation below to find
as:
Note that when the object returns to the same
point from which it was thrown, the displacement
Make the subject,
(height) is zero, i.e.
This implies that ,
(
)
Step 3. Calculate the velocity.
(
)
Either
or
√
NOTE
Alternatively,
The valid answer is 1.2 s as the time taken by the
ball to return to the same point from where it was
thrown.
√
as before.
[88] A ball is projected straight upwards with an
initial velocity of
from a point . How
long will it take for the object to return to the
same point from where it was thrown?
Solution
62
Shefiu S. Zakariyah
shefiuz@theiet.org
rocket continues to coast upwards in
NOTE
unpowered mode and reaches its maximum
Alternatively, the time taken for upward motion
height before finally falling back to the
can be calculated which is then multiplied by 2 to
ground. Determine the time interval, during
obtain the time to return to the same point from
which the rocket engine provides upward
which the object was thrown. This is because the
acceleration and the value of this acceleration.
time taken for upward motion is the same as that
Leave the answers correct to 2 significant
of the downward motion under gravity provided
figures.
the air resistance is disregarded.
Here is the working
Solution
Step 1. Write out the given values.
Note that the object returns because it has reached
its greatest height. At this height, the velocity is
zero, i.e.
. Thus,
Step 2. Calculate the time interval during which
the rocket engine provides upward
Therefore, the time to return is
acceleration.
Using
as before.
(
)
This implies that ,
Section 7.
Free fall Motion
(
(Advanced)
INTRODUCTION
I hope you have had a go with linear motion
under a free fall – great! It is time to delve more
into this type of motion by taking some
challenging questions, right?
Step 3. Calculate the upward acceleration of the
rocket during the burn phase.
[89] An amateur rocket is propelled vertically
Using
upwards from the ground during which the
rocket engine provides constant upward
This implies that ,
acceleration. At the instant the engine power
burns out, the rocket has risen to a height of
and acquired a velocity of
)
. The
63
Shefiu S. Zakariyah
shefiuz@theiet.org
Make
[90] A man on top of a building
the subject,
high
throws an object straight upwards with an
initial velocity of
. Take
to be
and find:
(i) the maximum height above the ground
reached by the body, and
Step 3. Calculate the velocity with which the
(ii) the velocity with which the body hits the
body hits the ground.
ground.
Solution
Step 1. Write out the given values.
√
(
)
NOTE
Alternatively,
√
(
)
as before.
[91] An object is projected straight upwards with
an initial velocity of
. If
is the time taken
to return to the point of projection (time of
flight),
the greatest height and
is the
acceleration due to gravity, show that
(i)
Step 2. Calculate the maximum height above the
ground reached by the body.
(ii)
At the height the ball is at rest. This implies that
the initial velocity is zero.
Solution
64
Shefiu S. Zakariyah
shefiuz@theiet.org
Step 1. Write out the given values.
proved.
[92] A body is projected vertically upwards with
an initial velocity of . Another body is
Step 2. Show that
projected with the same initial velocity
.
Using
()
seconds after the first. If
is the time when
the two bodies meet and
is the acceleration
due to gravity, show that
Remember that the time taken to reach the
maximum height from the point of projection is
equal to the time taken from the maximum height
Solution
to come back to the point of projection. That is
Step 1. Sketch the projection.
( )
Substituting (ii) in (i), we will have
(
)
This is the point where
the two bodies meets.
At this point, the height
above the point of
projection for the two
is equal.
This is because
(i) the final velocity is zero at the maximum
height,
, and
(ii) acceleration due to gravity is negative for an
object moving vertically upwards. Thus
Step 2. Provide tips.
Let the bodies be represented by
and
and
distances travelled by each be denoted with
and
of meeting,
proved.
Step 3. Show that
respectively as shown above. At the point
will have travelled
would have travelled for (
.
and
)
.
Step 3. Find the distances travelled by the
Using
bodies.
Using
Substitute the values noting the same as above,
, the distance are
we will have
and
(
Step 4. Express
65
)
(
in terms of ,
)
and .
Shefiu S. Zakariyah
shefiuz@theiet.org
At meeting point,
, therefore,
(
)
Step 2. Calculate the final velocity
(
above the ground.
)
At
(
)
(
above the ground, the new displacement
is
)
(
at height
We can use the following equation to find v
)
Now substitute the values taking note of the
signs,
(
Therefore,
)
√
proved.
[93] A metallic object is launched vertically
Since the final velocity is downward,
upwards with an initial velocity of
from a point on a tower
above the
[94] A ball falls freely (under gravity) from a 15-
ground. The object rises, then falls and strikes
storey building. The distance between floors
the ground. If upward direction is taken as
is constant. It takes
positive axis, calculate the velocity of the
object when it is
)(
to fall from the 15th
to the 14th floor and
above the ground.
to fall from the 14th
floor to the 13th. What is the height of the
building? Take
Solution
Step 1. Write out the given values.
to be
.
Solution
Step 1. Write out the given values.
Step 2. Calculate the distance between floors.
and (
Let the distance from the
be
).
(
. Similarly, the time
to (
taken to fall from the
)
initial velocity for journey from the
)
66
)
floor, the
to
floor and the final velocity for journey
from the
(
floor
For example, the distance from the
15th to 14th is denoted as
(
)
,
to (
(
)
)
, and
floor will be denoted by
(
)
respectively.
Shefiu S. Zakariyah
shefiuz@theiet.org
Let the distance between two consecutive floors
 Now let us calculate the journey between the
be . This can be calculated from equation (i) as
15th and 14th floor
(
)
(
)
Using
which implies that
(
)(
(
)
)
(
(
)(
NOTE
)
)
Alternatively,
()
is determined from equation (ii)
as
 In a similar manner, we can calculate the
(
)
(
)
journey between the 14th and 13th floor
Using
which implies that
as before.
(
)(
(
)
)
(
(
)
)
Therefore, if
( )
is the height of the building then
Since the distance between the floors is constant,
we can say that
[95] An object is thrown vertically upwards with
an initial velocity of
which means
(
)
(
which is
)
(
)
(
)
be
( )
above ground level. Taking
of the object above
the ground as a function of time .
between the 15th and the 14th floor is the same as
(ii) Use the expression in (i) to find the time
initial velocity of the subsequent journey, i.e. from
the object hits the ground and velocity
the 14th to the 13th floor. In other words,
with which it strikes the ground.
( )
Substitute equation (iv) in equation (iii), we will
Solution
have
(
(
)
)
(
)
(
(
)
(
)
(
)
)
(
(
to
,
(i) Express the height
We also know that the final velocity of the journey
from a ladder
Step 1. Write out the given values.
)
)
67
Shefiu S. Zakariyah
shefiuz@theiet.org
)
√(
( )(
)
( )
√
either
√
or
√
Step 2. Express the height of the object above the
Since time cannot be negative, it implies that the
ground as a function of time.
Let the distance covered after
time taken to reach the ground is
s from the
ladder be . Therefore
Step 4. Its velocity on striking the ground.
The velocity
()
of the object before it hits the
ground can be found using
( )
Substituting the given values in equation (i), we
Note that, for this case, the initial velocity will be
will have
(
taken as negative since its direction is opposite to
)
that of the final velocity. Thus,
From equation (ii), the expression for the height
(
)
(
)( )
above the ground as a function of time is
√
Step 3. Calculate the time the object strikes the
ground.
NOTE
The object hits the ground when the height is
Alternatively, the velocity
zero, thus
hits the ground
i.e.
using quadratic equation formula
√
as before.
we will have
68
of the object before it
Shefiu S. Zakariyah
shefiuz@theiet.org
Therefore,
[96] Two identical objects are made to collide in
the air such that one object moves vertically
upwards with a velocity of
and the
other moves vertically downwards with the
NOTE
same velocity. How far apart will the objects
be after
Alternatively, the distance apart can be found as
?
Solution
(
Step 1. Write out the given values.
Step 2. Calculate the distance apart after
Let the
)
(
)
as before.
.
be the distance apart between the two
objects after
. Also, let
and
denote the
[97] From the top of a
tree, a fruit falls freely
distance covered by the object thrown upwards
and at the same moment another fruit is
and downward respectively in
thrown downward with an initial velocity of
. It follows that
. How much later or earlier does the
For this case, we can use the formula below to
free-fall fruit reach the ground? Take
calculate the distance travelled.
.
Solution
 Object thrown upwards
( )
to be
Step 1. Write out the given values.
(
)( )
 Object thrown downward
Step 2. Calculate the time difference between
the two fruits.
( )
(
Let
)( )
and
be the time taken by the free-fall
fruit and the thrown fruit to reach the ground
respectively. Therefore, if is the time difference
69
Shefiu S. Zakariyah
shefiuz@theiet.org
between the thrown and free-fall object in
Since
reaching the ground, then
|
We can find both
and
|
()
using
It follows that the free-fall is slower and reaches
( )
the ground
 For the free-fall fruit, equation (ii) becomes
(
[98] A
)
later than the thrown object.
tall child throws a ball vertically
upwards from the top of a table with a
velocity of
. It takes
from the time
of being projected upwards before the ball
√
hits the ground. If the combined height of the
√
child and the table is
above the ground,
determine the height of the table. Take
 For the thrown fruit, equation (ii) becomes
(
.
)
Solution
Step 1. Write out the given values.
using the quadratic formula, we will have
√
( )(
√
)
( )
√
Therefore, either
√
or
√
Because time can only be positive, thus the only
value for
is
. Hence from equation (i)
|
|
70
to be
Shefiu S. Zakariyah
shefiuz@theiet.org
Step 2. Determine the height
.
NOTE
The time taken to reach the maximum height can
Alternatively, the time taken for the object to
be found using
reach the same point at which it is thrown can be
found using
which implies that
Note that at this point, the velocity is the same as
Since the velocity is zero at maximum height, it
the initial velocity but in opposite direction.
follows that
Therefore,
The distance AB can be obtained as
Again, if the time taken to travel from A to C is
Where
, thus
, then
Now let us calculate the distance from A to C
If the time taken to travel from B to C is
using
, then
where
, thus
Now let us calculate the distance from B to C
using
where
, thus
From this, the height of the table is
as before.
Hence, the height
is
[99] Two particles are projected vertically
upwards from exactly the same point on the
ground at
intervals. How high above the
ground do they meet if they both start with an
From this, the height of the table is
initial velocity
71
? Take
to be
.
Shefiu S. Zakariyah
shefiuz@theiet.org
Solution
Step 1. Write out the given values.
which becomes
Step 2. Calculate the distance travelled.
Thus
Let
and
be the initial speeds of the first and
(
)
(
)
(
)
second particles respectively. Thus
Now we need to find the height from either
Also, let
and
equation (ii)
be the time after the first and
second particles were projected vertically. It
follows that
(
)
(
)
(
)
(
)
()
The two particles meet when their height above
the ground is equal. If
or equation (iii)
is the height of the first
particle after projection, then
Substitute the values, we will have
(
as before.
)
[100]
( )
Similarly, if
In a tennis court, Joshua (aka Yuusha’)
hits a tennis ball with his racquet vertically up
is the height of the second particle
from a height of 1.20 m above the floor. The
after projection then
ball reached the floor after
Assuming the value of
Substitute the values, we will have
(
(ii) the greatest height above the floor
)
reached by the ball,
(iii) the speed of the ball on reaching the floor,
Substitute equation (i) into (ii)
)
(
, find:
(i) the speed with which Joshua hits the ball,
( )
(
to be
.
)
(iv) how high the ball bounces if it loses a
( )
quarter of its speed on hitting the floor.
The two particles meet when
that is, the expression on the right hand side of
equations (iii) and (iv) are equal, which implies
that
72
Shefiu S. Zakariyah
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Solution
(
)
Step 1. Sketch the graph.
(
)
Step 3. Calculate the greatest height above the
floor reached by the ball.
At greatest height, velocity is zero. Thus,
which implies that
(
)
If the greatest above the floor is denoted by
then
Step 4. Calculate the speed the ball hits the
floor.
The speed the ball hits the floor can be obtained
using
where
Step 2. Calculate the speed with which Joshua
and
are the initial velocity at the
greatest height, and the greatest height above the
hits the ball.
floor respectively.
The speed with which Joshua hits the ball is equal
For this case
to the initial velocity of the ball, which can be
substitute these values in the above equation to
found using
find
displacement is equal to the height above the floor
from which the ball is hit measured from the
floor. In other words,
Now substitute to have
(
)(
as
√
Note that when the ball hits the floor, the
(
and
)
)
73
. Now
Shefiu S. Zakariyah
shefiuz@theiet.org
NOTE
Alternatively, the speed the ball hits the ground
can be obtained using the same equation
where
is the speed Joshua hits the ball, which is
equal to
, and is the net displacement
which is equal to
END OF WORKED EXAMPLES
. Note that in this case we
are taking all upward values as negative and
downward variables as positive.
Now substitute these values in the above equation
to find
as
(
)
√
as before.
Step 5. Determine how high the ball bounces if it
loses a quarter of its speed.
If the ball loses a quarter of its speed, it implies
that the speed of rebound is ⁄ of the speed with
which the ball hits the floor. In other words, the
initial (bouncing) speed, , is
The height can be calculated using
where
is the height reached on bouncing and
is the final velocity; the latter is equal to zero in
this case. Thus,
(
)
(
)
(
)
(
)
(
)
74
Shefiu S. Zakariyah
shefiuz@theiet.org
Bibliography and Further Reading
1) Bolton, W. (2006). Engineering Science. 5th Edition. Oxford:Newnes.
2) Bryden, P., Berry, J., Graham, T. and Porkess, R. (2004). MEI Mechanics 1. 3rd
Edition. Abingdon:Hodder Murray.
3) Halliday, D., Resnick, R. and Walker, J. (2011). Fundamentals of Physics. 9th Edition.
John Wiley & Sons.
4) Hannah, .J. and Hiller, M. J. (1992). Applied Mechanics. 2nd Edition. Harlow:
Longman
5) Johnson, K., Hewett, S., Holt, S. and Miller, J. (2000). Advanced Physics for You.
Cheltenham: Nelson Thornes.
6) Okeke, P.N. and Anyakoha, M.W. (2000). Senior Secondary Physics. Revised Edition.
London: Macmillan Education.
7) Reid, D (2001). An Introduction to Engineering Mechanics. London: Palgrave
8) Rex, A. and Wolfson, R. (2009). Essential College Physics. 1st Edition. Addison Wesley.
9) Sadler, A.J. and Thorning, D.W.S. (1996). Understanding Mechanics. 2nd Edition.
Oxford: Oxford University Press.
10) Tuttuh-Adegun, M.R., Sivasubramaniam, S. and Adegoke, R. (1992). Further
Mathematics Project 2. Ibadan: NPS Educational.
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