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:
[email protected] |
[email protected]
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.
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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 ............................................................................................ 76
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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.
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∫ ( )
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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.
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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
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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, .
/
. /
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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
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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
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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
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[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
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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
)
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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
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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 acceleration.
Step 1. Write out the given values.
( )
Step 2. Calculate the velocity. Calculate the velocity at finish line.
25
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(
(
) (
) )
This implies that
√ [38] In a speed controlled zone, a car is initially travelling at a speed of (
Section 4.
)
before it begins
to accelerate uniformly at
. Calculate:
(i) the speed after
Linear Motion - Horizontal
,
(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 (
(
)
26
)
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(
)
Step 2. Calculate the acceleration. Using
as before.
Substituting the values, we will have
[39] In an electronics factory, a conveyor belt is used to move circuit boards. Initially at rest, a board moves from the production stage to the assembly stage with acceleration of
.
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.
Solution
Using
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
[40] The speed of a particle increases from to
in
and finally come to a
complete halt after
. Calculate the rate of change
. Determine:
(i) the rate at which it slowed down, and
and the time taken for this increment.
(ii) the distance covered during this period. Solution Solution
Step 1. Write out the given values.
Step 1. Write out the given values.
27
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[42] A passenger bus moves with a constant velocity along a straight road having three consecutive bus stops, , Step 2. Change the velocity from
to
.
to move from from
to
to . If
and . It takes and
to move
, find
.
Solution Step 1. Sketch the journey. Step 3. Calculate the retardation.
A
B
C
Step 2. Write out the given values.
Substituting the values, we will have
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.
28
.
)
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[43] A van skids to a halt from an initial speed of covering a distance of
stop?
. Find the Solution
acceleration of the van (assumed constant)
Step 1. Write out the given values.
and the time it takes to stop. Solution Step 1. Write out the given values.
Step 2. Calculate the acceleration.
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
29
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Emergency Road Sign when compared to a
Let ‘how much farther’ distance travelled by the
‘normal’ driver given that the former level of
drunk driver’s car be
alcohol would increase his / her RT to Take
, thus
?
as the ‘normal’ RT.
Solution Step 1. Write out the given values. NOTE Alternatively, we can say ( Step 2. Convert the initial velocity from
to
) (
)
.
as before.
Step 3. Calculate the distance covered by the normal driver before hitting the brake.
[46] Asmaa’ is cycling her bike initially at 1.5 m/s
Let the distance covered by the ‘normal’ driver’s car be
before she decides to accelerate at
, thus
.
What will be the time taken to cover where
a
is the reaction time for the ‘normal’
driver, therefore
on
straight road leading to her school.
Solution Step 1. Write out the given values.
Step 4. Calculate the distance covered by the drunk driver before hitting the brake. Let the distance covered by the drunk driver’s car be
, thus
Where
Step 2. Calculate the time. Using
is the reaction time for the drunk
driver, therefore Substitute above values to have ( Step 5. Calculate how much farther the drunk driver would travel.
30
)
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Divide through by 1.5, ( )( )
Now multiply through by 5, (
)(
)
Either
Step 4. Calculate the velocity.
or
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
accelerated at
(approx.) passes a signal at a level crossing 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 have Step 2. Convert
to
.
( )
Step 3. Calculate the acceleration.
Multiply through by 2,
Using
31
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(
)(
)
(
)
Either
() Its velocity at time
or
is (
) ( )
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
in equation
(i) as
time, i.e. 15 seconds earlier. (
[49] A particle moves along a straight line with acceleration
Therefore, either
. It starts its motion at
the origin with velocity
.
or
(i) Write down equations for its position and velocity at time
)
.
(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 Its position at time
seconds. is
32
.
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metres at time seconds is given by the
position (m)
position-time graph 30
expression
.
25
(i) Determine the position of the particle when times
20
, , ,
,
and .
(ii) Construct a table showing the position of
15 10
the particle at these times.
5
(iii) Draw a position- time graph. (iv) State the times when the particle is at the
0
0
2
4
6 time (s)
8
10
12
origin and describe the direction in which it is moving at those times. (v) Using the graph in (iv) or otherwise, find 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
6 4
, the position of the particle is ( )
( )
( )
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
33
( )
( )
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, and it is increasing or going in the
At
, the position of the particle is
positive direction. Step 6. Find the velocity at
( )
( )
( )
seconds.
Since the graph is not a straight line, we need to make a tangent to the curve at t = 4 as shown above. The velocity at this time is equal to the
At
, the position of the particle is ( )
( )
slope of the tangent at that point. From the graph, the slope of the tangent is
( )
Taking the values of Step 3. Construct a table.
and at two points, we can
find the gradient
The table showing the positions of the particle at the interval
(
)
is shown below.
( )
0
1
2
3
4
5
( )
0
-12
-32
-42
-24
40
38.3 Hence, the velocity at
is
correct to 2
significant figures.
Step 4. Sketch a graph of the position against time.
NOTE Alternatively, we can find the velocity using
Graph of position against time 50
differentiation as
40
position, x (m)
30
( )
20
Point 1
10 0
-10
0
1
2
3
4
5
(
6
)
-20 -30
The above is a velocity as a function of time. To
-40 -50
time (s)
find the velocity at
Point 2
in the expression as
Step 5. State the times when the particle is at the 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
, we simply substitute for
as before.
, and it is decreasing or going in the negative direction.
34
( )
( )
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[51] An object moving with a uniform acceleration covers distances
and
succeed in beating Faisal?
in the first two Solution
equal and consecutive intervals of time . Express
in terms of ,
Step 1. Write out the given values.
and .
Faisal Solution The initial velocity, , of the second stage of the motion is equal to the final velocity of the first Edward
stage. Using
and
Thus for
Step 2. Calculate the time it takes Faisal to cover
distance we have
the distance of
.
() 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.
Realising this and to beat Faisal, Edward
Faisal finished, Edward has 13.6 m distance to
decides to accelerate uniformly at
cover since he was 100 m away from the finish
until the end of the race while Faisal keeps on
line.
the same constant speed. Does Edward
NOTE
35
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E1 and accelerates uniformly for
Alternatively, we can find the time taken by
until it reaches a maximum speed of
Edward to reach the finish line as follows
.
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
after
leaving their respective terminals. The distance between the two terminals is
Using quadratic equation formula
and
√
and
represent the distances of T1
and T2 respectively from the International we have
Airport. Assuming the motion is considered ( )( ( )
√
)
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
36
it implies that
Shefiu S. Zakariyah
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or
Where
and
are the distances covered
during acceleration and constant velocity respectively. Thus {[
,
]
(
)-}
which implies that
For T2
*, Again, since
it implies that
-
,
*
or
,
( )( ) ]
{[
Similarly we will have
(
)-}
-+ +
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
{
for T2 after
(
)
.
The distance Airport E1 after
of train T2 from the International is (
The two trains meet at )
37
from terminal E1.
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[54] When John rows his boat, the two oars are both in water for water for
Step 3. Calculate the change in the boat’s speed
and then both out of the
for each 7 s cycle.
. This pattern is repeated for a
The change in the boat’s speed for each
cycle is
race lasting 1 minute.
the algebraic sum of the changes in and out of the
(i) Find the change in speed that takes place
water, which is (
in water if the boat accelerates at a constant rate of
)
(
)
( (
.
)
)
(ii) Find the change in speed that takes place out of water if the boat decelerates at a
The positive value indicates that the total change
constant rate of
in speed for each 7 s cycle is increase.
.
Step 4. Calculate the finishing speed.
(iii) Calculate the change in the boat’s speed for each
The speed with which John would finish his race
cycle.
is given by
(iv) What is John’s speed as he crosses the finishing line, if he starts the race from
Since he starts from the rest and takes
rest?
to
complete then Solution Step 1. Find the change in speed that takes place
But (
in water. If the acceleration in the water is
)
then
Thus (
)
[55] As soon as the traffic light changed to green, Abdullah accelerates uniformly with his Step 2. Find the change in speed that takes place
motorcycle at
out of water. If the deceleration out of the water is
for
. He then
maintains a steady velocity for a two-third of then
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
38
Shefiu S. Zakariyah
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rest if he then slows down at
?
We know that
Solution Step 1. Write out the given values. Where is the distance covered during a constant velocity, which is
, i.e. two-third of
.
Substituting the values, we will have
Step 2. Calculate the distance travelled during
The time spent on retardation can be found from
the acceleration. 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 Thus
a stop. 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
NOTE
as , then
This can be also solved using graphical method. This method is dealt with in section 5.
But,
39
Shefiu S. Zakariyah
[56] ,
and
[email protected]
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 between stations
is mid-way
[57] From a speed of
and . If the speeds with
which the train passes and
and
at a T-junction, Barrak
accelerates his car along a straight highway
are
road. He observes that he reaches a tunnel
respectively, what is the speed
from the junction
with which the train passes station .
and that he
crosses the tunnel, which is further
Solution
long, in a
.
The motion of the car is modeled by taking
Step 1. Sketch the journey.
A
)
the acceleration to be constant with a value of
B
C
. (i) By considering the part of the journey from the T-junction to the tunnel, show
Step 2. Calculate the speed at point . Let the distance between
and
that
be . It follows
(ii) Find another second equation involving
that
and .
()
(iii) By elimination method or otherwise,
Also, let the speed with which the train passes
solve for
be . This is equal to the final velocity of the journey between
and
of the journey between
and the initial velocity
in the two equations.
Step 1. Find an expression for the journey between the junction and the tunnel in terms of
journey
and
.
The distance between the T-junction and the tunnel can be calculated using
( ) For
and
Solution
and . Using
we can therefore write: For
.
journey Substituting the values, we will have ( )
(
)
Now equate the right-hand sides of equations (ii) and (iii)
Divide through by 12 ()
40
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Step 2. Find another expression involving
and
[58] A passenger bus starts from a stop and travels
’.
with a uniform acceleration of
The distance from the junction to the end of the
for
until it reached a maximum velocity. It
tunnel can be calculated using
travels with this maximum velocity for another
, the brake is then applied so that
a uniform retardation is obtained and then
For this case,
come to a halt at the next bus stop after
.
Sketch the velocity-time graph of this motion and use this to calculate the total distance between the two bus stops. Now substituting the values, we will have )
Step 1. Draw the graph. Vel. / (m/s)
(
Solution
Divide through by 18 ( ) Step 3. Find the values of
and
B
E
.
To find , subtract equation (i) from (ii),
C A
D
20
60
F
70
From equation (i), Step 2. Calculate the final velocity
’.
The slope of AB is equal to the acceleration, i.e.
( )
Section 5.
time / (sec)
Graphical Solution of One-
Step 3. Calculate the total distance travelled by
Dimensional Motion
the train. The total distance travelled during the journey is equal to the area of the trapezium ABEF
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.
(
41
)
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[59] A roller skater, Hawwah, starts off with an initial velocity of
Step 3. Fid an expression for the distance
before she gets to a
travelled as a function of time
sloppy surface where she gains speed at a constant rate. If after moving at
.
The distance (s) travelled after t seconds is equal to the area of the trapezium ABEF
she is already
,
(
)
(i) Find an expression of her speed and ( )
distance seconds after she started gaining speed.
Substitute equation (i) in equation (ii), we have
(ii) Given that the length of the slope is
,
(
what is her speed at the bottom of the
)
slope?
( )
Solution Step 1. Draw the graph.
NOTE
Vel. / (m/s)
Alternatively, the distance (s) travelled after seconds is E
C
(
)
B
Step 4. Find the speed at the bottom of the slope. A
D
F
The time taken to reach the bottom of the slope
time / (sec)
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
( )(
√ ( )
This implies that
Therefore, ()
42
)
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Step 2. Calculate the total journey time. or
The total journey time, , is given by
The slope of AB is equal to the acceleration, i.e. Therefore, the only valid value is
.
The final speed at the bottom of the slope can be found using equation (i) as (
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
E
. When the car
behind the front of the coach, Yaasir
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)
43
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accelerating correct to 2 significant figures.
of
before the driver decided on what to
do and applied a brake. The car then comes to Solution
rest with constant deceleration in a further
Step 1. Draw the graph.
distance.
Vel. / (m/s)
(i) Find the time that the driver takes to react to the situation. (ii) Using, a velocity-time graph, calculate the D
deceleration once the car starts braking.
Car
(iii) What is the stopping distance for another
C B
E
vehicle travelling at
Coach
if the reaction
time and the deceleration are the same as before? Solution
F
A
Step 1. Draw the graph.
time /(sec)
Step 2. Calculate the time
spent in Vel. / (m/s)
accelerating. The acceleration, , is equal to the slope of the line CD, thus
C
B
Case 2
Case 1
D
E
Therefore, A
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 (
)
Step 3. Calculate the deceleration. [62] A car is travelling at
The deceleration, a, is equal to the slope of the line EG, thus
when the driver
has to perform an emergency stop due to a sudden shock. The car has travelled a distance
44
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uniform velocity of
()
stationary car But the distance covered after applying the brake is equal to the area of the triangle EFG, thus ( (
and passed a
waiting to join the convoy.
Five seconds later, car at
starts and accelerates
. How far will
have to travel
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
Make
)
is travelling with a
45
the subject
)
Shefiu S. Zakariyah
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NOTE
Vel. / (m/s)
Alternatively, using figure below Since the two cars must have covered the same distance, it implies that , therefore (
)
Simplify the equation and solve for Car A
(
)(
) time / (sec)
Therefore, either
(
)
or
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. ( )
Make
the subject (
) (
)(
)
Since the two cars must have covered the same distance, it implies that , therefore (
)
Simplify the equation and solve for
(
)(
Therefore, either
or
46
)
Shefiu S. Zakariyah
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Vel. / (m/s)
Substitute the values of in either or to determine how far will car have to travel before it catches up car ,
C
as before. or
B
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.
D
G
H
time / (sec)
Step 2. Calculate the value of
.
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
continues with that speed for
F
A
[64] A motorist drives past a police patrol team with an over speed-limit of
E
, (
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, Step 3. Calculate the distance between the two
the patrol team started from rest
vehicles after stopping.
later, speed up and attained a speed of in
Distance covered by the motorist equals the area
. They directed the motorist to stop
of the trapezium ABEH
when they are abreast each other. As a result, the team also applied a brake and are brought
(
)
to rest under uniform retardation in another
. Assuming uniform
Distance covered by the patrol team is equal to
acceleration and retardation in both cases,
the area of the triangle DCG
using a suitable graph,
(
(i) find the value of ;
)
(ii) the distance between the two vehicles Therefore, the distance between the motorist’s car
after both have finally stopped.
and the patrol team car after both have finally Solution
stopped is
Step 1. Draw the graph.
(
47
)
Shefiu S. Zakariyah
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(
[
)]
[
]
This shows that the patrol team car will be ()
in front of the motorist’s car. [65] Three junctions ,
and
road. A motorcyclist passes junction speed of
Acceleration is the slope of the line BC given by
are on a straight with a
from where he accelerates at
a constant rate of
until he gets to
( )
junction . He therefore pulls up with a constant retardation of
Also, the deceleration is the slope of the line CE given by
and comes to a
complete halt at junction . Given that the total distance between junctions
and
is
, find: (i) the speed of the motorcycle at junction ,
( )
and (ii) the distance from junction
Substitute for
to junction .
and
(
Solution Step 1. Draw the graph.
)
(
)
( )
Vel. / (m/s)
Multiply through by 6 (
Junction Y
)
(
)
( )
This implies that
C
B
√
Junction X
A
in equation (i), thus
D
E
Junction Z
time /(sec)
Step 2. Calculate the value of
Step 3. Calculate the distance from
.
to .
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,
( Using equation (ii),
48
)
Shefiu S. Zakariyah
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)(
(
)(
( ( )(
) )
which implies that
)
[66] A train takes
or
to travel from
Stansted Airport to Cambridge. The train accelerates from rest at a rate of
for Step 3. Calculate the deceleration.
. It then travels at a constant speed before it is finally brought to rest in
The deceleration is the slope of the line EF, therefore
with a
constant deceleration. The motion is modelled as a linear motion on a straight railway track.
(
)
(i) Sketch a velocity-time graph for the which implies that
journey. (ii) Find the steady speed, the deceleration and the distance from Stansted Airport to Cambridge station.
(
Solution Step 1. Draw the graph.
)
Step 4. Calculate the distance from Stansted
Note: The time in the graph is in minutes, but this will be converted to seconds during the calculations where appropriate.
Airport to Cambridge.
Vel. / (m/s)
Let the distance covered by the train be , which is equal to the area of the trapezium ABEF. ,(
)
-
E
B
--
A
C
D
F
NOTE time / (min)
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
Step 2. Calculate the steady speed. Let the steady speed be , therefore
49
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Vel. / (m/s)
covered in stages 1, 2 and 3 respectively, then
where
E
B
C
D
F
A
time / (sec)
Step 2. Calculate the distance covered while accelerating. Let the distance covered during the acceleration be denoted as . This is equal to the area of the triangle ABC. Therefore, Therefore,
Step 3. Calculate the distance covered during the uniform speed. Let the distance covered during the uniform speed be denoted as . This is equal to the area of the rectangle BCDE. Therefore,
as before
[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 Step 1. Draw the graph.
50
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The deceleration, , is the slope of line FG. This is given as
NOTE ()
Alternatively, the total distance travelled, , is equal to the area of the trapezium ABEF. Therefore, (
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,
)
( )
as before. where [68] From an initial velocity of
,a
professional cyclist accelerates uniformly until he attains a maximum velocity of
.
He maintains this velocity for some time Hence,
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.
,
Solution
, (
) ]
-
,
)-
))]
,
-
This implies that
Vel. / (m/s)
Step 1. Draw the graph.
Thus, from equation (i) deceleration is C
F
( ) B
[69] A train accelerates uniformly from rest at a D
E
station
G
A
to a maximum speed of
.
The constant maximum speed is maintained
time / (sec)
for a period of time and the train then
Step 2. Calculate the deceleration.
51
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decelerates uniformly until it comes to a stop
Because the magnitude of the deceleration is double that of acceleration implies that
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: Equating (i) to (ii), we have
(i) the acceleration, in metre per second, and (ii) the time, in minutes, during which the train travels at its maximum speed. Solution
Therefore,
Step 1. Draw the graph. 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.
()
52
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(
Vel. / (m/s)
From the graph, the time for constant speed is
)
C
F
B
Therefore, time in minutes is
Stage 2
Stage 1
(
D
)
E
A
time / (sec)
Step 2. Calculate the time taken the car to reach
[70] On her way to Kuala Lumpur International
its maximum speed of
Airport, Katrina was initially driving at
If the car gains the acceleration is
. On getting to one end of the SMART (Stormwater Management and Road Tunnel),
every
. , it implies that
she began to increase her speed such that she gained
every
. She continued like
From the graph, the acceleration of the car is equal to the slope of the line BC, thus
this until reaching a maximum speed of which she maintained. Assuming the tunnel is modeled as a straight-line path, using a speed-time graph or otherwise,
where is the time the car reached its maximum speed. Therefore,
(i) Find the time taken to reach the maximum speed. (ii) If the tunnel is approximately 10 km long, find the time, in minutes correct to 2 significant figures, taken to drive through
Step 3. Calculate the time taken to drive through
the tunnel.
the tunnel.
(iii) Write down expressions for the speed of
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
the car as a function of time in seconds. Solution Step 1. Draw the graph.
where
53
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(
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
and
to reach point
. As he comes close to a traffic light at , the
But
brake is applied and the car gradually comes to rest in
Therefore,
. Draw a velocity-time graph for
the motion, and find: (i) the speed
of the car at
(ii) the acceleration a of the car when travelling from
to ,
(iii) the retardation of the car when (
)
travelling from
to , and
(iv) the total distance from Step 4. Determine expressions for the speed of
Solution
the car as a function of time in seconds.
Step 1. Draw the graph. Vel. / (m/s)
There are two stages of the motion, each has a different expression for the speed. Stage 1 :
to .
or
At this stage, the speed (v) as a function of time (t) is
C
D
B
Stage 2:
or
Stage 1
Stage 2
Stage 3
Stage 4
E A
At this stage, the speed (v) as a function of time (t) is
time / (sec)
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
straight road between two small villages. He
which implies that
moves with a constant acceleration of for
where he attained a velocity of
.
54
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or
Hence, Step 3. Calculate acceleration of the car when travelling from B to C. The slope of BC is equal to the acceleration of the car between B and C, therefore [72] From a petrol station, Benjamin increases his But
speed at a constant rate to
, thus
in the first
of his journey. On seeing a temporary speed limit on the road, he then decreases uniformly to
or
in a further
,
remaining constant for some times before he decelerates uniformly from this speed to a complete stop in
. Assuming the motion is
modeled to be a straight line, find the distance
Step 4. Calculate the distance between A and E.
covered if the whole journey takes
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
.
Solution
Vel./ (m/s)
Step 1. Draw the graph.
where
B
E
F
Stage 2
Stage 1 C
Stage 3 D
Stage 4 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
55
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correct to 2 significant figures, covered by the driver.
where
Solution
Vel. / (m/s)
Step 1. Draw the graph.
(
I
) B
E
F
(
)
Stage 1
Stage 2
Stage 3
C
D
Stage 4
G
Stage 5 J
H
A
time / (min.)
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
where
for
before he begins to slow down 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
56
)
(
)
Shefiu S. Zakariyah
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(
)
(
) Step 2. Calculate the distance.
(
Substituting the values, we will have
)
(
)( )
Therefore,
Step 3. Calculate the velocity. Substitute the values, we will have
Section 6.
Free fall Motion (Basic)
[75] A light weight stone is thrown vertically 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 Step 1. Write out the given values.
from and down to the ground; it thus follows that the time taken to reach the maximum height is half of this, i.e.
57
.
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[77] A helium-filled balloon is thrown straight
(ii) At maximum height the ball comes to rest. In
upwards with an initial velocity of
other words, final velocity for the upward
.
Assuming the air resistance is negligible,
motion is zero.
calculate the greatest height reached and the
(iii) Height is the same as the distance in the
time taken to reach it. Take
equation of motion. We will use these tips frequently in this and
.
Solution
subsequent questions.
Step 1. Write out the given values.
Step 2. Calculate the height.
Substituting the values, we will have ( )
( (
Step 2. Calculate the greatest height reached. Using
)
)
it implies that , (
)
[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.
58
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Solution
Because
, we will have
Step 1. Write out the given values. This implies that √ Substitute the values, we will have √ Step 2. Calculate the time. [80] A package is dropped from a helicopter Because
moving upwards at
, we will have
. If it takes
before the package reaches the ground, how high above the ground, correct to 2 s. f., was Make
the subject of the formula
the package when it was released? Solution Step 1. Write out the given values.
√ 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.
59
)(
)
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[81] An object falls off from the edge of a blade of
[83] A fountain is designed such that water can be
an off-shore wind turbine. If the tower is
projected vertically upwards to a height of
, high find the velocity with which the
. Determine the speed at which the water
object hits the surface of the water below it.
must be leaving the fountain nozzle, leaving
Leave the final answer in surd form.
the answer in surd form.
Solution
Solution
Step 1. Write out the given values.
Step 1. Write out the given values.
Step 2. Calculate the initial velocity. When the water reaches a height of
Step 2. Calculate the velocity.
, the
velocity is zero, i.e. This implies that , (
)(
)
This implies that , (
)( )
√ √ √
[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.
. Determine:
(i) the initial velocity of the parcel, (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.
60
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Provided the air resistance is negligible, the initial
(
)( )
velocity of the parcel is the same as that of the balloon. In other words, √ Step 3. Calculate the time. [86] An object at rest is subjected to a free fall from Substituting the values, we will have (
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.
)
Either
or
Step 2. Calculate the height.
Since time cannot be negative, it follows that the
Since
, 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
Solution
upwards with an initial velocity of
Step 1. Write out the given values.
the floor is
. 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 ,
61
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NOTE Alternatively,
√ 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 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
62
)
)
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into this type of motion by taking some challenging questions, right? [89] An amateur rocket is propelled vertically upwards from the ground during which the The valid answer is 1.2 s as the time taken by the
rocket engine provides constant upward
ball to return to the same point from where it was
acceleration. At the instant the engine power
thrown.
burns out, the rocket has risen to a height of and acquired a velocity of
NOTE
. The
Alternatively, the time taken for upward motion
rocket continues to coast upwards in
can be calculated which is then multiplied by 2 to
unpowered mode and reaches its maximum
obtain the time to return to the same point from
height before finally falling back to the
which the object was thrown. This is because the
ground. Determine the time interval, during
time taken for upward motion is the same as that
which the rocket engine provides upward
of the downward motion under gravity provided
acceleration and the value of this acceleration.
the air resistance is disregarded.
Leave the answers correct to 2 significant figures.
Here is the working
Solution Note that the object returns because it has reached
Step 1. Write out the given values.
its greatest height. At this height, the velocity is zero, i.e.
. Thus,
Therefore, the time to return is
Step 2. Calculate the time interval during which the rocket engine provides upward acceleration.
as before.
Using (
Section 7.
Free fall Motion
)
This implies that ,
(Advanced)
(
INTRODUCTION I hope you have had a go with linear motion under a free fall – great! It is time to delve more
63
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Step 3. Calculate the upward acceleration of the rocket during the burn phase. Using This implies that ,
[90] A man on top of a building
high
throws an object straight upwards with an initial velocity of
. Take
to be
and find:
Step 2. Calculate the maximum height above the
(i) the maximum height above the ground
ground reached by the body.
reached by the body, and
At the height the ball is at rest. This implies that
(ii) the velocity with which the body hits the
the initial velocity is zero.
ground.
Make
Solution
the subject,
Step 1. Write out the given values.
Step 3. Calculate the velocity with which the body hits the ground.
√ (
64
)
Shefiu S. Zakariyah
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(
NOTE Alternatively,
)
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
√ (
)
as before. proved. [91] An object is projected straight upwards with an initial velocity of
. If
Step 3. Show that
is the time taken
Using
to return to the point of projection (time of flight),
the greatest height and
.
is the
Substitute the values noting the same as above,
acceleration due to gravity, show that
we will have
(i) (ii) Solution 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
65
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proved. [93] A metallic object is launched vertically upwards with an initial velocity of from a point on a tower
This is the point where the two bodies meets. At this point, the height above the point of projection for the two is equal.
above the
ground. The object rises, then falls and strikes the ground. If upward direction is taken as positive axis, calculate the velocity of the object when it is
Step 2. Provide tips.
above the ground.
Solution
Let the bodies be represented by
and
and
Step 1. Write out the given values.
distances travelled by each be denoted with and
respectively as shown above. At the point
of meeting,
will have travelled
would have travelled for (
and
)
.
Step 3. Find the distances travelled by the bodies. Using
, the distance are
and ( Step 4. Express
)
(
)
in terms of ,
At meeting point,
and . Step 2. Calculate the final velocity
, therefore, (
)
(
above the ground.
) At
(
)
( (
at height
above the ground, the new displacement
is
)
We can use the following equation to find v
)
Now substitute the values taking note of the signs, (
Therefore,
66
)(
)
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√ which implies that
Since the final velocity is downward,
( [94] A ball falls freely (under gravity) from a 15-
)(
(
)
)
(
(
)
)
( )
storey building. The distance between floors
Since the distance between the floors is constant,
is constant. It takes
we can say that
to fall from the 15th
to the 14th floor and
to fall from the 14th which means
floor to the 13th. What is the height of the building? Take
to be
(
.
)
(
) (
)
(
)
( )
Solution
We also know that the final velocity of the journey
Step 1. Write out the given values.
between the 15th and the 14th floor is the same as initial velocity of the subsequent journey, i.e. from
Step 2. Calculate the distance between floors. and (
Let the distance from the be to
14th
is denoted as
)
initial velocity for journey from the )
)
Substitute equation (iv) in equation (iii), we will have
floor, the
(
to
(
floor and the final velocity for journey to (
from the (
( )
. Similarly, the time to (
taken to fall from the
(
floor
) . For example, the distance from the
(
15th
)
the 14th to the 13th floor. In other words,
,
(
)
)
) )
(
(
)
(
)
(
)
)
( (
)
)
Let the distance between two consecutive floors be . This can be calculated from equation (i) as
15th and 14th floor
(
Using
which implies that )( )
(
)
respectively.
Now let us calculate the journey between the
(
(
floor will be denoted by
, and
(
)
(
)
(
)(
)
) ()
In a similar manner, we can calculate the journey between the 14th and 13th floor Using
67
)
(
)
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NOTE Alternatively,
is determined from equation (ii)
as (
)
(
)
as before.
Therefore, if
is the height of the building then Step 2. Express the height of the object above the ground as a function of time. Let the distance covered after
[95] An object is thrown vertically upwards with an initial velocity of which is be
ladder be . Therefore
from a ladder
above ground level. Taking
s from the
to
()
,
(i) Express the height
( )
of the object above
Substituting the given values in equation (i), we
the ground as a function of time .
will have
(ii) Use the expression in (i) to find the time
(
the object hits the ground and velocity
)
with which it strikes the ground. From equation (ii), the expression for the height Solution
above the ground as a function of time is
Step 1. Write out the given values. Step 3. Calculate the time the object strikes the ground. The object hits the ground when the height is zero, thus i.e. using quadratic equation formula √ we will have
68
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)
√(
( )(
)
[96] Two identical objects are made to collide in
( )
the air such that one object moves vertically
√
upwards with a velocity of either
and the
other moves vertically downwards with the √
same velocity. How far apart will the objects be after
or
?
Solution √
Step 1. Write out the given values.
Since time cannot be negative, it implies that the time taken to reach the ground is Step 2. Calculate the distance apart after
Step 4. Its velocity on striking the ground. The velocity
Let the
of the object before it hits the
. Also, let
and
denote the
distance covered by the object thrown upwards and downward respectively in
Note that, for this case, the initial velocity will be taken as negative since its direction is opposite to
)
(
. It follows that
For this case, we can use the formula below to
that of the final velocity. Thus, (
be the distance apart between the two
objects after
ground can be found using
calculate the distance travelled. )( ) Object thrown upwards
√ ( )
NOTE Alternatively, the velocity
.
(
)( )
of the object before it
hits the ground Object thrown downward
( ) as before.
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(
)( )
Shefiu S. Zakariyah
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Therefore,
between the thrown and free-fall object in reaching the ground, then | We can find both
and
|
()
using ( )
NOTE Alternatively, the distance apart can be found as
For the free-fall fruit, equation (ii) becomes (
(
)
(
)
)
√
√
For the thrown fruit, equation (ii) becomes
as before.
( [97] From the top of a
)
tree, a fruit falls freely
and at the same moment another fruit is thrown downward with an initial velocity of . How much later or earlier does the free-fall fruit reach the ground? Take
using the quadratic formula, we will have
to be
√
. ( )(
√
)
( )
Solution Step 1. Write out the given values.
√ Therefore, either √
or
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
Because time can only be positive, thus the only
respectively. Therefore, if is the time difference
value for
is
. Hence from equation (i) |
70
|
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Step 2. Determine the height Since
.
The time taken to reach the maximum height can be found using
It follows that the free-fall is slower and reaches the ground [98] A
which implies that
later than the thrown object. tall child throws a ball vertically
Since the velocity is zero at maximum height, it
upwards from the top of a table with a velocity of
. It takes
follows that
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
The distance AB can be obtained as
to be
. Where
Solution
, thus
Step 1. Write out the given values.
If the time taken to travel from B to C is
, then
Now let us calculate the distance from B to C using
where
, thus
Hence, the height
is
From this, the height of the table is
71
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Solution
NOTE
Step 1. Write out the given values.
Alternatively, the time taken for the object to reach the same point at which it is thrown can be found using
Step 2. Calculate the distance travelled. Note that at this point, the velocity is the same as
Let
the initial velocity but in opposite direction.
second particles respectively. Thus
and
be the initial speeds of the first and
Therefore, Also, let
and
be the time after the first and
second particles were projected vertically. It follows that () The two particles meet when their height above
Again, if the time taken to travel from A to C is
the ground is equal. If
, then
is the height of the first
particle after projection, then
Now let us calculate the distance from A to C
Substitute the values, we will have
using
(
) ( )
where
, thus
Similarly, if
is the height of the second particle
after projection then
Substitute the values, we will have (
From this, the height of the table is
) ( )
Substitute equation (i) into (ii)
as before.
(
)
(
)
( )
The two particles meet when
[99] Two particles are projected vertically upwards from exactly the same point on the ground at
that is, the expression on the right hand side of
intervals. How high above the
ground do they meet if they both start with an
equations (iii) and (iv) are equal, which implies
initial velocity
that
? Take
to be
.
72
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(
)
(
)
Solution
which becomes
Step 1. Sketch the graph.
Thus (
)
Now we need to find the height from either equation (ii) (
)
(
)
(
)
(
)
or equation (iii)
as before. [100]
In a tennis court, Joshua (aka Yuusha’) Step 2. Calculate the speed with which Joshua
hits a tennis ball with his racquet vertically up
hits the ball.
from a height of 1.20 m above the floor. The ball reached the floor after Assuming the value of
to be
The speed with which Joshua hits the ball is equal
.
to the initial velocity of the ball, which can be
, find:
found using
(i) the speed with which Joshua hits the ball, (ii) the greatest height above the floor reached by the ball,
Note that when the ball hits the floor, the
(iii) the speed of the ball on reaching the floor,
displacement is equal to the height above the floor
(iv) how high the ball bounces if it loses a
from which the ball is hit measured from the
quarter of its speed on hitting the floor.
floor. In other words, Now substitute to have ( (
73
)( )
)
Shefiu S. Zakariyah
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(
)
(
)
NOTE Alternatively, the speed the ball hits the ground can be obtained using the same equation
Step 3. Calculate the greatest height above the where
floor reached by the ball.
is the speed Joshua hits the ball, which is
equal to
At greatest height, velocity is zero. Thus,
, and is the net displacement
which is equal to
. Note that in this case we
are taking all upward values as negative and
which implies that (
downward variables as positive.
)
Now substitute these values in the above equation to find If the greatest above the floor is denoted by
as (
then
)
√ Step 4. Calculate the speed the ball hits the as before.
floor. The speed the ball hits the floor can be obtained using
Step 5. Determine how high the ball bounces if it loses a quarter of its speed.
where
and
are the initial velocity at the
If the ball loses a quarter of its speed, it implies
greatest height, and the greatest height above the
that the speed of rebound is ⁄ of the speed with
floor respectively. For this case
which the ball hits the floor. In other words, the and
. Now
initial (bouncing) speed, , is
substitute these values in the above equation to find
as
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
(
)
(
)
(
)
(
)
(
)
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END OF WORKED EXAMPLES
75
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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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