Linear Motion Explained with Worked Examples

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

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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.


What Next 3.

equation to use to find .

[4] Given , ,

Write the final answer. in a problem, decide which


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.


What Next 2.

Simplify the expression on the

What Next 3. [2] Given , ,

(

)

right hand side.

What Next 1.


Write the final answer.

What Next 2.


in a problem, decide which What Next 3.


[5] Given , , Solution



The equation that connects the four variables and Solution


The equation that connects the four variables and What Next 1.


What Next 2.



right hand side. What Next 3.

What Next 1.


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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.



What Next 2.


What Next 3.


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



the subject of the

Solution

formula.


What Next 2.



What Next 3.


What Next 4. [7] Given , ,

What Next 1.


Make

the subject of the

formula.



What Next 2.


What Next 3.


Solution What Next 4.




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.


What Next 3.


[10] Find

[8] Given , ,

,

and

.

right hand side. What Next 4.

when


Solution


Step 1. Decide on the formula.


We need to use ()

Solution

Step 2. Solve for


.

Substituting the given values in equation (i), we


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


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


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 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).


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


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 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


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


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


a velocity of

. He then accelerates

uniformly at a rate of Calculate his final velocity. Step 2. Calculate the initial velocity. (


)

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

.

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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?


(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 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


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


.

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 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


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.


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


. 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 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.


[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

.


( )( )

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.


[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

.

(

)


Solution

Using


Substituting the values, we will have Step 2. Calculate the velocity. .

/

.

/


(

)

√ [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



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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



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)


and the time it takes to stop. Solution Step 1. Write out the given values.


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


[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.


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


passes the signal. Solution Step 1. Write out the given values.


Substituting the values in above equation, we will have Step 2. Convert

to

.

( )


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


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)


[50] A particle moves such that its position

33

( )

( )

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, and it is increasing or going in the

  At


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


( )

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 ,


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


√

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

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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

.


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


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)


( ) 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

)

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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


(

47

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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

(


)

(

)

( )

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

)

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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.

(


)

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,


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)


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,


( ) 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.


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)


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)


(

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.

(


)

(

)( )

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


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 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.


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 2. Calculate the initial velocity. When the water reaches a height of


, 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


. 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


to be

.



)(

)

√ Step 2. Calculate the initial velocity.

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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 (

)(


)

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


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 ,

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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.


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.


Step 3. Calculate the velocity. ( Either √

or

62

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)

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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


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 3. Calculate the velocity with which the body hits the ground.

√ (

64

)

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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


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


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


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

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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 √


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.

69

(

)( )

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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

√

. ( )(

√

)

( )


√ 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


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

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Solution

NOTE


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


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

.

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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

)( )

)

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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

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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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Linear Motion Explained with Worked Examples

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