Applied Maths – commonly asked questions

Make sure to try out the Transition Year module which begins on page 11 of this booklet

Table of contents

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Introduction: What is Applied Maths?

Some commonly asked questions Why study Applied Maths? Do I need to be doing Physics to study Applied Maths? Who shouldn’t study Applied Maths? How likely is it that I will get A? How long is the course? What is the format of the paper in the leaving cert exam? What textbook do I need? What is the paper like at Ordinary Level? Is it possible to study Applied Maths in college? I want to study Maths in college – should I take Applied Maths

now?

Testimonies

Related Careers

Applied Mathematics Syllabus

Introduction

Unlike most other subjects there is no Applied Maths option for Junior Cert so for you the student it is quite difficult to know just what the subject is like at Leaving Cert level.

The best way to get a feel for what the subject is to attend an Applied Maths class; we usually hold a series of classes for interested students in January/Febrauary of 4th year for TY who are interested in finding out more about the subject. Here we start off looking at some typical Ordinary Level questions and then tackle some Higher Level questions. Students are generally pleasantly surprised that they are able to do Higher Level questions after only one or two classes of an introduction.

Here however we try to answer some of your questions in advance.

It is a fascinating subject which deals with solving real-life problems using mathematical models. It overlaps with both the Mathematics course as well as the Physics course. In particular the new Project Maths syllabus is very similar in its approach to Applied Maths. The emphasis is on using different mathematical models to solve everyday problems.

Unlike any other subject in the Leaving Cert (with the possible exception of the new Project Maths), Applied Maths is about problem-solving, so if you like a challenge then Applied Maths is for you.

The maths is all based around Physics problems

Applied Maths will instil skills which will last a lifetime; how to analyse a problem, how to represent the problem mathematically, how to solve the maths, and then how to interpret your answer so that it makes sense when applied to real life situations.

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Why study Applied Maths? If you like (and are good at) Maths

If you are thinking about studying Engineering in college

It complements the Maths Course and enables students to obtain the bonus points.

Looks good on your CV (see the Careers section).

See the testimonies at the end of this booklet from other students around the country who studied Applied Maths.

Do I need to be good at Maths to study Applied Maths?Many students are put off doing the subject because they feel that they need to be brilliant at maths. You do need to be reasonably good at maths if you intend taking it at Higher Level, but you certainly don’t need to be an ‘A’ student. If you are a comfortable ‘B’ student then you should have little difficultly lasting the pace.The best indicator at this stage is to check if you can follow the examples in the second section of this booklet and see if you can do the questions that follow.

There is a high Maths content in the course so studying Applied Maths will give you a better understanding of some parts of the Honours Maths course – especially Trigonometry, Calculus (Differentiation and Integration) and Vectors  In studying Applied Maths you will improve on your mathematical skills in areas such as trigonometry, geometry, vectors, differentiation and integration.

What does the name Applied Maths actually mean?The subject actually has a very confusing title – after all you can apply maths to almost any situation. The subject should be called Mathematical-Physics; it’s all about analysing a physics problem and solving it using maths equations.You may have often heard the phrase “it’s not rocket science”; well this subject is rocket science. We study (among many other things) the force required to launch a rocket into orbit and analyse the forces which act on it while it travels through the atmosphere (it’s not as bad as it seems, honest!).

Do I need to be doing Physics to study Applied Maths?It is not necessary to take Physics as a Leaving Cert subject in order to do Applied Maths. There is an overlap, but it’s not as great as you might think. If anything, doing Applied Maths will help you in Physics much more than the other way around in that you will develop a deeper understanding of many of the concepts which only get covered superficially in Physics.

I want to do Business; wouldn’t I be just wasting my time doing Applied Maths?It mightn’t seem obvious but employers in Business and Finance are always looking for graduates from the field of Applied Maths – they already have lots of employees who know about business, what they don’t have is enough people who can offer solutions to unusual problems.Many Applied Mathematicians get jobs in the business world because they have analytical and problem-solving skills which can be applied to the money markets and/or the stock-exchange.

It is also (relatively) straightforward to get a business (or business management) degree once you have an Science/Engineering degree if you so choose. It is not however straightforward to do it the other way around.

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Who shouldn’t study Applied Maths?This subject doesn’t suit students who just like learning things off by heart.

In fact the questions are designed to catch out those very students and whether that is fair or not is a moot point - you are being warned about it now so if you don’t like it you know what to do. You cannot come out of an Applied Maths exam and say ‘we never did that question before – sir you never covered it with us in class’. It is my job to ‘train’ you to tackle problems which you haven’t come across before.

So Applied Maths suits people who like solving puzzles (we like to make it sound more impressive so we call it ‘problem solving’). This means being able to think for yourself, and because almost all of your secondary-school education encourages you to ‘learn the right answer off by heart’ it can make a lot of students uncomfortable. The ability to problem-solve is however a very important skill and is highly-valued by many employers. It is one of the reasons why you often see politicians and business people on the news saying that the country needs more scientists and engineers.

As the name suggests, the course is mathematical in nature, and you do need to be very good at maths to be successful at higher level. In general you should be averaging a ‘B’ or higher, and just because you got an ‘A’ in Maths in the Junior Cert does not necessarily mean that you will be comfortable with the subject, so pay attention to the problems we will cover below to see if you can follow them.

The material itself is similar to that which you would study in Engineering in university, so even if you are not great at maths but want to be an engineer you should consider taking up the subject. Even if you end up taking the Ordinary Level exam at the end it will still stand to you at third level.

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How long is the course?The course itself is fairly short (see the syllabus at the end). There are ten topics but only six questions to do on the exam itself so there is enormous choice. In fact normally we only cover about 8 topics, and try to ensure we know them very well rather than trying to cover too much. It means that we are normally finished the course by Christmas in sixth year, so there’s a lot of time for revision from then on. The Christmas exam is itself usually a full mock paper.

What is the format of the paper in the leaving cert exam?Time: 2½ hoursOrdinary Level: six questions to be answered out of nine Higher Level: six questions to be answered out of ten (about 25 minutes per question).

We generally aim to cover 8 (or at most 9) topics (where each topic corresponds to a different question); it’s more advisable to spend your time becoming competent in 8 questions rather than spending time on others and this still allows for ample choice on the day itself.

The exam began in 1970 so there are over 40 years of exam questions. We will cover all of these for each topic (many of them are repetitive) so there should be very little on your leaving cert paper that you wouldn’t already have seen.

What is the paper like at Ordinary Level?Applied Maths at the Ordinary Level is probably the easiest subject on the leaving cert curriculum. It is very short and can easily be covered in one year or less. The questions vary very little from year to year so with a just a little practice it should be easy to nail the A.

What textbook do I need?There are a number of textbooks available for studying Applied Maths:1. Fundamental Applied Mathematics (Oliver Murphy); a new addition came out in 2012.2. Applied Mathematics Leaving Certificate Foundation Mechanics for Third Level (Kevin Conliffe).3. Applied Mathematics:  A Comprehensive Course for Leaving Certificate, 2nd Edition (Dominick

Donnelly). It can be ordered directly from the author from his website here: http://appliedmathematics.ie/

I don’t use a textbook however. I have a booklet for each topic we cover. This has a short introduction to the topic but after that it’s all exam questions, arranged in order of difficulty.You don’t even need to buy exam papers because all exam questions are included in these notes (and anyway exam-papers only go back 10 years or so while we you will receive 40 years worth of questions).Because some students like to work ahead I also provide a separate booklet containing all the answers.These solutions/marking schemes are all available from www.thephysicsteacher.ie

How likely is it that I will get A at Higher Level?27% of students on average receive an A in the Leaving Cert.

I used to think that the subject was very difficult because the questions that come up on the Leaving Cert exam were completely different every year. Then I noticed that almost all the questions in the exam papers in recent years were just variations on older questions (some of them 30 years older!). So now our students cover all questions back to 1970 – that’s 40 years worth of questions.It may just be a co-incidence, but one of our students recently got the top mark in the country in 2011.

Any student who can get an A Maths should certainly be getting an A in Applied Maths (but with much less effort!)

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Is it possible to study Applied Maths in college?You can study Applied Maths itself at third level. It is known as ‘Mathsphysics’ in NUI universities and as ‘Theoretical Physics’ in Trinity College.

I want to study Maths in college – should I take Applied Maths now?Applied Maths is an invaluable subject for those who plan to study pure maths in university. Indeed many of those who have studied maths at university say that Applied Maths was a more important preparation than Maths itself!

I want to study Engineering in college – should I take Applied Maths now?If you are considering studying any kind of engineering in college, Applied Maths is very important; all engineering students have to study Applied Maths in first year in college and you will have a head start if you have the Leaving Cert course done beforehand.

Related Links YouTube – just enter “applied maths” which will bring up a wide variety of different videos Irish Applied Maths Teachers Association: http://www.iamta.ie/Careers PortalScoilnet

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Reasons you should consider doing applied maths

From Brendan Doheny, former chairman of the Irish Applied Maths Teachers’ Association (IAMTA) Many students study Applied Maths for the benefit of their future maths results.

These are students who are chasing the bonus 25 points in honours Maths and who will find maths easier as a result of studying Applied Maths.

Apart from being a terrific subject (which takes work, like any other), about of 3 in every 10 students will achieve an A1.

Some universities (like UCC) are accepting a C1 in either Maths OR Applied Maths.

Many students study Applied Maths to improve their problem solving skills and so produce better HPAT results to ensure they get into medicine.

The single greatest obstacle for engineering students is passing Applied Maths in college - not a problem for those who studied it in school.

The text in Italics is taken from quotes on the iamta.ie website.

Applied maths has a reputation to be compared to Shrek! To those that don’t know the beast it seems like a big ugly ogre, but really it is quite a lean course.

As few schools offer Applied Maths as a scheduled class, most that take it on do so outside of school hours. This can be stressful and tiring at first but the end result is worth any hardships suffered.

There are 10 questions on the paper and the candidate has only to answer six questions. NO other paper offers this much choice.

Getting the actual correct answer only gives you a small amount of the marks.

In total the recall work is minimal possibly 30 formulae in total and the overlap between the courses less than half are unique to Applied Maths Syllabus

Applied Maths was my favourite subject at school. It was a course completely unlike any other. Lateral and logical thinking are needed for it and solving problems is something I have always been interested in. The subject doesn’t involve enormous amounts of theory that you have to learn off by heart, like in biology, but just encourages you to think and apply principles.

Is there any subject that you struggle with, any language that you will only manage to do ordinary level in? Then Applied Maths can help you make your points higher!

Although no points in the Leaving Cert ever come easy, the Applied Maths examiners love to give away marks for writing down basic information and applying simple rules. Applied Maths may seem like a tough subject at first, but given a little time and a good bit of practice it becomes a tremendously satisfying and rewarding subject.

For those that like to experience the success of solving problems, applied maths is more satisfying than sudoku. If you have always been the student that enjoyed solving the problem maths that were in the primary books, then Applied Maths is for you.

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I soon discovered that Applied Maths is a hugely under-rated subject with a lot of benefits. Firstly, for those people who, like me, have lazy tendencies, and can think of better ways to spend an afternoon than memorising dates until you can’t even remember what year it is any more, Applied Maths is the ideal subject. There is almost no theory and very little writing involved, which is something I greatly appreciated when trying to write poetry essays every weekend.

If you are doing Higher Level Maths, this course compliments it! The Points may be the same but the amount of equations, topics, theorems and methods to recall is less. Plenty of practice in trig identities and geometry, the vectors question is a doosy.

Applied Maths allows the student to solve real life problems through logical reasoning and creative thought, and in my opinion that is something that no other Leaving Cert subject offers. As a student it seems daunting to take on another subject when school is already difficult and hectic as it is. However, Applied Maths added very little to my work load. Subjects such as Maths and Physics overlap with Applied Maths in many areas such as integration, vectors and mechanics.

If you are doing Physics then this will sort out about 40% of the entire physics paper for you! If you are doing both .....

Applied Maths also made life easier on a number of levels. Whole sections of Maths seemed easier having covered them the year before. Quadratic equations, integration and trigonometry benefited purely because we were using them so often. The mechanics section of physics posed little difficulty for many of us as we knew our UVAST inside and out and in other ways Applied Maths always encouraged you to look outside the box for an easier way of doing things – which is invaluable whatever you’re doing. Although it needed attention, with a little effort it’s definitely a manageable course with relatively little learning off. In short it’s a gift to anyone mathematically inclined who enjoys a challenge

It will help you if you want to pursue a career in Engineering, Architecture or any other technical subject. I am currently studying Engineering in U.C.D. one of the core modules of the course is Mechanics, which is little more than Leaving Certificate Applied Maths

Applied Maths has advantages which I really came to appreciate in those last few days and weeks before the exam,There was very little theory to be learnedIt was a predictable subject with lots of choice.And most importantly I had two days off after having finished all my other exams in which to try and learn everything.

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Related Careers

Who would want to employ somebody who has studied Applied Maths?In short, anybody who would like an employee who can solve problems, so basically everybody.

Applied Maths is like Engineering for beginners. Also Architects must know some applied maths. Degree courses in the Building industry contain applied maths modules.

Studying Applied Maths enables students achieve employment and career prospects at the top end of the market in very diverse areas.

The following represent just some of the areas which students in up in after studying Applied Maths. Actuary Airlines...designing efficient rosters Anthropologist...dating fossils Architecture Astronomy Business Chemistry Computer Programming...animated films, computer games Currency Exchange companies Designer...computer aided design, nano-technology Economics Education Engineering environmental studies Forensics Genetics...dna codes Hardware design...iphones, speakers etc Information Technology Investment banking Maths Medicine Metereology Pharmaceuticals Physics Planning traffic flow systems for big cities Political studies Robotics Science Statistician in the Central Statistics Office or in a casino The army ballistics division

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APPLIED MATHEMATICS SyllabusOrdinary and Higher Level Courses

Knowledge of the relevant parts of the Mathematics course is assumed. N.B. Those parts of the syllabus which are printed in italics belong to the Higher Level course only.

The Higher Level course includes the Ordinary Level course treated in greater depth.

1. Motion of a particle. Displacement, velocity as vectors. Applications of the vector addition law. Description of vectors in terms of unit perpendicular vectors. Elementary treatment of relative motion.

2. Newton’s laws. Mass, momentum. Acceleration and force as vectors. Units and dimensions.

3. Motion in a straight line under uniform acceleration e.g. motion under gravity, motion on smooth and rough inclined planes. Work, potential energy, kinetic energy, power. Application of energy conservation. Motion of connected particles.

4. Equilibrium of a particle under concurrent forces, including friction.

5. Centre of gravity of simple bodies and systems of particles Moments and couples. Equilibrium of a rigid body or bodies.

6. Liquid pressure. Thrust on a horizontal surface. Archimede’s Principle.

7. Projectiles. Projectiles on inclined plane.

8. Angular velocity. Uniform motion in a circle without gravitational forces. Conical pendulum. Circular orbits.

9. Conservation of momentum. Collisions. Direct collisions, elastic (0 < e ≤ 1) and inelastic (e = o). Oblique collisions of smooth elastic spheres in two dimensions.

10. Simple harmonic motion of a particle in a straight line. (Application of simple harmonic motion to include the simple pendulum.)

11. Motion of a rigid body about a fixed axis:

a. Calculation of moments of inertia for a rod, rectangular lamina, circular lamina and compound bodies formed of those. (Sphere is excluded). Application of parallel and perpendicular axes theorems, with proofs done as classwork. Idea of radius of gyration. Application of the conservation of energy principle to a rotating body.

b. (b) Application of the principle of angular momentum: rate of change of angular momentum about a fixed axis equals the total external moment about that axis. Compound pendulum. Simple applications.

12. Ordinary differential equations and applications: (a) first order, variables separable; (b) Second order reducing to type

In conclusion

Applied Maths is not an easy subject, but it is a wonderfully rewarding subject to take on if you are good at maths and enjoy challenging yourself. It is interesting, challenging, educational, and it is good for careers.

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Applied Maths: Transition Year ModuleMost problems we will encounter in Applied Maths can be split into three or four parts:1. Read in the problem in English2. Translate the English into maths3. Solve the maths4. Sometimes you may need to interpret your answer, i.e. convert the information back into english to show

that you understand the significance of your answer.

Before we tackle the problems we need to have a couple of tools in our bag. We begin by looking at the concept of acceleration.

Acceleration is defined as the rate of change of velocity with respect to time

Consider an object which is falling through the air. Its instantaneous velocity after one, two, three and four seconds is given in the table.We can see that with every second that passes the velocity increases by 10 m/s.Alternative ways of writing this are that the velocity increases by 10 m/s per second or 10 m/s/s or 10 m/s2 or 10 m s-2.The unit of acceleration is the metre per second squared (m s-2, or m/s2)

Incidentally, this rate of falling through the air is known as acceleration due to gravity and is given the symbol g. It is also not exactly 10 m s-2 and in fact it changes with altitude (among other things). We usually take it to be 9.8 m s-2.We will encounter problems which involve g later on in the module.

Equations of MotionWhen an object (with initial velocity u) moves in a straight line with constant acceleration a, its displacement s from its starting point, and its final velocity v, change with time t.Note that both v and u are instantaneous velocities.

The following three equations tell us how these quantities are related:

v = final velocityu = initial velocitya = accelerations= displacement (not distance)t = time

To Derive v = u + at

acceleration= change∈velocitytime taken acceleration= final velocity−initial velocity

timetaken

a= v−u

t v = u + at

We could derive the other two equations in a similar fashion (and you will actually need to know how to do this for Leaving Cert Physics) but for now we will just accept that they are valid. They are also in the log tables if you forget them.

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v = u + at

s = ut + ½ at2

v2 = u2 + 2as

Acceleration= change∈velocitytime taken

Time Velocity1 second 10 m/s2 seconds 20 m/s3 seconds 30 m/s4 seconds 40 m/s

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Procedure for solving problems using equations of motion

1. Write down v, u, a, s and t underneath each other on the left hand side of the page; fill in the quantities you know and put a question mark beside the quantity you are looking for.

2. Decide which of the three equations has only one unknown in it.

3. Substitute in the known values in to this equation and solve to find the unknown.

4. If an object starts from rest then its initial velocity (u) is 0.

5. If an object comes to a stop then its final velocity (v) is 0.

6. All times must be in seconds before using them in the equations (e.g. 4 minutes = 360 seconds).

ExamplesExample 1:A car accelerates from 20 m s−1 to 50 m s−1 in 5 seconds. What is its acceleration?

Solution:v = 50u = 20a = ?s = t = 5Pick equation no. 1: v = u + at 50 = 20 + a (5) 30 = 5a a = 6 m s-2

Example 2:A car is travelling at a velocity of 10 m s-1 when the driver hits the brakes.Calculate how far the car will travel before coming to rest if the deceleration is 5 m s-2?

Solution:Here the car was decelerating so the acceleration is written as –5 m s-2.

v = 0u = 10a = -5s = ?t = Pick equation no. 3: v2 = u2 + 2as 0 = 102 + 2(–5)s –100 = –10s s = 10 m

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v = u + at

s = ut + ½ at2

v2 = u2 + 2as

Velocity-time graphsYou will often be asked to draw a velocity-time graph.

Use a ruler to draw the axes; then put velocity on the y-axis and time on the x-axis.

Now put in as much information as you can on the graph itself.

Your graph might look like the following:

The area under a velocity-time graph corresponds to the distance travelled.

For example in this graph the distance travelled

area of triangle 1 + area of rectangle 2 + area of triangle 3 = ½ (5)(15) + (20)(15) + ½ (3)(15)

= 37.5 + 300 + 22.5

= 360 m

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Applied Maths Exam Questions (Ordinary Level) with worked solutions

If there are different stages to the cars journey then you must treat each stage separately (i.e. be careful not to mix values for one stage with values for another when using the equations of motion).

2007 OLA car travels from p to q along a straight level road.It starts from rest at p and accelerates uniformly for 5 seconds to a speed of 15 m/s.It then moves at a constant speed of 15 m/s for 20 seconds.Finally the car decelerates uniformly from 15 m/s to rest at q in 3 seconds.(i) Draw a speed-time graph of the motion of the car from p to q.(ii) Find the uniform acceleration of the car.(iii) Find the uniform deceleration of the car.(iv)Find |pq|, the distance from p to q.(v) Find the speed of the car when it is 13.5 metres from p.

Solution (i)

(ii) v = 15u = 0a = ?s = t = 5Use v = u + at 15 = 0 + 5 a a = 3 m s-2

(iii) v = u + atv = 0u = 15a = ?s = t = 3Use v = u + at 0 = 15 + 3 a a = -5 m s-2

(iv) distance = area of first triangle + area of middle rectangle + area of second triangle= ½ (5)(15) + (20)(15) + ½ (3)(15)= 37.5 + 300 + 22.5

= 360 m

(v) v = u = 15a = 3s = 13.5t = Use v2 = u2 + 2as = 0 + 2(3)(13.5) v2 = 81 m v = 9 m s-1

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2012 OLA car travels along a straight level road.It passes a point P with a speed of 8 m s−1 and accelerates uniformly for 12 seconds to a speed of 32 m s−1.It then travels at a constant speed of 32 m s−1 for 7 seconds.Finally the car decelerates uniformly from 32 m s−1 to rest at a point Q.The car travels 128 metres while decelerating.Find(i) the acceleration(ii) the deceleration(iii) |PQ|, the distance from P to Q(iv)the speed of the car when it is 72 m from Q {this last part is quite tricky – see how you get on}

SolutionBegin by drawing a velocity-time graph to represent the motion

Note that the initial velocity here is not 0.

(i) v = u + at32 = 8 + a(12)a = 2 m s-2

(ii) v2 = u2 + 2as02 = (32)2 + 2a(128)a = - 4 m s-2

(iii) There are three distinct sections here; when the car is accelerating, when it is travelling at constant speed, and finally when it is decelerating In each case we can use s = ut + ½ at2

s1 = 8(12) + ½ (2)(144)s1 = 240 m

s2 = 32 × 7 + ½ (0)(7)2

s2 = 224 m

s3 = 128 m

|PQ| = 240 + 224 + 128= 592 m

Can you see how you could do this much more quickly by drawing a velocity-time graph?

(iv)v2 = u2 + 2as(0)2 = u2 + 2(-4)(72)

u = 24 m s-1

Can you see why this makes sense by looking at the velocity-time graph?Could you use the area under a section of the velocity-time graph to get the same answer?

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2011 OLThe points P and Q lie on a straight level road.A car passes P with a speed of 10 m s-1 and accelerates uniformly for 6 seconds to a speed of 22 m s-1.The car then decelerates uniformly to a speed of 18 m s-1 and travels 80 m during this deceleration.The car now maintains a constant speed of 18 m s-1 for 3 seconds and then passes Q.Find (i) the acceleration(ii) the deceleration(iii) |PQ|, the distance from P to Q(iv)the average speed of the car, correct to one decimal place, as it moves from P to Q.

{Formula for average speed is: Average speed = totaldistance

total time }

Solution

(i) v = u + at 22 = 10 + a(6) a = 2 m s-2

(ii) v2 = u2 + 2as (18)2 = (22)2 + 2a(80) a = -1 m s-2

(iii) s = ut + ½ at2

s1 = 10(6) + ½ (2)(36)s1 = 96 m

s2 = 80 m

s = ut + ½ at2

s3 = 18(3) + 0= 54 m

|PQ| = 96 + 80 + 54= 230 m

Or again, use a velocity-time graph to solve more quickly

(iv)Average speed = totaldistance

total time

So we need to find the total time. We know that t1 = 6 seconds, and t3 = 3 seconds, so we need to find t2

Use v = u + at

t2 = v – u

a t2 = 18 – 22−1 t2 = 4 secs

Average speed = totaldistance

total time = 230

6+4+3 = 17.7 m s-1

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2008 OLFour points a, b, c and d lie on a straight level road.A car, travelling with uniform retardation, passes point a with a speed of 30 m/s and passes point b with a speed of 20 m/s.The distance from a to b is 100 m. The car comes to rest at d.Find (i) the uniform retardation of the car(ii) the time taken to travel from a to b(iii) the distance from b to d(iv)the speed of the car at c, where c is the midpoint of [bd].

Solution

(i) v2 = u2 + 2as 202 = 302 + 2(a)(100) -500 = 200aa = - 2.5 m s-2

(ii) v = u + at 20 = 30 – 2.5 tt = 4 s

(iii) v2 = u2 + 2as 02 = 202 + 2(-2.5)(s) s = 80 m

(iv)We know that the total distance is 80 m, so the midpoint would be 40 m from b.So we can use v2 = u2 + 2as v2 = 202 + 2(-2.5)(40) v2 = 200

v = 200 or 14.1 m s-1

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2009 OL3 points p, q and r lie on a straight level road.Two cars, A and B, are moving towards each other on the road.Car A passes p with speed 3 m/s and uniform acceleration of 2 m/s2

and at the same instant car B passes r with speed 5 m/s and uniform acceleration of 4 m/s2.A and B pass each other at q seven seconds later.Find (i) the speed of car A and the speed of car B at q.(ii) |pq| and |rq|, the distances A and B have moved in these 7 s.(iii) Car A stops accelerating at q and continues on to r at uniform speed.

Find, correct to one place of decimals, the total time for car A to travel from p to r.

Solution

(i) v = u + atvA = 3 + 2(7)vA =17 m s-1

v = u + atvB = 5 + 4(7)vB =33 m s-1

(ii) s = ut + ½ at2

sA = 3(7) + ½ 2(49)sA =70 m

s = ut + ½ at2

sB = 5(7) + ½ 4(49)sB =133 m

(iii) Car A now has to travel a distance of 133 m. His speed will be the speed he had when he reached point q, so his speed = 17 m s-1

So we can use s = ut + ½ at2 133 = 17(t) + 0 t = 7.8 s

Total time = 7 + 7.8 = 14.8 s

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2010 OLA car travels along a straight level road.It passes a point P at a speed of 12 m s-1 and accelerates uniformly for 6 seconds to a speed of 30 m s-1.It then travels at a constant speed of 30 m s-1 for 15 seconds.Finally the car decelerates uniformly from 30 m s-1 to rest at a point Q.The car travels 45 metres while decelerating.Find (i) the acceleration(ii) the deceleration(iii) |PQ|, the distance from P to Q(iv)the average speed of the car as it travels from P to Q.

Solution

(i) v = u + at30 = 12 + a(6)a = 3 m s-2

(ii) v2 = u2 + 2as0 = (30)2 + 2a(45)a = - 10 m s-2

(iii) s = ut + ½ at2

s1 = 12(6) + ½ (3)(36)s1 = 126 m

s2 = ut + ½ at2

s2 = 30 (15) + 0s2 = 450 m

s2 = 45 m {you were told this in the question}

|PQ | = 126 + 450 +45 = 621 m

(iv)Average speed = totaldistance

total time

We know that the total distance is 621 m, so we need to find the total time. We know that t1 = 6 seconds, and t2 = 15 seconds, so we need to find t3

Use v = u + at t3 = v – u

a t3 = 0– 30−10 t3 = 3 secs

Average speed = totaldistance

total time = 621

6+15+3 = 25.875 m s-1

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Ordinary Level Exam Questions2006 OLA car travels along a straight level road.It passes a point p at a speed of 10 m/s and accelerates uniformly for 5 seconds to a speed of 30 m/s.It then moves at a constant speed of 30 m/s for 9 seconds.Finally the car decelerates uniformly from 30 m/s to rest at point q in 6 seconds.Find (i) the acceleration(ii) the deceleration(iii) pq , the distance from p to q(iv)the average speed of the car as it travels from p to q.

2005 OLA particle travels from p to q in a straight line. It starts from rest at p and accelerates uniformly to its maximum speed of 20 m/s in 10 seconds. The particle maintains this speed of 20 m/s for 15 seconds before decelerating uniformly to rest at q in a further 20 seconds.(i) Draw a speed-time graph of the motion of the particle from p to q.(ii) Find the uniform acceleration of the particle.(iii) Find the uniform deceleration of the particle.(iv)Find pq, the distance from p to q.(v) Find the average speed of the particle as it moves from p to q, giving your answer in the form a/b where

a, b ∈ N.

2004 OLThree points a, b and c, lie on a straight level road such that ab=bc= 100 m.A car, travelling with uniform retardation, passes point a with a speed of 20 m/s and passes point b with a speed of 15 m/s.(i) Find the uniform retardation of the car.(ii) Find the time it takes the car to travel from a to b, giving your answer as a fraction.(iii) Find the speed of the car as it passes c, giving your answer in the form p q , where p, q ∈ N.(iv)How much further, after passing c, will the car travel before coming to rest?Give your answer to the nearest metre.

2003 OLA car travels from p to q on a straight level road. It passes p with a speed of 4 m/s and accelerates uniformly to its maximum speed of 8 m/s in 4 seconds. The car maintains this speed of 8 m/s for 6 seconds before decelerating uniformly to rest at q.The car takes 12 seconds to travel from p to q.(i) Draw a speed-time graph of the motion of the car from p to q.(ii) Find the uniform acceleration of the car.(iii) Find the uniform deceleration of the car.(iv)Find pq, the distance from p to q.

(v) Another car travels the same distance from p to q in the same time of 12 seconds.This car starts from rest at p and accelerates uniformly to its maximum speed of v m/s and then immediately decelerates uniformly to rest at q.Find v, the maximum speed of this car, giving your answer as a fraction.

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2002 OLA train stops at stations P and Q which are 2000 metres apart. The train accelerates uniformly from rest at P, reaching a speed of 20 m/s in 10 seconds. The train maintains this speed of 20 m/s before decelerating uniformly at 0.5 m/s2 coming to rest at Q. (i) Find the acceleration of the train. (ii) Find the time for which the train is decelerating. (iii) Find the distance and the time for which the train is travelling at constant speed. (iv)Draw an accurate speed-time graph of the motion of the train from P to Q.

2001 OLTwo points, p and q, lie on a straight stretch of level road. Car A passes the point p with a speed of 2 m/s travelling towards q and accelerating uniformly at 2 m/s2. As car A passes p, car B passes the point q with a speed of 1 m/s travelling towards p and accelerating uniformly at 3 m/s2. The two cars meet after 10 seconds. (i) Find the speed of each car when they meet. (ii) Find the distance each car has travelled during these 10

seconds.

(iii) Suppose now that the speed of car A when passing point p is u m/s instead of 2 m/s, while the speed of car B passing point q and the acceleration of each car remain unchanged. If the time taken for the two cars to meet in this case is 8 seconds, find the value of u.

2000 OLA car is travelling on a straight stretch of level road pq. The car passes the point p with a speed of 5 m/s and accelerates uniformly to its maximum speed of 20 m/s in a time of 6 seconds. The car continues with this maximum speed for 30 seconds before decelerating uniformly to rest at q in a further 4 seconds.(i) Draw a speed-time graph of the motion of the car from p to q.

Hence, or otherwise, find(ii) the uniform acceleration of the car(iii) the uniform deceleration of the car(iv)| pq |, the distance from p to q .

(v) Another car, with acceleration and deceleration the same as in (i) and (ii) above, starts from rest at p and accelerates uniformly to its maximum speed of 25 m/s. It continues with this maximum speed for a certain time and then decelerates uniformly to rest at q .How long does it take this car to go from p to q?

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Answers to Ordinary Level Exam Questions: 2006 - 2000Full solutions (marking schemes) for all of these questions are available from the exam material section of thephysicsteacher.http://thephysicsteacher.ie/exammaterialappliedmaths.html

2006 (i) Acceleration = 4 m s-2

(ii) a = -5 m s-2

(iii) Distance = 460 m(iv)Average speed = 23 m s-1

2005(i) See diagram(ii) a = 2 m s-2

(iii) a = -1 m s-2

(iv)s = 600 m(v) Average speed = 40/3 m s-1

2004 (i) a = - 0.875 m s-2

(ii) t = 40/7 s(iii) v = 52 m s-1

(iv)s = 29 m

2003(i) See diagram (ii) a = a m s-2

(iii) a = -4 m s-2

(iv)s = 80 m(v) v = 40/3 m s-1

2002 (i) a = 2 m s-2 (ii) t = 40 s(iii) t = 75 s(iv)See diagram

2001(i) vA = 22 m s-1, vB = 31 m s-1

(ii) sA = 120 m, sB = 160 m(iii) u = 14 m s-1

2000(i) See diagram(ii) a = 2.5 m s-2

(iii) a = -5 m s-2

(iv)s = 715 m

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Higher Level Questions

Type 1: Common Initial Velocity

Here the acceleration is constant throughout and we are given information about different stages.Usually we are given information on two consecutive sections of an objects motion.We need to get an equation for both and solve, but to do this the variables (particularly u) must represent the same number for both equations.The only way to do this is to make the second equation represent the first two stages. This means the distance s must correspond to the distance from the very beginning and t must correspond to the time from the very beginning for both equations.

The relevant equation will be s = ut + ½ at2 (because the variables involved will be distance and time).

These questions have been arranged in order of increasing difficulty.

1996 (a)A particle starts from rest and moves in a straight line with uniform acceleration. It passes three points a, b and c where |ab| = 105 m and |bc| = 63 m. If it takes 6 seconds to travel from a to b and 2 seconds to travel from b to c find(i) its acceleration (ii) the distance of a from the starting position.

The key to question is as follows:|ab| = 105 m: 105 = u(2) + ½ (a)(2)2

|ac| = 168 m: 168 = u(5) + ½ (a)(5)2

And solve the simultaneous equations as normalCan you see why it would be incorrect to use 63 = u(3) + ½ (a)(3)2??

Answer: u = 7 m s-1, a = 3.5 m s-2, s = 7 m

Similar-style questions1974A sprinter runs a race with constant acceleration k throughout. During the race he passes four posts a, b, c, d in a straight line such that |ab| = |bc| = |cd| = 36 m. If the sprinter takes 3 seconds to run from a to b and 2 seconds to run from b to c, find how long, to the nearest tenth of a second, it takes him to run from c to d.

Answer: u = 8.4 m s-1, a = 2.4 m s-2, time = 1.6 seconds {use the “-b” formula for this last part}

2003 (a) The points p, q and r all lie in a straight line. A train passes point p with speed u m/s. The train is travelling with uniform retardation f m/s2. The train takes 10 seconds to travel from p to q and 15 seconds to travel from q to r, where | pq| = | qr | = 125 meters.(i) Show that f = 1/3 (ii) The train comes to rest s meters after passing r. Find s, giving your answer correct to the nearest metre.

Answer: u = 14.17 m s-1, a = 1/3 m s-2, s = 51 m

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1993 (a)A particle moving in a straight line travels 30 m, 54 m and 51 m in successive intervals of 4, 3 and 2 seconds.(i) Verify that the particle is moving with uniform acceleration(ii) Draw an accurate speed-time graph of the motion.

Answer: u = 1.5 m s-1, a = 3 m s-2 See diagram

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