AQA A-level Mathematics for A-level Year and AS is available as a Whiteboard eTextbook and Student eTextbook Whiteboard eTextbooks are online interactive versions of the printed textbook that enable teachers to: ● Display interactive pages to their class ● Add notes and highlight areas ● Add double-page spreads into lesson plans Student eTextbooks are downloadable versions of the printed textbooks that teachers can assign to students so they can: ● Download and view on any device or browser ● Add, edit and synchronise notes across two devices ● Access their personal copy on the move Important notice: AQA only approve the Student Book and Student eTextbook The other resources referenced here have not been entered into the AQA approval process To find out more and sign up for free trials visit: www.hoddereducation.co.uk/dynamiclearning Integral A-level Mathematics online resources Our eTextbooks link seamlessly with Integral A-level Mathematics online resources, allowing you to move with ease between corresponding topics in the eTextbooks and Integral These online resources have been developed by MEI and cover the new AQA A-level Mathematics specifications, supporting teachers and students with high quality teaching and learning activities that include dynamic resources and self-marking tests and assessments Integral A-level Mathematics online resources are available by subscription to enhance your use of this book To subscribe to Integral visit www.integralmaths.org AQA A-level Mathematics For A-level Year and AS Authors Approval message from AQA The core content of this digital textbook has been approved by AQA for use with our qualification This means that we have checked that it broadly covers the specification and that we are satisfied with the overall quality We have also approved the printed version of this book We not however check or approve any links or any functionality Full details of our approval process can be found on our website We approve print and digital textbooks because we know how important it is for teachers and students to have the right resources to support their teaching and learning However, the publisher is ultimately responsible for the editorial control and quality of this digital book Please note that when teaching the AQA A-level Mathematics course, you must refer to AQA’s specification as your definitive source of information While this digital book has been written to match the specification, it cannot provide complete coverage of every aspect of the course A wide range of other useful resources can be found on the relevant subject pages of our website: aqa.org.uk Sophie Goldie Val Hanrahan Cath Moore Jean-Paul Muscat Susan Whitehouse Series editors Roger Porkess Catherine Berry Consultant editor Heather Davis Bibliography Hachette UK’s policy is to use papers that are natural, renewable and recyclable products and made from wood grown in sustainable forests The logging and manufacturing processes are expected to conform to the environmental regulations of the country of origin Orders: please contact Bookpoint Ltd, 130 Park Drive, Milton Park, Abingdon, Oxon OX14 4SE Telephone: (44) 01235 827720 Fax: (44) 01235 400454 Email education@bookpoint.co.uk Lines are open from a.m to p.m., Monday to Saturday, with a 24-hour message answering service.You can also order through our website: www.hoddereducation.co.uk ISBN: 978 4718 5286 © Sophie Goldie,Val Hanrahan, Jean-Paul Muscat, Roger Porkess, Susan Whitehouse and MEI 2017 First published in 2017 by Hodder Education, An Hachette UK Company Carmelite House 50 Victoria Embankment London EC4Y 0DZ www.hoddereducation.co.uk Impression number 10 Year 2021 2020 2019 2018 2017 All rights reserved Apart from any use permitted under UK copyright law, no part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying and recording, or held within any information storage and retrieval system, without permission in writing from the publisher or under licence from the Copyright Licensing Agency Limited Further details of such licences (for reprographic reproduction) may be obtained from the Copyright Licensing Agency Limited, Saffron House, 6–10 Kirby Street, London EC1N 8TS Cover photo © Tim Gainey/Alamy Stock Photo Typeset in Bembo Std, 11/13 pts by Aptara®, Inc Printed in Italy A catalogue record for this title is available from the British Library Contents Getting the most from this book Prior knowledge Problem solving 1.1 Solving problems 1.2 Writing mathematics 1.3 Proof Problem solving: Mountain modelling Surds and indices 2.1 2.2 Using and manipulating surds Working with indices Quadratic functions 3.1 3.2 3.3 Quadratic graphs and equations The completed square form The quadratic formula Equations and inequalities 4.1 4.2 Simultaneous equations Inequalities Coordinate geometry 5.1 5.2 5.3 5.4 5.5 Working with coordinates The equation of a straight line The intersection of two lines The circle The intersection of a line and a curve Problem solving: Integer point circles v Polynomials vii 12 16 19 20 24 32 33 42 47 53 54 59 65 66 71 78 80 87 92 Practice questions: Pure mathematics 96 Trigonometry 99 6.1 6.2 6.3 6.4 6.5 Trigonometric functions Trigonometric functions for angles of any size Solving equations using graphs of trigonometric functions Triangles without right angles The area of a triangle 100 104 112 118 126 7.1 7.2 7.3 Polynomial expressions Dividing polynomials Polynomial equations Graphs and transformations 8.1 8.2 8.3 8.4 The shapes of curves Using transformations to sketch curves Using transformations Transformations and graphs of trigonometric functions The binomial expansion 9.1 9.2 130 131 140 142 148 149 155 164 167 172 Binomial expansions Selections 173 180 Practice questions: Pure mathematics 186 10 Differentiation 190 10.1 The gradient of the tangent as a limit 10.2 Differentiation using standard results 10.3 Tangents and normals 10.4 Increasing and decreasing functions, and turning points 10.5 Sketching the graphs of gradient functions 10.6 Extending the rule 10.7 Higher order derivatives 10.8 Practical problems 10.9 Finding the gradient from first principles Problem solving: Proofs 11 Integration 11.1 Integration as the reverse of differentiation 11.2 Finding areas 11.3 Areas below the x axis 11.4 Further integration 191 194 198 201 206 210 213 218 221 226 229 230 234 238 241 iii 12 Vectors 247 19 Kinematics Running head verso Contents 12.1 Vectors 12.2 Working with vectors 12.3 Vector geometry 247 253 259 13 Exponentials and logarithms 13.1 13.2 13.3 13.4 13.5 264 Exponential functions Logarithms The exponential function The natural logarithm function Modelling curves 265 268 273 278 280 Practice questions: Pure mathematics 288 14 Data collection 291 14.1 Using statistics to solve problems 292 14.2 Sampling 297 15 Data processing, presentation and interpretation 306 15.1 15.2 15.3 15.4 15.5 15.6 Presenting different types of data Ranked data Discrete numerical data Continuous numerical data Bivariate data Standard deviation 16 Probability 16.1 Working with probability Problem solving: Alphabet puzzle Problem solving: Estimating minnows 17 The binomial distribution 17.1 Introduction to binomial distribution 17.2 Using the binomial distribution 308 312 317 324 335 342 350 351 368 370 372 373 377 18 Statistical hypothesis testing using the binomial distribution 383 18.1 The principles and language of hypothesis testing 18.2 Extending the language of hypothesis testing Practice questions: Statistics iv 385 391 399 19.1 19.2 19.3 19.4 The language of motion Speed and velocity Acceleration Using areas to find distances and displacement 19.5 The constant acceleration formulae 19.6 Further examples 20 Forces and Newton’s laws of motion 20.1 20.2 20.3 20.4 20.5 Force diagrams Force and motion Types of forces Pulleys Applying Newton’s second law along a line 20.6 Newton’s second law applied to connected objects Problem solving: Reviewing models for air resistance 21 Variable acceleration 21.1 Using differentiation 21.2 Finding displacement from velocity 21.3 The constant acceleration formulae revisited Problem solving: Human acceleration 403 403 406 411 415 421 426 434 434 440 442 447 450 457 468 472 473 475 479 484 Practice questions: Mechanics 487 Dataset 490 Answers* 492 Index 561 *Please note that the marks stated on the example questions are to be used as a guideline only, AQA have not reviewed and approved the marks Getting the most from this book Mathematics is not only a beautiful and exciting subject in its own right but also one that underpins many other branches of learning It is consequently fundamental to our national wellbeing This book covers the content of AS Mathematics and so provides a complete course for the first of the two years of Advanced Level study The requirements of the second year are met in a second book Between 2014 and 2016 A-level Mathematics and Further Mathematics were very substantially revised, for first teaching in 2017 Major changes include increased emphasis on Q Problem solving Q Proof Q Use of ICT Q Modelling Q Working with large data sets in statistics This book embraces these ideas Chapter is on problem solving and this theme is continued throughout the book with several spreads based on the problem solving cycle In addition a large number of exercise questions involve elements of problem solving; these are identified by the PS icon beside them The ideas of mathematical proof and rigorous logical argument are also introduced in Chapter and are then involved in suitable exercise questions throughout the book The same is true of modelling; the modelling cycle is introduced in the first chapter and the ideas are reinforced through the rest of the book Questions which involve an element of modelling are identified by the M icon The use of technology, including graphing software, spreadsheets and high specification calculators, is encouraged wherever possible, for example in the Activities used to introduce some of the topics in Pure mathematics, and particularly in the analysis and processing of large data sets in Statistics A large data set is provided at the end of the book but this is essentially only for reference It is also available online as a spreadsheet (www.hoddereducation.co.uk/AQAMathsYear1) and it is in this form that readers are expected to store and work on this data set, including answering the exercise questions that are based on it Places where ICT can be used are highlighted by a T icon Throughout the book the emphasis is on understanding and interpretation rather than mere routine calculations, but the various exercises nonetheless provide plenty of scope for practising basic techniques The exercise questions are split into three bands Band questions (indicated by a green bar) are designed to reinforce basic understanding, while most exercises precede these with one or two questions designed to help students bridge the gap between GCSE and AS Mathematics; these questions are signposted by a icon These include a 'thinking' question which addresses a key stumbling block in the topic and a multiple choice question to test key misconceptions Band questions (yellow bar) are broadly typical of what might be expected in an examination: some of them cover routine techniques; others are design to provide some stretch and challenge for readers Band questions (red bar) explore round the topic and some of them are rather more demanding In addition, extensive online support, including further questions, is available by subscription to MEI’s Integral website, http://integralmaths.org (Please note that these external links are not being entered in an AQA approval process.) In addition to the exercise questions, there are five sets of questions, called Practice questions, covering groups of chapters All of these sets include identified questions requiring problem solving PS , mathematical proof MP , use of ICT T and modelling M The book is written on the assumption that readers have been successful in GCSE Mathematics, or its equivalent, and are reasonably confident and competent with that level of mathematics.There are places where the work depends on knowledge from earlier in the book and this is flagged up in the margin in Prior knowledge boxes This should be seen as an invitation to those who have problems with the particular topic to revisit it earlier in book At the end of each chapter there is a summary of the new knowledge that readers should have gained v Running head Getting the most from thisverso book Two common features of the book are Activities and Discussion points These serve rather different purposes The Activities are designed to help readers get into the thought processes of the new work that they are about to meet; having done an Activity, what follows will seem much easier The Discussion points invite readers to talk about particular points with their fellow students and their teacher and so enhance their understanding Another feature is a Caution icon , highlighting points where it is easy to go wrong The authors have taken considerable care to ensure that the mathematical vocabulary and notation are used correctly in this book, including those for variance and standard deviation, as defined in the AQA specification for AS Level in Mathematics In the paragraph on notation for sample variance and sample standard deviation (page 344), it explains that the meanings of ‘sample variance’, denoted by s2, and ‘sample standard deviation’, denoted by s, are defined to be calculated with divisor (n – 1) In early work in statistics it is common practice to introduce these concepts with divisor n rather than (n – 1) However there is no recognised notation to denote the quantities so derived Students should be aware of the variations in notation used by manufacturers on calculators and know what the symbols on their particular models represent Answers to all exercise questions and practice questions are provided at the back of the book, and also online at www.hoddereducation.co.uk/AQAMathsYear1 Full step-by-step worked solutions to all of the practice questions are available online at www.hoddereducation.co.uk/AQAMathsYear1 All answers are also available on Hodder Education’s Dynamic Learning platform (Please note that these additional links have not been entered into the AQA approval process.) Finally a word of caution This book covers the content of AS Level Mathematics and is designed to help provide readers with the skills and knowledge for the examination However, it is not the same as the specification, which is where the detailed examination requirements are set out So, for example, the book uses a data set about cycling accidents to give readers experience of working with a large data set Examination questions will test similar ideas but they will be based on different data sets; for more information about these sets readers should consult the specification Similarly, in the book cumulative binomial tables are used in the explanation of the output from a calculator, but such tables will not be available in examinations Individual specifications will also make it clear how standard deviation is expected to be calculated So, when preparing for the examination, it is essential to check the specification Catherine Berry Roger Porkess vi Prior knowledge This book builds on GCSE work, much of which is assumed knowledge The order of the chapters has been designed to allow later ones to use and build on work in earlier chapters The list below identifies cases where the dependency is particularly strong The Statistics and Mechanics chapters are placed in separate sections of the book for easy reference, but it is expected that these will be studied alongside the Pure mathematics work rather than after it Q The work in Chapter 1: Problem solving pervades the whole book Q Chapter 3: Quadratic equations and graphs requires some manipulation of surds (chapter 2) Q Chapter 4: Equations and inequalities uses work on solving quadratic equations (chapter 3) Q Chapter 5: Coordinate geometry requires the use of quadratic equations (chapter 3) and simultaneous equations (chapter 4) Q Chapter 6: Trigonometry requires some use of surds (chapter 2) and quadratic equations (chapter 3) Q Chapter 7: Polynomials builds on the work on quadratic equations (chapter 3) Q Chapter 8: Graphs and transformations brings together work on quadratic graphs (chapter 3), trigonometric graphs (chapter 6) and polynomial graphs (chapter 7) Q Chapter 9: The binomial expansion builds on polynomials (chapter 7) Q Chapter 10: Differentiation draws on a number of techniques, including work on indices (chapter 2), quadratic equations (chapter 3), coordinate geometry (chapter 5) and polynomial graphs (chapter 8) Q Chapter 11: Integration follows on from differentiation (chapter 10) Q Chapter 12: Vectors builds on coordinate geometry (chapter 5) Q Chapter 13: Logarithms and exponentials builds on work on indices (chapter 2) Q Chapter 15: Data processing, presentation and interpretation follows on from data collection (chapter 14) Q Chapter 17: The binomial distribution draws on ideas from probability (chapter 16) and the binomial expansion (chapter 9) Q Chapter 18: Hypothesis testing uses ideas from probability (chapter 16) and the binomial distribution (chapter 17) Q Chapter 19: Kinematics requires fluency with quadratic equations (chapter 3) and simultaneous equations (chapter 4) Q Chapter 20: Forces ties in with work on vectors (chapter 12), although these two chapters could be covered in either order Q Chapter 21: Variable acceleration uses differentiation (chapter 10) and integration (chapter 11) vii Acknowledgements The Publishers would like to thank the following for permission to reproduce copyright material Questions from past AS and A Level Mathematics papers are reproduced by permission of MEI and OCR Question on page 322 is taken from OCR, Core Mathematics Specimen Paper H867/02, 2015 The answer on page 541 is also reproduced by permission of OCR Practice questions have been provided by Chris Little (p288–290), Neil Sheldon (p399–402), Rose Jewell (p487–489), and MEI (p96–98, p186–189) p.309 The smoking epidemic-counting the cost, HEA, 1991: Health Education Authority, reproduced under the NICE Open Content Licence: www.nice.org.uk/Media/Default/About/Reusing-ourcontent/Open-content-licence/NICE-UK-Open-Content-Licence-.pdf; p.310 Young People Not in Education, Employment or Training (NEET): February 2016, reproduced under the Open Government Licence www.nationalarchives.gov.uk/doc/open-government-licence/version/3/; p.334 The World Bank: Mobile cellular subscriptions (per 100 people): http://data.worldbank.org/indicator/IT.CEL.SETS.P2; p.340 Historical monthly data for meteorological stations: https://data.gov.uk/dataset/historic-monthlymeteorological-station-data, reproduced under the Open Government Licence www.nationalarchives gov.uk/doc/open-government-licence/version/3/; p.341 Table 15.26 (no.s of homicides in England & Wales at the start and end of C20th): https://www.gov.uk/government/statistics/historical-crime-data, reproduced under the Open Government Licence www.nationalarchives.gov.uk/doc/open-governmentlicence/version/3/; p.362 Environment Agency: Risk of flooding from rivers and the sea, https://floodwarning-information.service.gov.uk/long-term-flood-risk/map?map=RiversOrSea, reproduced under the Open Government Licence www.nationalarchives.gov.uk/doc/open-government-licence/version/3/ Photo credits p.1 © Kittipong Faengsrikum/Demotix/Press association Images; p.19 © Randy Duchaine/Alamy Stock Photo; p.32 © Gaby Kooijman/123RF.com; p.53 © StockbrokerXtra/Alamy Stock Photo; p.65 © polifoto/123RF.com; p.99 (top) © Sakarin Sawasdinaka/123RF.com; p.99 (lower) © Rico Koedder/123RF.com; p.130 © Edward R Pressman Film/The Kobal Collection; p.148 © ianwool/123RF com; p.174 © ullsteinbild/TopFoto; p.180 © Emma Lee/Alamy Stock Photo; p.190 © Bastos/Fotolia; p.229 © NASA; p.247 © Graham Moore/123RF.com; p.264 © BioPhoto Associates/Science Photo Library; p.291 (top) © Jack Sullivan/Alamy Stock Photo; p.291 (lower) © Ludmila Smite/Fotolia; p.297 © arekmalang/Fotolia; p.350 (top) © molekuul/123.com; p.350 (lower) © Wavebreak Media Ltd/123RF com; p.372 (top) © Tom Grundy/123RF.com; p.372 (lower) © George Dolgikh/Fotolia; p.383 © Fotoatelie/Shutterstock; p.403 ©Tan Kian Khoon/Fotolia; p.406 © Volodymyr Vytiahlovskyi/123RF.com; p.434 © Stocktrek Images, Inc./Alamy Stock Photo; p.437 © Dr Jeremy Burgess/Science Photo Library; p.438 © Herbert Kratky/123RF.com; p.440 © Matthew Ashmore/Stockimo/Alamy Stock Photo; p.445 © NASA; p.450 © V Kilian/Mauritius/Superstock; p.452 © scanrail/123RF.com; p.472 © FABRICE COFFRINI/AFP/GettyImages; p.484 © Peter Bernik/123RF.com Every effort has been made to trace all copyright holders, but if any have been inadvertently overlooked, the Publishers will be pleased to make the necessary arrangements at the first opportunity viii Number of (mm) Mid Fref× away from m point x quency f midpoint −4 −24 96 −3 ⩽ (l) < −1 −2 12 −24 48 −1 ⩽ (l) < 58 0 ⩽ (l) < 17 34 68 ⩽ (l) < 28 112 100 (i) 14 − x 0.14 (ii) σ 1.8 324 (iii) representative value for the data set in this case In this case the standard deviation and semi-IQR are both relatively large suggesting a varied set of data Throwing a single, normal, die Discussion point (page 363) You would work out the probability that it is not flooded in years and then subtract this from ∑ x = 33819.03 (iii) n = 202, So for years ∑ x = 3196.285, P(the street is flooded in at least one year) = − P(it is not flooded in years) = − 29 ≈ 30 6.4 The risk is about once in every or years Mean = 15.823 g, standard deviation = 3.978 Estimated numbers of patients are as follows: within sd of the mean 2170, so 68.8% - very close to 68% within sds of the mean 2990, so 94.9% - extremely close to 95% within sd of the mean 3126 so 99.2% - very close to 99.75% (These values are rounded and answers may vary according to how the data is cleaned.) The mean is 30 with a standard deviation of 20 (rounded); the median is 22 with a semi-IQR of 16 The mean takes extreme values into account, so the very young and very old can make a big difference to the mean – this is demonstrated by the large standard deviation The median with the smaller semi-IQR may give a better (iv) Discussion point (page 361) (ii) ∑ x = 53748.02 Chapter 16 Answers −5 ⩽ (l) < −3 The estimates in (i) and (ii) assumed that all of the values in a group were the same value, the midpoint The sample mean and standard deviation from the raw data was calculated using the actual measurements Both students have used the same data so the variation in plant height within the sample is the same The different figures for the standard deviation are as a result of the different units of measurement chosen by the two students In Alisa’s case the standard deviation is 7.46 mm whilst Bjorn’s standard deviation is 0.746 cm; these are equivalent distances (ii) Mean 58, standard deviation 14.9 (iii) (a) 58 + × 14.9 = 87.8 < 96 (b) Outlier (i) The box shows that the central part of the distribution is very nearly symmetrical about the median.The whiskers show that the distribution is skewed to the right (ii) 50–60 16 × 10 = 160 60–70 51 × 10 = 510 70–80 80 × 10 = 800 Total 1470 81.7 ± × 15.7 so 50.3 to 113.1 (iv) 3152 − (16 × 9.7 + 51 × 10 + 80 × 10 + 81 × 10 + 48 × 10 + 21 × 10 + 3.1 × 8) = 162 (v) There are more very heavy people.The boxplot shows the distribution is skewed and this can also be seen on the histogram with more very high values than very low ones (i) Total weight is 1963.951 g (iii) fx2 ( ) Exercise 16.1 (page 363) P(F ∩ G) = 0.31 and P(G ∩ F´) = 0.27 0.88 66 534 , assuming each faulty torch has only one fault A TRUE – 90 = 120 UNCERTAIN – not enough information but it seems unlikely C TRUE – the expected number is 80 = 2 30 fledglings D UNCERTAIN – not enough information.You don’t know how many of the existing adult females and the female fledglings will have died by next year E TRUE – of 80 is 40 B and 40 = 120 545 (i) (ii) Answers (iii) (iv) (i) (ii) (iii) (iv) (v) (vi) (i) (ii) (iii) (i) (ii) (iii) 13 36 = 52 13 13 16 = 52 13 12 100 = 25 53 100 45 100 = 20 42 21 100 = 50 56 14 100 = 25 100 = 20 0.4 0.5 2, 2.2 0.2 0.6 0.26 The readings are independent Approximately 60 = 0.32 Approximately 19 19 = 0.1 × (iv) Consecutive journey times are independent of each other However, the same roadworks could cause delays to both journeys, so the times would not be independent 2000 1995 2000 Lose £100, if all tickets are sold 25p 2500 0.35 They might draw 0.45 0.45 k = 0.4 (i) (ii) (iii) 11 12 r (iv) (v) (i) (ii) (iii) (iv) (i) (i) Exercise 17.1 (page 376) 0.4 First die + 6 7 8 9 10 10 11 10 11 12 (ii) (iii) (iv) P(X = r) Formula P(X = 2) ( 52 )( 13 ) ( 23 ) P(X = 3) = 12 36 The different outcomes are not all equally probable P(X = 4) 0.2 Left-handed 0.008 0.8 Not left-handed 0.2 Left-handed 0.992 0.8 Not left-handed Not colour-blind 0.0016 0.0064 0.2064 0.7936 in 2.6 (i) 0.020 (ii) 0.65 (iii) 0.99 15 (i) (a) 365 15 16 17 20 (b) 365 (c) (ii) 330 365 55 730 000 = 0.000 075 Heat presents the greater risk (i) k = 0.08 (iii) 18 r P(X = r) 0.2 0.24 0.32 0.24 (ii) r 10 11 Let Y represent the number of chicks P 0.351 04 0.449 28 0.184 32 0.015 36 (Y = r) 3 4 3 10 5 ()() 15 64 (i) (ii) (iii) (iv) q ⎛ 20⎞ 13 ⎜ ⎟⎠ 5 ⎝ p 2 Colour-blind n ( 46 )( 41 ) ( 43 ) ( 73 )( 23 ) ( 13 ) (104 )( 51 ) ( 45 ) P(X = 4) 14 60 P(X = r) 0.1 0.2 0.3 546 13 Chapter 17 19 (ii) (iii) 60 10 0.3 (b) 0.35 (ii) (a) Second die 0.271 0.214 0.294 (i) 0.146 (ii) Poor visibility might depend on the time of day, or might vary with the time of year If so, this simple binomial model would not be applicable (i) (ii) (iii) (iv) (i) 0.246 (ii) Exactly heads (i) (a) 0.058 (b) 0.198 (c) 0.296 (d) 0.448 (ii) (i) (a) 0.264 (b) 0.368 (c) 0.239 (d) 0.129 12 Exercise 17.2 (page 380) The score you expect to get is the one with the highest probability and is a value of the data The expectation of the score is a statistic and may not take a value that is a possible data item (i) (a) 0.000129 (b) 0.0322 (c) 0.402 (ii) and are equally likely (i) (ii) 0.388 (iii) 0.323 (i) (a) 0.240 (b) 0.412 (c) 0.265 (d) 0.512 (e) 0.384 (f) 0.096 (g) 0.317 (ii) Assumption: the men and women in the office are randomly chosen from the population (as far as their weights are concerned) (i) (a) 81 (b) 81 24 (c) 81 32 (d) 81 (ii) 40 s (i) He must be an even number of steps (ii) (iii) (iv) (v) (i) (ii) (iii) (i) (ii) (iii) from the shop (The numbers of steps he goes east or west are either both even or both odd, since their sum is 12, and in both cases the difference between them, which gives his distance from the shop, is even.) 0.00293 At the shop 0.242 0.0735 2.7 0.267 (a) 0.349 (b) 0.387 (c) 0.194 0.070 0.678 Chapter 18 Discussion point (page 384) Assuming both types of parents have the same fertility, boys born would outnumber girls in the ratio : In a generation’s time there would be a marked shortage of women of childbearing age Discussion point (page 388) It does seem a bit harsh when the number came up more often than the others.You cannot actually prove a result with statistics but even trying to show it beyond reasonable doubt, in this case using a 5% significance level, can be difficult This is particularly so if the sample size is small; it would have been better is the die had been thrown a lot more than 20 times Exercise 18.1 (page 390) The p-value is calculated in a hypothesis test and compared with the significance level.The value of p is the default probability that is used to the calculation of the p-value H0 : p = H1 : p > (i) Null hypothesis: p = 0.25; alternative hypothesis: p > 0.25 (ii) 0.0139 (iii) 5% (iv) Yes 0.1275 Accept H0 0.0547 > 5% Accept H0 H0: probability that toast lands butter-side down = 0.5 H1: probability that toast lands butter-side down > 0.5 0.240 Accept H0 0.048 Reject H0 There is evidence that the complaints are justified at the 5% significance level, though Mr McTaggart might object that the candidates were not randomly chosen 0.104 Accept H0 Insufficient evidence at the 5% significance level that the machine needs servicing (i) 0.590 (ii) 0.044 (iii) 0.0000712 (iv) 0.0292 (v) H0: P(long question right) = 0.5; H1: P(long question right) > 0.5 (vi) No Answers Assumed the probability of being born in January 31 This ignores the = 365 possibility of leap years and seasonal variations in the pattern of births throughout the year The three possible outcomes are not equally likely: ‘one head and one tail’ can arise in two ways (HT or TH) and is therefore twice as probable as ‘two heads’ or ‘two tails’ (ii) Discussion point (page 393) X⩽4 Exercise 18.2 (page 396) A critical value is a single value at one end of the distribution that marks the start of the critical region 547 Answers 548 at that end The critical region may consist of several values For a 2-tail test there is a critical value and a critical region at both ends of the distribution, even if one is empty H0 : p = H1 : p ≠ (i) Henry is only interested in whether tails is more likely so he should carry out a 1-tail test Mandy is interested in knowing about any bias so should carry out a two-tail test (ii) Henry: for 29 tails p = 0.8987 and for 30 tails p = 0.9405 so 30 is the critical value Mandy: for 18 tails p = 0.0325 and for 19 tails p = 0.0595 For 30 tails p = 0.9405 and for 31 tails p = 0.9675 so her critical values are 18 and 31 (ii) Henry will accept any number of tails up to and including 29; Mandy will accept any number of tails between and including 19 and 30 (iii) Henry will say the coin is biased if there are 30 or more tails (iv) Mandy will say the coin is biased if there are 18 or fewer tails or if there are 31 or more tails (v) Henry has only looked at whether tails is more likely; he has no way of knowing if heads is more likely Mandy will be able to say whether either heads or tails is more likely and therefore whether the coin is biased 10 11 12 13 P(X ⩽ 7) = 0.1316 > 5% Accept H0 P(X ⩾ 13) = 0.0106 < 2% Reject H0 P(X ⩾ 9) = 0.0730 > 2 % Accept H0 P(X ⩾ 10) = 0.0139 < 5% Reject H0, but data not independent P(X ⩾ 6) = 0.1018 > 5% Accept H0 (i) 0.0395 < 5% Reject H0 (ii) 0.0395 > 2 % Accept H0 ⩽1 or > males ⩽1 or > correct Critical region is ⩽3 or ⩾13 letter Zs (i) 20 (ii) 0.0623 (iii) Complaint justified Practice questions (page 399) 27 0.753 = 64 (or 0.421 875) [2] (ii) 0.753 + 0.23 + 0.053 = 0.43 [2] (i) Pie charts (and variants such as this chart) are used to represent proportions of a total In this case there is no such total as not all goals scored are shown [1] This type of chart makes comparisons difficult, particularly when the quantities are similar to one another [1] (ii) A bar chart would make for easier visual comparison [1] In this case there is little to be gained from having any sort of chart The list of goal scorers and their numbers of goals is sufficient [1] NB: other sensible answers are possible (i) Simple random sampling is sampling in which all possible samples of the required size have an equal chance of selection [1] In this case simple random sampling would not be possible because there is no sampling frame That is, there is no list of all customers from which a sample could be constructed [1] (ii) In opportunity sampling the interviewer(s) would interview any convenient customers until a total of 200 had been reached [1] In quota sampling the interviewer(s) would be given target numbers of interviewees in different groups, e.g 100 male, 100 female; equal numbers under and over 40 years of age [1] Quota sampling is preferable [1] Because it attempts to make the sample representative of the target population [1] (i) Eruptions typically last between 1.5 and 5.5 minutes, with intervals between eruptions typically being between 40 and 95 minutes [1] There appear to be two different types of eruption, short and long Similarly there are short and long intervals between eruptions [1] Short eruptions are typically followed by short intervals, long eruptions by long intervals [1] (i) magnitude; it looks bigger because of the scale.) [2] (iv) There would be some merit in fitting a straight line model to the data set [1] There is a slight downward trend [1] Interpolation or extrapolation would be meaningless as the data varies considerably from year to year [1] (i) Mean 93.1, standard deviation 13.26 [3] (ii) Mean 92.1, standard deviation 8.11 [3] (iii) The households in the South East eat less sugar than households in the North West [1] The weekly sugar consumption of households in the North West varies more than for households in the South East [1] (iv) The difference in the mean mass of sugar eaten per person per week supports this view However, the difference in the standard deviation suggests that there are other factors involved besides location [2] NB: other sensible answers are possible (i) (1, 2), (2, 1), (2, 2) [1] 36 equally likely outcomes so the probability is = 12 [1] 36 36 k= [1] 11 P(X = 6) = 11k = 36 [1] (iii) P(X = Y ) = k2 + (3k)2 + (5k)2 + (7k)2 + (9k)2 + (11k)2 = 286k2 [1] (ii) (i) (ii) (iii) (iv) P(X ≠ Y ) = − 286k2 (= 0.77932) [1] A and B are equally likely to win [1] So P(B beats A) = 0.390 to d.p [1] Expected number is 0.15 × 25 = 3.75 cars [1] Use X ~ B(25, 0.15) to find P(X ø 3) [1] P(X ø 3) = 0.471 (to d.p.) [1] H0: p = 0.15, H1: p < 0.15, where p is the proportion of cars with unsafe tyres [2] Use X ~ B(50, 0.15) to find P(X ø 5) = 0.219 [1] Observe that 0.219 > 0.05 so the observed result is not in the 5% critical region, hence insufficient evidence that p has reduced [1] Use X ~ B(100, 0.15) to find P(X ø 7) = 0.012 > 0.01 and P(X ø 6) = 0.005 < 0.01 [2] Hence the possible values of k are 0, 1…,6 [1] Answers The data points separate out into two fairly distinct clusters Within each cluster there is only a slight tendency for longer eruptions to be followed by longer times to the next eruption [1] (iii) The histogram of Length of eruption would be bimodal [1] The histogram of Time to next eruption would show little or no bimodality [1] (iv) The mean, 3.5 minutes, represents a length of eruption that is very unlikely to occur So in that sense it could be misleading [1] However, it could be useful in other ways For example, it is a figure that would be needed to estimate the proportion of the total time for which the geyser was erupting [1] NB: other sensible answers are possible (i) The description of the year involves two years [1] Some data may be included in both of 2005–06 and 2006 [1] (ii) The data varies with no particular pattern over the years [1] Removing the data point for 2008 gives a slight downward trend [1] The variation in weekly cheese consumption is only about 10 g overall [1] (iii) It is unlikely that this data point represents a real change in consumption; almost certainly this is just a random fluctuation in the data (The actual change is of very small (ii) Chapter 19 Discussion point (page 404) −4, 0, −5 (i) +4 (ii) −5 Discussion point (page 405) The marble is below the origin Exercise 19.1 (page 405) Position is where you are relative to a fixed origin, displacement is where you are relative to where you 549 started from Both can be described by vectors 0m (i) +1 m (ii) +2.25 m (i) 3.5 m, m, 6.9 m, m, 3.5 m, m (ii) m, 2.5 m, 3.4 m, 2.5 m, m, −3.5 m (iii) (a) 3.4 m (b) 10.3 m (i) m, m, −0.25 m, m, m, m, 12 m Answers (ii) t = 1, t = 0, t = 1.5 –2 t=4 t=5 12 x (m) 10 (iii) m, −2 m, −2.25 m, −2 m, m, m, 10 m (iv) 14.5 m (i) m, −16 m, −20 m, m, 56 m (ii) 56 –20 –16 and returns to pass the starting point at B, continuing past to C, where it changes direction again, returning to its initial position at D (ii) An oscillating ride such as a swing boat Speed (km h−1) 24 Constant speed, infinite acceleration (i) (a) m, m (b) m (c) m (d) m s−1, m s−1 (e) m s−1 (f) m s−1 (ii) (a) 60 km, km (b) −60 km (c) 60 km (d) −90 km h−1, 90 km h−1 (e) −90 km h−1 (f) 90 km h−1 (iii) (a) m, −10 m (b) −10 m (c) 50 m (d) OA: 10 m s−1, 10 m s−1; AB: m s−1, m s−1; BC: −15 m s−1, 15 m s−1 (e) −1.67 m s−1 (f) 8.33 m s−1 (iv) (a) km, 25 km (b) 25 km (c) 65 km (d) AB: −10 km h−1, 10 km h−1; BC: 11.25 km h−1, 11.25 km h−1 (e) 4.167 km h−1 (f) 10.83 km h−1 10.44 m s−1, 37.58 km h−1 20.59 km h−1 1238.71 km h−1 40 km h−1 (i) 32 km h−1; (ii) 35.7 km h−1 (i) (a) 56.25 km h−1 (b) 97.02 km h−1 (c) 46.15 km h−1 (ii) The ratio of distances must be in the ratio 10 :3 (i) 4.48 km h−1, 36.73 km h−1, 18.32 km h−1; 26.15 km h−1 (ii) B finishes first 16 Discussion point (page 412) 10, 0, −10 The gradient represents the velocity Discussion point (page 408, top) The graph would curve where the gradient changes Little effect over this period Discussion point (page 408, bottom) +5m s−1, m s−1, −5m s−1, −6m s−1 The velocity decreases at a steady rate Exercise 19.2 (page 409) (iii) 40 Position (m) 20 Time (s) –20 –40 (iv) t = moving backwards, t = moving forwards height above river (m) height above ground (m) (ii) 2000 positive direction 100 time positive direction (i) time 0 300 400 Time (s) (i) The person is waiting at the bus stop (ii) It is faster time The ride starts at t = At A it changes direction 550 200 12 (iii) height above bridge (m) (iii) 10 11 40 A speed–time graph records the magnitude whereas a velocity–time graph shows the direction as well A speed–time graph is always on or above the time axis whereas a velocity–time graph may go below it Graph A Speed (m s−1) 10 (i) Discussion point (page 407) (iv) 9.15 9.30 9.45 Time (i) (ii) (iii) D B, C, E A Exercise 19.3 (page 413) Approximately 460 m 10 15 20 25 30 35 Time (s) 20 m s−1, 40 m s−1, 10 m s−1 (i) Exercise 19.4 (page 418) (ii) 40 30 Speed (m s−1) 10 15 20 20 Time (s) –1 10 –2 20 m s−1 7.5 s m s−2 m s−1 a = 3t t=5 Discussion point (page 418) No, as long as the lengths of the parallel sides are unchanged, the trapezium has the same area –0.4 (i) (ii) (i) (ii) (iii) (iv) (i) Discussion point (page 417) 0.4 (ii) Acceleration (m s−2) It represents the displacement 20 (iii) (i) 40 60 Time (s) 1550 m Displacement (m) 150 100 50 Velocity (m s−1) 2 Time (s) (ii) (a) v (m s−1) 40 10 15 Time (s) 20 (ii) Velocity (m s−1) Time (s) (b) a (m s−2) The area under a speed–time graph represents distance travelled whereas the area under a velocity–time graph represents the displacement from the starting point The acceleration is 0.8 m s–2 (i) (A) 0.4 m s−2, m s−2, m s−2 (B) −1.375 m s−2, −0.5 m s−2, m s−2, m s−2 (ii) (A) 62.5 m (B) 108 m (iii) (A) 4.17 m s−1 (B) 3.6 m s−1 (i) Enters motorway at 10 m s−1, accelerates to 30 m s−1 and maintains this speed for about 150 s Slows down to a stop after a total of 400 s (ii) Approx 0.4 m s−2, −0.4 m s−2 (iii) Approx 9.6 km, 24 m s−1 (i) Speed (m s−1) 20 Answers Either an acceleration in the opposite direction, or a deceleration Graph B (i) (a) 0.8 m s−2 (b) −1.4 m s−2 (c) 0.67 m s−2 (d) m s−2 (e) 0.5 m s−2 Discussion point (page 416) (iv) Acceleration (m s−2) 20 10 15 20 25 30 35 Time (s) Time (s) 10 Discussion points (page 415) –2 (iii) +0.4 m s−2, m s−2, −0.4 m s−2, m s−2, −0.4 m s−2, m s−2, +0.4 m s−2 20 45 They are the same (i) (ii) (iii) (ii) 100 200 300 Time (s) 3562.5 m 551 10 (i) Answers Speed (m s−1) 30 16 35 seconds after A starts, 9450 m from the start (i) 11.125 s (ii) 10 s (i) (ii) Velocity (m s−1) 20 20 17 10 Discussion point (page 422) 10 0 (ii) (i) 20 40 60 Time (s) 11 558 m Speed (m s−1) 60 40 100 200 300 Time (s) (v) F 30 000 (ii) (iii) (iv) 100 150 Time (s) 60 s 6600 m v = 20 + 0.5t, ⩽ t ⩽ 60; v = 50, t ⩾ 60 Velocity (m s−1) 10 000 (ii) D B A 100 300 500 700 900 Time (s) 462.5 s 40 s (i) 10 m s−1, 0.7 s 200 400 600 Time (s) 5 Speed (m s−1) 40 Time (s) (iii) 6.25 m s−2 20 15 (ii) 20 40 Time (s) 13 m s−1 16 552 (iv) (i) (ii) (iii) (iv) (i) (u + v )t v2 = u2 + 2as (vi) s = ut + at 2 (vii) v2 = u2 + 2as (viii) s = vt − at 10 (i) s = (v) Velocity (m s−1) 15 15 m s−1, −1 m s−2, 8.66 km ) s = ut + at2, because it is the only one with those four letters s = (u + v)t (i) 22 m s−1 (ii) 120 m (iii) m (iv) −10 m s−2 (i) v2 = u2 + 2as (ii) v = u + at (iii) s = ut + at (iv) (ii) 10 E C 12 13 14 20 20 000 (i) ( + at ) × t Exercise 19.5 (page 425) 20 50 (2u s = u + 21 at × t 337 s BC: decelerates uniformly, s = ut + at 2 CD: stopped, DE: accelerates uniformly Discussion point (page 424) (ii) a = −0.5 m s−2, 2500 m −1 v = 26 m s −1 , −2 u = 13 m s , (iii) 0.2 m s , 6250 m t = 5, s = 100 m (iv) 325 s (ii) (i) Distance (m) s= 33.9 m 7.5 m 1.2 m s−1 10 s a = −0.4 m s−2 A: 17 30 seconds after starting; B: 16 27 seconds after starting 10 11 9.8 m s−1, 98 m s−1 4.9 m, 490 m s, speed and distance after 10 s, both over-estimates 2.08 m s−2, 150 m, assume constant acceleration 4.5 m s−2, m −8 m s−2, s a = −0.85 m s−2, 382 m 3.55 s (i) s = 16t − 4t2 v = 16 − 8t (ii) (a) s (b) s (i) (ii) (iii) (iii) Position (m) (v) (i) Underestimate 16 m sí 10 m sí 30 m hb hs R Answers 12 are sitting on, weight of a cup and the normal reaction from the table, a ladder leaning against a wall and the reaction of the wall are some examples Graph B 15 m sí Time (s) 13 hS = 15t − 4.9t2 hb = 30 − 4.9t2 t = 2s 10.4 m 5.4 m s−1 −4.4 m s−1 m s−1 gain m s−1 Too fast 2.94 m 43.75 m (ii) 15.15 = u + 5a (iii) 14.4 m s−1 (iv) No, distance at constant a is 166.5 m 22.7 m s−1 14 − 12 m s−2, 40 m Velocity (m sí) 16 12 –4 Time (s) –8 10 11 12 –12 –16 Speed (m sí) 16 12 15 16 Time (s) Discussion point (page 427) u = −15.0 No Exercise 19.6 (page 429) Acceleration = m s–2 and time = s 15 = aT 604.9 s, 9.04 km (i) v = + 0.4t (ii) s = 2t + 0.2t2 (iii) 18 m s-1 No, 10 m behind (i) h = + 2t − 4.9t2 (ii) 1.13 s (iii) 9.08 m s−1 (iv) t greater, v less (i) 11.75 m s−1 (ii) 8.29 m (iii) 12.75 m s−1 (iv) 5.31 m 17 (ii) (iii) (iv) (v) (i) (ii) (iii) (iv) (v) W T W R 2.5 m 13 m s−1, 91 m s−2, 9.5 m from the bus 107.5 m Chapter 20 W R P F W WBook Discussion point (page 437) RBook The reaction between the chair and the person acts on the chair The person’s weight acts on the person only Discussion point (page 438) RBook WBook RTable R1 R2 Vertically up Exercise 20.1 (page 439) In these diagrams,W represents a weight, R a normal reaction with another surface,T the tension, F a friction force, D drag and P another force Your weight and the normal reaction from the chair you W (i) R (ii) R P W W 553 10 (i) (ii) R Answers P P R F W 11 D D W D W W 12 (i) (ii) P P W W R R P W P W 13 W F2 R2 R1 F1 W (ii) (a) R = W1 (b) R1 + R2 = W2 + R (i) R = W, (ii) W > R, W − R, down (iii) R > W, R − W up (i) No (ii) Yes (iii) Yes (iv) No (v) Yes (vi) Yes (vii) Yes (viii) No Forces are required to give passengers the same acceleration as the car (i) A seat belt provides a backwards force (ii) The seat provides a forwards force on the body and the head rest is required to make the head move with the body Discussion point (page 443) 14 The first one, with the pencil in tension R R Exercise 20.3 (page 446) W Discussion point (page 440, left) To provide forces when the velocity changes Discussion point (page 440, right) The friction force was insufficient to enable his car to change direction at the bend Exercise 20.2 (page 442) False.A resultant force may decelerate an object moving in the opposite direction, such as a ball that is hit up into the air W W + mg (i) T A a a m s–2 70 g (ii) (iii) (i) a T2 T1 A B a T1 a C T2 2g 3g A: T1 − 2g = 2a B: T2 − T1 = 5a C: 3g − T2 = 3a (iii) a = m s−2, T1 = 22 N, T2 = 27 N (iv) N (ii) T (ii) (iii) (iv) (i) (ii) 11 T = 750 N Tension = 44.4 N 170 N 0.625 m s−2 25 000 N × 104 N 12 500 N Reduced to 10 000 N 0.25 m s−2 T1 = kN, T2 = kN 1000 N (v) (i) 13 m 10 12 Time (s) 40 000g N (iii) (iv) (i) (ii) 10 R T1 N Speed (m s−1) 12 (ii) (iii) (iv) (v) (i) (ii) 500 N T2 N T2 N 50 Time (s) 0.8 m s−2 40 s−1, –1.33 m s−2 66.4 kN 90 kg 2.26 m s−2, 60.3 N 0.56 m above ground 1000 N T1 N T N a msí2 556 (iv) 600 N m s-2 Stationary for s, accelerating at m s−2 for s, at constant speed for s, decelerating at m s−2 for s, stationary for s Speed (m s−1) T A: T − 0.98 = 0.1a B: 1.96 − T = 0.2a (iii) 3.27 m s−2, 1.31 N (iv) 1.11 s 2.8 m s−2, 14 N (ii) R (i) Mg Rp = RL = 490 N, T = 4900 N (iii) Rp = RL = 530 N, T = 5300 N B 0.2 g 0.25 m s−2, T1 = 1.5 kN in tension, T2 = −1.5 kN in thrust.The second locomotive is now pushing rather than pulling back on the truck (ii) a 0.1 g 10 mg Exercise 20.6 (page 462) (iii) 15 000 N 13 14 13 m s−2, 14.9 N 1.23 kg (ii) T = 5200 N (i) (ii) 17 (compression) (b) T = 10340 N (tension) (i) 0.1 m s−2 (ii) 27.5 kN (iv) 1.1 kN (S1 − Sn ) (n − 1) M (i) (ii) Time Velocity Exercise 21.2 (page 479) 15 0 1.5 Time –15 The acceleration is the gradient of the velocitytime graph (iv) The acceleration is constant; the velocity decreases at a constant rate (i) v = 18t2 − 36t − a = 36t − 36 (iii) Exercise 21.1 (page 474) 2 –15 0.81 s 0.42 s Velocity is the rate of change of displacement with respect to time and the gradient of a displacement–time graph Differentiating is finding the rate of change 6t2 + (i) (a) v = − 2t (b) 10, (c) 1, 11 (ii) (a) v = −4 + 2t (b) 0, −4 (c) 2, −4 (iii) (a) v = 3t2 − 10t (b) 4, (c) 0, and , −14.5 (i) (a) a = (b) 3, (ii) (a) a = 12t − (b) 1, −2 (iii) (a) a = (b) −5, v=4+t a=1 –10 Chapter 21 –5 (ii) Time –36 Velocity 50 30 10 –10 0.5 1.5 2.5 Time –30 The acceleration is the gradient of the velocitytime graph; the velocity is at a minimum when the acceleration is zero (iv) It starts in the negative direction, v is initially −6 and decreases to (iii) (iii) r = 73 t − 5t + (i) 20 15 10 5 36 dr a dt dt ∫ r = t – 5t + c (i) r = 2t2 + 3t (ii) r = t − t + t + Speed Acceleration −24 when t = 1, before increasing rapidly to 0, where the object turns round to move in the positive direction (i) −16 m s−1, m s−2 (ii) 25.41 m 12 m s−2 when t = 1, −12 m s−2 when t = Answers 16 a = −10 T (iii) –0.186 m s−2 (iv) (a) T = –1660 N 15 v = 15 − 10t Acceleration 10 000 kg 200 N (i) (ii) 10 Time 85 m When t = 972 m 4.47 s 119 m (i) v = 10t + 23 t − 13 t , (ii) (i) (ii) (i) (ii) x = 5t + 21 t − 12 t (ii) v = + 2t − t x = + 2t + 3t , − 61 t (iii) v = −12 + 10t − 3t , x = − 12t + 5t − t Discussion points (page 479) Case (i) s = ut + 21 at ; v = u + at; a = 4, u = In the other two cases, the acceleration is not constant 557 Activity 21.1 (page 479) Substituting in at = v − u ➁ gives s = ut + 21 (v − u )t + s0 s = (u + v )t + s0 Answers ➂ Substituting v − u = at and v + u = 2t ( s − s0 ) ⇒ (v − u )(v + u ) = at × ( s − s0 ) t 2 v − u = 2a( s − s0 ) ➃ Substituting u = v − at in ➁ gives s = (v − at )t + 21 at + s0 s = vt − 2 at + s0 This gives the displacement of the ball, which is (i) −3 m, −1 m s−1, m s−1 (ii) (a) s (b) 2.15 s (iii) (v) Position (iii) 14 10.5 3.5 0 –3.5 (i) (ii) (iii) (iv) (v) (i) (ii) (iii) t Velocity ➄ Exercise 21.3 (page 480) Displacement, area under a velocity–time graph, ∫ v dt ; a displacement–time graph, d r ; dt acceleration, gradient of a velocity–time graph, d v dt v = u + at (i) v = 15 − 10t (ii) 11.5 m, m s−1, m s−1; 11.5 m, −5 m s−1, m s−1 12 (iv) (i) (ii) Speed –3 0 3s t v (m sí) 9.6 10 t (s) 0 2.4 4.8 7.2 9.6 t (s) 11.52 m 11.62 s Elizabeth by 0.05 s and 0.5 m (vi) Andrew wins (iii) (iv) (v) 0.5 1.5 t 0.5 1.5 t Acceleration 558 Velocity (iv) 0.5 1.5 2.5 t 10 t (ii) v (m sí) 15 10 Object starts at O and moves towards A, slowing down to when t = 1, then accelerates back towards B reaching a speed of m s−1 when t = (v) m (i) Andrew: 10 m s−1, Elizabeth: 9.6 m s−1 Position –5 Speed 15 t v = + 4t − t x = 4t + 2t − 13 t The object starts at and moves in the negative direction from to −3, which is reached when t = 2.15 It then moves in the positive direction with increasing speed 15 –15 (iv) 10 t Velocity t –4 2 15 0.5 1.5 2.5 t a (iii) v Position 10 –2 velocity, gradient of The object starts at the origin and moves in the positive direction with increasing speed reaching a maximum speed of m s−1 after s 0, 10.5, 18, 22.5, 24 The ball reaches the hole at s v = 12 − 3t m s−1 a = −3 m s−2 v = 3t2 − 3, a = 6t t=1 t v 90 80 70 60 50 40 30 20 10 (i) 10 20 30 40 50 60 70 80 90 t 12 13 14 15 16 17 s(0) = s ( 23 ) = − 272 , distance = 27 m 18 m s−1, 0.5 m s−2 15 s (iii) 41 m Time at maximum velocity 68.5 = = 7.61 s [1] Total time = + 7.61 = 14.61 s (14.6 to s.f.) [1] (iii) Displacement = (i) (ii) Practice questions (page 487) s = 2800 m, u = 0, v = 70 m s−1, a = , t = ? s = 21 (u + v)t 2800 = 21 (0 + 70)t [1] 2800 t = 35 = 80 s [1] x = + 2.lt − 0.07t v = dx = 4.2t − 0.02lt [1] dt When t = 7, v = 4.2 × − 0.21 × 72 = 19.11 m s−1(19.1 m s−1 to s.f.) [1, 1] 2F1 − 3F2 + F3 = [1] 2(7i − 2j) − 3(9i − 3j)+ F3 = [1] −13i + 5j + F3 = [1] F3 = 13i − 5j (i) Diagram showing weight for both objects [1] Tension with arrows and labels [1] Normal reaction and friction in correct directions [1] (ii) T = F and T = 1.25g [1] F = 1.25g (= 12.3 N to s.f.) [1] (iii) Inextensible string [1] (iv) N2L for block: T − F = 5a [1] N2L for hanging mass: 2g − T = 2a [1] Solving simultaneous equations 2g − 1.25g = 7a [1] a = 1.05 m s−2 [1] (i) Displacement = area under graph [1] d = × × = 31.5 m [1] (ii) Distance to go = 100 − 31.5 = 68.5 m [1] ⎡ t3 t4 ⎤ 0.9 t − 0.l t d t = 0.9 − 0.1 ∫0 ⎢⎣ ⎥⎦ Answers 10 11 Cara is in free fall until t = 10 s, then the parachute opens and she slows down to terminal velocity of m s−1 (ii) 1092 m (iii) 8.5 m s−2, 1.6t − 32, m s−2, 16 m s−2 2s (i) 40 m (ii) r = when t = and 10 (iii) 25 − 5t (iv) 62.5 m (v) r = for t = and t = 10 for both models In Michelle’s model, the velocity starts at 25 m s−1 and decreases to −25 m s−1 at t = 10 The teacher’s model is better as the velocity starts and ends at (i) (a) 112 cm (b) 68 cm (ii) 4t, 16 (iii) 2t2, 16t − 32 (iv) 89 cm less (i) PQ: train speeds up with gradually decreasing acceleration QR: train travels at constant speed ST: train slows down with constant deceleration (ii) a = −0.000 025t + 0.05 (iii) 50 m s−1 (iv) m s−1 (v) 111 km (i) −24 + 18t − 3t2, 2, (ii) −2, (iii) 28 4 15 m, m s-1 (i) 12.15 m (ii) 13.85 s (iii) 26.33 s s(2) = 2, and 23 s, [1, 1] = 32.4 m [1] (iv) EITHER distance to cover = 100 − 32.4 = 67.6 m [1] 67.6 Time = 10.8 = 6.26 s [1] Total time = + 6.26 = 12.26 s (12.3 to s.f.) which is less than 14.61 s for Sunil, [1] so Mo will win [1] OR distance Mo covers by the time Sunil finishes [1] = 32.4 + 10.8 × (14.6 − 6) = 125 m which is beyond the finish, [1,1] so Mo will win [1] (v) Distance = 100 m [1] 100 = (12.26 + (12.26 − 7))v [1] 200 v = 17.52 = 11.4 m s−1 [1] v (m s–1) 12 10 6 (i) 10 15 t (s) Returns when s = u = 25, v =, a = −g, t = ? s = ut + 21 at2 s = 25t – 4.9t2 = t(25 – 4.9t) = t = 0, 5.1 s [1] [1] 559 ... www.hoddereducation.co.uk/AQAMathsYear1 Full step-by-step worked solutions to all of the practice questions are available online at www.hoddereducation.co.uk/AQAMathsYear1 All answers are also available on Hodder... subscription to enhance your use of this book To subscribe to Integral visit www.integralmaths.org AQA A- level Mathematics For A- level Year and AS Authors Approval message from AQA The core content... Pure mathematics 288 14 Data collection 2 91 14 .1 Using statistics to solve problems 292 14 .2 Sampling 297 15 Data processing, presentation and interpretation 306 15 .1 15.2 15 .3 15 .4 15 .5 15 .6