Grade 11 Physics Module 2: Motion (Kinematics) PDF

DEPARTMENT OF EDUCATION GRADE 11 PHYSICS MODULE 2 MOTION (KINEMATICS) PUBLISHED BY FLEXIBLE OPEN AND DISTANCE EDUCATION PRIVATE MAIL BAG, P.O. WAIGANI, NCD FOR DEPARTMENT OF EDUCATION PAPUA NEW GUINEA 20...

DEPARTMENT OF EDUCATION GRADE 11 PHYSICS MODULE 2 MOTION (KINEMATICS) PUBLISHED BY FLEXIBLE OPEN AND DISTANCE EDUCATION PRIVATE MAIL BAG, P.O. WAIGANI, NCD FOR DEPARTMENT OF EDUCATION PAPUA NEW GUINEA 2017 Writer Tommy Sariaman Content Editors Science Department Subject Review Committee Language Editor Dr. Mirzi. L. Betasolo Course Format Editor Elizabeth. W. Aimundi GR 11 PHYS M2 TITLE GRADE 11 PHYSICS MODULE 2 (MOTION) KINEMATICS IN THIS MODULE YOU WILL LEARN ABOUT: 11.2.1: CHARACTERICS OF MOTION 11.2.2: GRAPHS OF MOTION 11.2.3: LINEAR MOTION 11.2.4: TWO DIMENSIONAL MOTION 1  GR 11 PHYS M2 ACKNOWLEDGEMENT & ISBN Acknowledgement We acknowledge the contribution of all Lower and Upper Secondary teachers who in one way or another helped to develop this Course. Our profound gratitude goes to the former Principal of FODE, Mr. Demas Tongogo for leading FODE towards this great achievement. Special thanks are given to the staff of the Science Department of FODE who played active roles in coordinating writing workshops, outsourcing of module writing and editing processes involving selected teachers of Central Province and NCD. We also acknowledge the professional guidance and services provided through-out the processes of writing by the members of: Science Subject Review Committee-FODE Academic Advisory Committee-FODE Science Department- CDAD This book was developed with the invaluable support and co-funding of the GO-PNG and World Bank. DIANA TEIT AKIS Principal-FODE. Flexible Open and Distance Education Papua New Guinea Published in 2017 ©Copyright 2017, Department of Education Papua New Guinea All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means electronic, mechanical, photocopying, recording or any other form of reproduction by any process is allowed without the prior permission of the publisher. Printed by the Flexible, Open and Distance Education ISBN 978-9980-89-540-0 National Library Services of Papua New Guinea 2  GR 11 PHYS M2 CONTENTS TABLE OF CONTENTS Page Title………………………………………………………………………………….......................................... 1 Acknowledgement & ISBN….…………………………………………………………………………………... 2 Contents……………………………………………………………………………………….………………….…….. 3 Secretary’s Message…………………………………………………………………………………….….……… 4 MODULE 11.2: MOTION (KINEMATICS) 5 - 130 Course Introduction………………………………………………………………………………………...….….. 5 Learning Outcomes…………………………………………………………………………………………………. 6 Time Frame…………………………………………………………………………………………………………….. 7 11.2.1: Characteristics of Motion……………………………………………………..………….……. 8 - 37 Distance, Displacement, Speed, Velocity…………………………..…….………………. 9 - 22 Acceleration………………………………………………………………………………….………... 23 - 37 11.2.2: Graphs of Motion……………………………………………………………………...……..….. 38 - 69 Distance Time Graph……………………………………………………………………….……… 39 - 42 Velocity Time Graph……………………………………………………………………….………. 42 - 59 The Ticker-Tape Timer…………………………………………………………………….……… 6 60 - 69 11.2.3: Linear Motion…………………………………………………………………………………..……. 70 - 89 Equations of Linear Motion…………………………………………………………….………. 70 - 79 Free Fall Motion…………………………………………………………………………….……….. 80 - 89 11.2.4: Two Dimensional Motion………………………………………………….…….……………… 90 - 116 Projectile Motion…………………………………………………………………………………..…9 90 - 104 Circular Motion…………………………………………………………………………………..…… 1 105 - 112 Rotary Motion………………………….………………………………………………………….….. 112 - 117 Summary……………………………………….…………………………………………………………………..…… 118 - 119 Answers to Learning Activities…………………………………………………………………….………….. 120 - 136 References……………………………………………………………………………………………………..………. 137 Quantities, Symbols and Units………………………………………………………………………………… 138 3  GR 11 PHYS M2 ACKNOWLEDGEMENT & ISBN SECRETARY’S MESSAGE Achieving a better future by individual students, their families, communities or the nation as a whole, depends on the kind of curriculum and the way it is delivered. This course is part of the new Flexible, Open and Distance Education curriculum. The learning outcomes are student-centred and allows for them to be demonstrated and assessed. It maintains the rationale, goals, aims and principles of the National Curriculum and identifies the knowledge, skills, attitudes and values that students should achieve. This is a provision by Flexible, Open and Distance Education as an alternative pathway of formal education. The Course promotes Papua New Guinea values and beliefs which are found in our constitution, Government policies and reports. It is developed in line with the National Education Plan (2005 – 2014) and addresses an increase in the number of school leavers affected by lack of access into secondary and higher educational institutions. Flexible, Open and Distance Education is guided by the Department of Education’s Mission which is fivefold;  To develop and encourage an education system which satisfies the requirements of Papua New Guinea and its people  To establish, preserve, and improve standards of education throughout Papua New Guinea  To make the benefits of such education available as widely as possible to all of the people  To make education accessible to the physically, mentally and socially handicapped as well as to those who are educationally disadvantaged The College is enhanced to provide alternative and comparable path ways for students and adults to complete their education, through one system, two path ways and same learning outcomes. It is our vision that Papua New Guineans harness all appropriate and affordable technologies to pursue this program. I commend all those teachers, curriculum writers, university lecturers and many others who have contributed so much in developing this course. UKE KOMBRA, PhD Secretary for Education 4  GR 11 PHYSICS M2 INTRODUCTION MODULE 11.2 Motion (Kinematics) Introduction In this module, you are introduced to motion (kinematics) without considering what caused it. There are fundamental principles governing motion, and they can be used to explain the motion of an object. Motion of objects and people can be quite complex, so we usually simplify it in a number of ways. In this module, you will study straight line motion and then you will look at motion under constant acceleration. Motion in two dimensions will be studied with the use of vectors. This module has four (4) four topics. The first topic explains the characteristics of motion. It describes the different types of motion that we see and experience every day. The next topic is all about reading and plotting graphs. Motion in straight line, help us calculate the distance, displacement, acceleration speed, and velocity of an object moving along the surface or falling downwards. The equations of motion, you will study under linear motion can be applied in two dimensional motion. The projectile, circular and rotary motions are the topics that explain moving objects in the circular path. The learning activities and their answers are given right after you study each topic to help understanding it better. However, there may be some words that may be slightly difficult. If you cannot understand a word or words, please do consult your tutor. This module is also a follow up to the first module on Measurement. It requires reasoning skills and your knowledge of mathematics to help you solve word problems. You will also need to have a scientific calculator to assist you in your calculations. Always remember to take note of any equation and formulas that you come across in the module for use in other modules. The more you use them you will go through the module with ease. At the end of the module, you will find a table of units containing some quantities with their corresponding units that should help to solve word problems. The picture below shows examples of motions we see every day. Busy traffic Cyclists moving around a bend 5  GR 11 PHYS M2 OUTCOMES Learning Outcomes After going through this module, you are expected to:  describe the characteristics of linear motion including distance, displacement, speed, velocity, acceleration, instantaneous and average speed, velocity and acceleration, and calculate these quantities from a given set of experimental data.  emphasise the correct units of the quantities described.  perform vector analysis on displacement, velocity and acceleration.  draw and interpret distance–time, speed–time, velocity–time, displacement–time and acceleration–time graphs.  conduct ticker timer experiments for uniform and accelerated motions and plot the relevant graphs.  describe the motion of an object (uniform and non-uniform) from a ticker tape and record graphs.  determine unknown parameters from graphs of motion; for example, area under displacement-time graph represents average velocity of the moving object.  solve worded problems with the aid of diagrams.  derive and explain the following equations of linear motion using graphs where necessary.  apply the equations to calculate parameters such as distance, displacement, speed, velocity, acceleration and time.  translate the above equations of linear motion to one-dimensional vertical motion of an object under the influence of gravity, as the motion of this object, is an example of a linear motion.  apply equations of motion to one-dimensional free-fall motion (vertical) and solve problems associated with this motion.  draw graphs of motion in two dimensions from a given set of experimental data and worked examples on a two-dimension projectile motion.  describe the projectile motion in two separate directions, vertical and horizontal directions, and emphasise the essential facts about the motion along each direction.  derive the relevant equations of motion along each direction.  solve word problems on projectile motion.  describe quantities of circular motion and their units including angular displacement (), velocity () and acceleration ().  write the equations of circular motion analogous to the equations of linear motion.  explain the relationships between linear and angular quantities.  solve worked examples on circular motion. 6  GR 11 PHYS M2 TIME FRAME Time Frame Suggested allotment time: 10 weeks This module should be completed within ten (10) weeks. If you set an average of 3 hours per day, you should be able to complete the module comfortably by the end of the assigned week. Try to do all the learning activities and compare your answers with the ones provided at the end of the module. If you do not get a particular question right in the first attempt, you should not get discouraged but instead, go back and attempt it again. If you still do not get it right after several attempts, then you should seek help from your friend or even your tutor. DO NOT LEAVE ANY QUESTION UNANSWERED. 7  GR 11 PHYS M2 (MOTION) KINEMATICS 11.2.1 Characteristics of Motion There are different types of motion occurring around us all the time. Some examples are the movement of people walking, athletes running on tracks, cars being driven along roads, aeroplanes flying in the sky, footballs being kicked, compact discs rotating on CD players, trains travelling along tracks, mail being sorted and basketball falling through the ring and so on. This movements form an important part of everyday life. Figure 1 Different types of motion Movement involves a change in position in a certain time. Therefore, quantities like distance, time, speed and acceleration with their measurements must be considered when describing the motion of an object. Our knowledge of Physics enables us to analyze the motion of an object precisely. For example, aeroplane pilots need to know their exact position in the air, how long it will take to fly to their destination, how much fuel to take and the effect of the wind on their speed and direction of flight. Manufacturers of motor vehicles need to know whether their vehicles will stop within certain limits, accelerate and use fuel at acceptable rates, withstand collision and have sufficient power to climb steep hills and overtake other vehicles. Traffic accident investigation also requires knowledge of motion. Physics can be used to reconstruct an accident and to determine the speed and direction of the vehicles before the collision. In the study of motion, we find ways to specify the position of the object at any time. Graphing is the most common method used to explain the characteristics of motion. The geographical system of the north, south, east and west is also convenient to use. A number line such as the x-axis marked in units allows us to specify the position along a straight line. The zero can represent a starting point, with the positive and negative numbers giving the direction. The positions of a point on the line are described by a number, a unit, and a sign to indicate the direction. The direction is the line itself indicating which way, positive or negative along the line. We usually use direction to make sense of how the motion is going. As direction is involved, we say that position is a vector quantity. A common symbol for position is x where the half arrow shows that we are dealing with a vector quantity. 8  GR 11 PHYS M2 (MOTION) KINEMATICS Displacement is a change in position in a certain direction. It is a change that is obtained by subtracting or adding one vector from or to another, the result is a vector. The average speed of a moving object is the distance travelled divided by the time taken. Therefore, displacement divided by the time taken gives the average displacement per unit time which is the velocity. Velocity is a vector quantity because the information about direction is in the displacement. Distance, Displacement, Speed and Velocity Motion in one dimension was described using the quantities displacement (s), time (t), velocity (v) and acceleration (a). It is also true for motion in two dimensions but the vector nature of some of the quantities needs further clarification. It is also important to redefine the quantities covered in the last unit regarding their vector or scalar nature. Distance and displacement are both quantities describing length. However, a distance is a scalar while displacement is a vector, requiring a direction. Both quantities use the symbol, s, and both units of measurements are in metres. Example, d = 10m: Distance s =10m east: Displacement Distance is the length of actual path travel from the starting point to the finishing point. It has a magnitude (size) but no direction. A girl walks 5 metres from point A to point B in the classroom and backs to point A another 5 metres. The distance travelled by the girl from point A to B and from B to A is 10 metres, but does not give the girl’s direction. Since distance is a scalar quantity and does not give direction, the total length she travelled is 10 metres (5m +5m =10m). B 5m A B 5m A Figure 2 Girl walking in a 5m length 9  GR 11 PHYS M2 (MOTION) KINEMATICS Displacement is the distance in a straight line in a certain direction from the starting point. It has magnitude (size) and direction. To understand this, we can simply study the diagram in Figure 3 below. Consider a girl walking west (point B) 5m in the classroom and then back to east (point A) 5 metres. The displacement of the girl will be zero. This is because the girl walks 5m in the west direction which is negative and the east direction to be positive. This gives us negative 5m plus positive 5m east which will cancel out giving us a net displacement of zero. (-5 +5 = 0 m) B 5m A B 5m A Figure 3 Girl walking 5m West and then 5m East Displacement is a vector that joins the initial position to the final position of an object. distance B displacement A Figure 4 Displacement and distance illustration Displacement is that vector from the beginning position to the end position of a motion. 10  GR 11 PHYS M2 (MOTION) KINEMATICS Example 1 A body moves from A to C via B. It has travelled a total distance of 16m. Distance = 10m+ 6m N = 16m B Resultant = 2 10 + 6 2 10m 6m 2 R = 100m+ 36m A 12m C R2 = 136m Displacement representation R= 11.67m East The distance travelled is 16m. However, its displacement from the starting position is 12m to the East or displacement is equal to 12m east. The displacement is calculated using Pythagoras theorem which you will come across later in the unit. Displacement is a vector quantity. For straight line motion, positive displacement shows forwards motion while a negative displacement shows backward motion. Displacement can also be referred to by compass direction. Positive displacement is always to the right and negative displacement is always to the left. Now consider a person running around a circular track. The circumference of the circular track is always the distance travelled, which is calculated by using c = 2r and the displacement would be the diameter of the track. This is because the displacement is always in a straight line in a certain direction. In the Figure 5, the circular track is 400m and the radius is 63.69m, thus giving us the diameter of 127.38 metres. Using the formula above as 400/2x3.14 =63.69 x 2 =127.38m. Therefore the diameter is 127.38m which is now the runner’s displacement if he runs halfway around the track. If the runner starts and completes the race, then the displacement is zero. D = 127.38m start finish Figure 4 Displacement and distance representation in circular track 11  GR 11 PHYS M2 (MOTION) KINEMATICS We will use letter d to represent the total distance travelled and s for the change in displacement. The change in displacement is defined by; s = final displacement – initial displacement If the motion consists of many parts, then the change in displacement is the sum of the displacement in each part of the motion. Take for example, if an object starts at an origin let us say 15m and then by -5m then by 2m, the change in displacement is 15 – 5 + 2 = 12m. The final displacement is 12 + 0 = 12m. Measurement of time Time is used to measure speed and velocity. Time, t, is a scalar and is measured in seconds. For example, t = 2min 10s which is equal to 130s. Speed and velocity In the last unit, you learnt about the difference between scalar and vector quantities. Now that we are studying motion, you expect to meet more vector quantities because object has motions in particular directions. We use speed as a scalar which has a magnitude (size) and velocity as a vector that has magnitude and direction. So velocity is both the speed and direction of the motion of an object. We can use some terms to describe the motion of a motor vehicle. The terms commonly used include speed, velocity, acceleration and distance. The everyday use of these terms is often different from the scientific usage. For example, most people use the words speed and velocity to mean the same thing. But to the scientist, the two terms have slightly different meanings. When we talk about an object’s speed, we have no idea about its direction since it is a scalar. We can say the speed is the time rate of change of distance. Velocity, on the other hand, is the time rate of change of displacement. Both quantities use the symbol, v, and both units of measurement are in metres per second, ms-1. For example v = 5ms-1: Speed v = 5 ms-1 East: Velocity.  Speed is the distance travelled in unit time.  Velocity is the distance travelled in particular direction in unit time. Velocity is a vector and is a derived quantity. It is related to the quantities displacement and time and is build up from the fundamental quantities of length and time. 12  GR 11 PHYS M2 (MOTION) KINEMATICS The average velocity of a moving object is found by dividing total displacement by the total time taken. s vav t = From the equation, the dimensions of the average velocity are LT-1. While the average velocity is a vector quantity, and therefore has direction and magnitude, average speed is a scalar quantity and is found by dividing the total distance by the total time. Average velocity Two very simple required measurements for the calculation of an average velocity are:  displacement(s) of an object and  time (t) for this displacement to occur. Displacement, s, is the straight line distance from a reference position. The average velocity is given the symbol vav. Mathematically, average velocity is the displacement undergone by an object in unit time. In everyday situations, for example in a car, the units of velocity are kilometres per hour (kmh-1). However, SI units for velocity are metres per second (ms-1) as displacement has unit of measures in metres (m) and the time taken in seconds(s). The table below shows the average velocity of some things you are familiar with in both ms-1 and kmh-1. These velocities are useful for future comparison. Movement m/s Km/h Snail 0.003 0.01 Fast walker 2 7.2 Runner 5 18 Bicycling 14 50.4 Fast car 36 129 Sound in air 340 1224 Jet plane 600 2160 Earth around the Sun 30 000 108 000 Light and radio waves 300 000 000 1 080 000 000 Table 1 Object’s movement and their speed If a car takes 10 seconds to travel along a straight road for 300 metres, it means that the car travels at an average speed of 30 metres per second (30m/s or 30ms-1). When an object moves, it changes its position. The speed at which it moves depends on how far it moves and how long it takes. 13  GR 11 PHYS M2 (MOTION) KINEMATICS The speed of a moving body is the distance travelled in unit time. When we talk about an object’s speed, we have no idea about its direction (so speed is a scalar quantity). Distance d Speed = s= Time t Here, v is the speed, s is the distance and t is the time taken. Speed has a variety of units, for example, kilometres per hour (kmh-1) or in millimetres per second (mms-1). However, the correct S.I. unit for speed is the metre per second (ms-1). Total distance travelled Average speed = Total time taken d speedav = t The average velocity of a moving object is found by dividing total displacement by the total time taken. Total displacement Average velocity = Total time taken s velocityav = t = Definition of velocity is defined as speed in a particular direction. For example, 40kmh-1 East or 30kmh-1 West. Velocity is a vector quantity. Distance travelled in a particular direction Velocity = Time taken Displacement Velocity = Time s v= t A car is travelling at a constant speed. Its velocity changes each second. 3km C B N 2km 4km W E displacement A D S 5km Figure 5 Car travelling at various velocities 14  GR 11 PHYS M2 (MOTION) KINEMATICS As with speed, we can also use the term average velocity to describe the motion of an object like a car. Total displacement s Average velocity = = Total time t Where (delta) represents ‘change in‘, t means change in time. The speed of a car is sometimes given in kilometres per hour (km/h) or kmh-1. 1000 1 1km / h = = m/s 60 x 60 36 Example 2 A car travels from one city to another 96km away in 1.2 hours. What is the car’s average speed in: a) km/h? b) m/s? Solution Distance covered 96 a) Average speed = = = 80km/h Time taken 1.2 80m/s b) 80 km/h = = 22.2m/s 3.6 Example 3 A person rides a bicycle 2km east and then 2km north. The trip takes 2 hours. displacement 2km north  =45 o 2km east Bicycle’s route Find the, a) person’s average speed and b) average velocity 15  GR 11 PHYS M2 MOTION (KINEMATICS) Solution Total distance 4 a) Average speed = = = 2km/h Time taken 2 Total displacement 2.82 b) Average velocity = = = 1.41km/h North East Time taken 2 The diagram below shows a car travelling at constant speed. Its velocity changes each time it turns a corner because the direction of its motion changes. The velocity of the car measured at a particular moment is called the instantaneous velocity of the car. v = 20m/s east v = 20m/s south v = 20m/s west Figure 6 Car’s route Small distance travelled in a stated direction Instantaneous velocity = Small time taken taken Small distance travelled in a stated direction s (in metres) Instantaneous velocity (v) = Small time taken t (in seconds) Sometimes, when velocity is not uniform, we are concerned with the velocity of an object at a particular instant. This is called instantaneous velocity of an object at that instant. To find the instantaneous velocity of an object, we find its average velocity during a small period of time that we are interested in. The small period of time is written as t, and the small displacement is called s. We try to make t and s as small as possible. Distance moved in a stated direction is often replaced by the term displacement. Thus, the instantaneous velocity is the rate of change of displacement. 16  GR 11 PHYS M2 MOTION (KINEMATICS) In a modern car, a device within the transmission produces a series of electrical pulses, which are sent to a calibrated device that translates the pulses into the speed of the car. This information is then displayed on the car in the form of a deflected speedometer needle or a digital readout. In older cars, speedometers were linked mechanically to the transmission. Here, a car's speedometer needle is pointing to zero, meaning that the car is at a standstill. The speedometer measures instantaneous speeds or speeds at that instant. Figure 7 Speedometer of modern car Instantaneous speed is speed in a small time interval and can be plotted on a graph such as the one given below. SPEED TIME GRAPH 30 25 C D 20 speed 15 E (m/s) 10 A 5 0 B 0 5 10 15 20 time in minutes From the graph, the readings of the speeds at 2.5 seconds and 10.5 seconds are given below. a) 2.5 seconds = 0 b) 10.5 seconds. Speed = 20m/s 17  GR 11 PHYS M2 MOTION (KINEMATICS) Changing units In many situations, the person solving the problem is required to change the units of a given quantity. The following conversions are commonly required. 1 kilometre = 1000 metres 1 minute = 60 seconds 1 hour = 60 minutes 1 hour = 60 x 60 = 3600 seconds Imagine a velocity of 20kmh-1 is to be converted to ms-1. Here, one unit should be converted at a time. Now 1 hour = 3600s So, 20 000mh-1 = 20 000m in 3600 s = 20 000/3600 = 5.55ms-1 In the laboratory you may require these conversions: 1mm = 10-3m (11000 = 0.001= 10-3m) 1cm = 10-2m (1100 =0.01 = 10-2m) Now check what you have just learnt by trying out the learning activity below! Learning Activity 1 40 minutes Answer the following questions on the spaces provided. 1 Suppose you walk a distance of 4m to the blackboard in your classroom and then back 3m. (a) What distance have you covered? (b) What is your displacement? 2. James walks 3km north, turns and walks 4km east, then walks 3km south. What; a) will be the distance he has covered? 18  GR 11 PHYS M2 MOTION (KINEMATICS) b) is his displacement? 3. A train covers 400m at a constant speed of 8ms-1. What is the time taken to travel that distance? 4. Charlie walks in a straight line at a steady speed of 2ms-1 for 15 seconds. Work out the distance Charlie travelled from the starting point after 10 seconds of walking. 5. A person rides a bicycle to a shop. The person travels 300m north along a straight road and then travels east for another 400m. The trip to the shop takes 2 minutes. Find the person’s average a) speed. b) velocity. 6. A PMV travels two kilometres from one bus stop to another East of it in 20 minutes. Calculate the: a) distance travelled. 19  GR 11 PHYS M2 MOTION (KINEMATICS) b) displacement. c) speed. d) velocity. 7. The graph shows a motion of a body. s (m) 50 t (s) 20 What is the velocity of the body? 8. A boy walks 5km North East and then another 5km South East. It takes 2 hours. Calculate: a) The distance he travels. 20  GR 11 PHYS M2 MOTION (KINEMATICS) b) His displacement from his starting position. c) His speed. d) The velocity. 9. A runner races around a circular track of radius 80m. a) Work out the distance the runner covers if the start and the finish line are the same? (use circumference = 2r) b) Find the displacement as he crosses the finish line? c) If the runner in the above question completes one circuit in 60 seconds, what is the average speed? d) What is the average velocity if the runner races halfway around the field in 30 seconds? 21  GR 11 PHYS M2 MOTION (KINEMATICS) 10. An aeroplane flies North 200km for 2 hours and then East 200kmh-1 for 1 hour. a) What is the total distance flown? b) Find the total displacement. c) Calculate the average speed. d) Work out its average velocity. Thank you for completing learning activity 1. Now check your work. Answers are at the end of the module. 22  GR 11 PHYS M2 MOTION (KINEMATICS) Acceleration Whenever the velocity of an object is changing, it has acceleration. The object could be speeding up, slowing down or changing direction. In each of this case, its velocity is changing. The equations of motion you shall come across later in the chapter describe the motion of objects that have a constant acceleration in a straight line. Figure 8 Cars travelling at different speeds We have seen cars increasing their speed and slowing down on a busy road. We must know that the change of speed is called acceleration. When we want to measure acceleration we need to know how much time it takes for a certain change of speed or velocity, and we define acceleration as follows: Acceleration, (a) is the change of velocity (v) over period of time (t). So, we calculate the acceleration from the change of velocity which occurs in unit time. Many objects move at a constant velocity. The velocity of a car increases when it starts moving from rest and decreases on an application of a brake. Cars can thus speed up (accelerate) or slow down (decelerate). If the change in velocity is measured in metres per second and the time in seconds, then the acceleration is measured in metres per second squared that is ms-2. A car slowing down is said to decelerate. In other words, it will have a negative acceleration. Acceleration is a vector quantity. The term retardation is sometimes used instead of deceleration. You may at times encounter certain problems dealing with calculating acceleration of linear motion. For example, if the words deceleration and retardation are used in describing the motion, then the acceleration will have a minus sign in front of the value. Final velocity – Initial velocity Acceleration = Time 23  GR 11 PHYS M2 MOTION (KINEMATICS) Change in velocity v Acceleration = = Time t v a = (measured in ms-2) t v-u a = t Example 1 A train increases speed from 4m/s to 10m/s in 3 seconds. Find the acceleration. Final velocity – Initial velocity Acceleration = Time 10 - 4 a = = 2ms-2 3 Example 2 A taxi travelling at 15m/s slows down to 5m/s in 2 seconds. Find its acceleration. Final velocity – Initial velocity Acceleration = Time 5 - 15 -10 a = = = -5ms-2 2 2 Note: This worked example shows that the taxi is decelerating or slowing down. Example 3 A large truck increases velocity from 6m/s to 10m/s in 2 seconds. Find its acceleration. Final velocity – Initial velocity a= Time 10 - 6 4 a = = = 2ms-2 2 2 24  GR 11 PHYS M2 MOTION (KINEMATICS) Example 4 A train traveling at 20m/s slows down to 10m/s in 2 seconds. Find its acceleration. Final velocity – Initial velocity a= Time 10 - 20 -10 a = = = -5ms-2 2 2 Note: The negative sign here indicates that the car is slowing down. You can also describe it by saying that the car is decelerating at 5ms-2 or it has an acceleration of -5ms-2. Adding other vectors Often an object can be given two velocities at once. An aircraft, for example, can be given a velocity by its engines and another by the wind. The two velocities then are added to give the aircraft its actual velocities. When adding a vector, the size and direction both affect the result. If the engine speed is 4m/s and the wind speed is 3m/s, the aircraft’s actual speed can work out to be 1m/s or 7m/s or 5m/s. 3m/s 4m/s Into the wind 3m/s 4m/s With the wind Figure 9 Adding vector into the wind and with the wind 25  GR 11 PHYS M2 MOTION (KINEMATICS) Cross wind Flying into the cross –wind In the first second, the engine of the aircraft drives at 4m/s east and at the same time, the wind blows north at 3m/s. The aircraft finishes up at A, 4m/s east and 3m/s north from its starting point. So the diagonal line OA, as shown in the solution below is the aircraft’s actual velocity vector. 4m/s 3m/s wind Figure 10 Resolving vector across the wind R is the resultant vector you get if you add X and Y (see diagram in the solution below). To find the sum of the two velocity vectors we use the Pythagoras theorem. This is because the two vectors form a right angle triangle. R2 = X2 + Y2 The diagram illustrates how Pythagoras theorem is used to calculate the resultant velocity of the plane when encountering a cross wind. Solution Y A OA2 = Y 2 + X2 OA2 = 32 + 42 -1 3ms 2 OA = 16 + 9 OA2 = 25  O -1 X -1 4ms OA = 5ms OA = 5ms-1 N 53.10 E 26  GR 11 PHYS M2 MOTION (KINEMATICS) Solution The angle R makes with X from tan  = Y/X = opposite /adjacent opposite tan  = adjacent Y tan  = X 3 tan  = 4 tan  = 0.75  = 36.90 Now press the shift button followed by tan button on your calculator. You should arrive with the angle indicated above as the angle it makes with the horizontal component. However, your true bearing is calculated and written together with the resultant velocity like what is shown above. The true bearing is: 900 - 36.90 = 53.10 Adding other vectors is carried out in the same way as forces. These vectors include velocity and displacement. The example below can be resolved into components as well. Example 1 A pilot aims his plane due east at 50m/s while the wind is blowing north 20m/s. The resultant speed and direction of the plane are found from the diagonal of the parallelogram. N R -1 20ms A D C  =22 0 -1 E 50ms B The working out for this example is shown on the next page. It explains how you can find the resultant and the true bearing. 27  GR 11 PHYS M2 MOTION (KINEMATICS) Solution opposite Resultant = 202 + 502 tan  = adjacent C Resultant = 400 + 2500 tan  = B 20 Resultant = 2900 tan  = 50 Resultant = 53.9ms-1 tan  = 0.4 -1 0 Resultant = 53.9ms N 68 E  = 220 Resolving into two components In some cases, it is necessary to split a single vector into two parts. It is called the components of the vector. The method is the reverse of an addition of forces called the resolution. Sometimes, it is useful to reverse the parallelogram method of adding vectors. In the diagram above, the resultant (54m/s) can be resolved into two components, (20m/s and 50m/s). The diagram below shows how a single vector (R) has two components, horizontal (H) and vertical (V). The two components taken together have exactly the same effect as single vector(R). V R  H Figure 11 Resultant vector showing two components V and H The magnitudes of the two components are given by: Horizontal component of vector is H= R cos  Vertical component of vector is V = R sin  Using cosine ratio for horizontal component H cos  = R R cos220 = H 53.9 x 0.927=50ms-1 28  GR 11 PHYS M2 MOTION (KINEMATICS) Using sine ratio for vertical component V sin  = R 0 R sin22 = V 53.9 x 0.374 =20ms-1 Note: the calculated answers are rounded to the nearest whole number. Adding velocities (a) 5km/h velocity of the train at 80km/h Figure 12 Illustration of adding velocities at same direction The diagram above shows a train moving at a velocity of 80km/h in a direction along the railway track to the right. This is the velocity of the train relative to the track and how an observer would see as the train went past. A man inside the train can walk along the train either to the right or to the left at a velocity of 5km/h. How fast is the man on the train moving about the railway track? Clearly the direction in which he walks makes a difference because if he walks in the same direction with the train, his velocity adds to the velocity of the train. On the other hand, if he walks opposite the train’s direction, his velocity is subtracted from the train’s velocity. We usually take the velocity to the right as positive and velocity to the left as negative. b) velocity of the train at 80km/h 5km/h towards the left Figure 13 Illustration of subtracting velocities, at opposite direction 29  GR 11 PHYS M2 MOTION (KINEMATICS) It gives the train a velocity of +80km/h in two cases. The man has a velocity of +5km/h if he went towards the right and he has -5km/h if he went towards the left. Therefore, we add the velocities of the train and the man in both cases to find their resultant or combine the velocity relative to the track. a) resultant, v = 80km/h + 5km/h = 85km/h to the right. b) resultant, v = 80km/h – 5km/h = 75km/h to the right. Arrows always represent the direction of vectors. Adding velocities by parallelogram law Two velocities that are not in the same straight line can be added using the same parallelogram law that we use for adding forces. Example 2 A car travels 60ms-1 East and then 80ms-1 South. What is the resultant velocity of the car? Suppose two velocities are represented in size and direction by the sides of a parallelogram (drawn to scale). The resultant velocity is represented in size and direction by a diagonal drawn from where the two velocities act. Use a scale of 1cm = 20m/s v1 = 60m/s (3cm long)  Resultant= 100m/s v2 = 80m/s (4cm long) (5cm long) R Resultant using Pythagoras Theorem Using Pythagoras theorem for velocity at right angles (90o), we can solve for the resultant, which is; Resultant = 802 + 602 Resultant = 6400 + 3600 Resultant = 10 000 Resultant = 100ms-1 30  GR 11 PHYS M2 MOTION (KINEMATICS) Resultant =100ms-1 at 53o to the East. opposite tan  = adjacent V2 tan  = V1 80 tan  = 60 tan  = 1.33  = 530 The Pythagoras theorem is the analytical method of finding the resultant of the force. An arrow represents a vector. The length of an arrow represents a magnitude and a direction of the vector shown by the direction of the arrow. The size of the arrow measures from the arrow head to the tail of the arrow. This method will be discussed next. Combining vectors Rules for adding vectors Step 1. Set up a direction reference, e.g., cardinal points. N W E S Direction reference Step 2. Calculate a suitable scale. Vector diagrams should not be small because this will reduce its accuracy. Step 3. Draw the first vector keeping its length and direction accurate. Show the direction of the vector with an arrow head in the centre or at the end. Step 4. From the head of the first vector, draw the second vector. You must be accurate with its length and its direction. Step 5. The resultant or the vector sum is the line joining the tail of the first to the head of the second vector. The magnitude of the sum is the length of the vector, and the direction is the direction of the vector. This process can be used in adding two or more vectors. Arrow represents two or more vectors can be combined to produce a single resultant vector. 31  GR 11 PHYS M2 MOTION (KINEMATICS) Now let us consider an aeroplane flying East at a velocity of 120km/h. The wind also blows east at 20km/h. The resultant velocity is found by drawing the two vectors to the same scale and adding head to tail. The resultant is 140km/h and has the same direction as the two velocities. 120km/h east 10km/h east 140km/h east + = 120 + 20 = 140 Figure 14 Direction of aeroplane flying East The second case is where an aeroplane is flying eastward at 120km/h with the wind blowing towards the west at 20km/h. The head wind will slow down the plane. The resultant velocity is found by drawing the two vectors at the same scale and adding head to tail. The resultant velocity of the plane will be 100km/h east. The resultant simply has the same direction as the greater velocity. 120km/h east 20km/h west 100km/h east + = 120 + (- 20) = 140 Figure 15 Direction of aeroplane flying East with Wind flowing West An aeroplane flying eastward at 120km/h encounters a strong wind blowing to the north at 90km/h. To find the resultant we draw a scaled diagram and use trigonometry involving a triangle. In this case, the vectors are added head to tail and the resultant is found by joining the starting point to the finishing point as shown in the diagram on the next page. 32  GR 11 PHYS M2 MOTION (KINEMATICS) Scale 1cm =20km/h Resultant = 150km/h 90km/h  wind 120km/h Figure 16 Direction of aeroplane flying East with Wind blowing North The resultant is the diagonal to the parallelogram. This is constructed by using the component velocity vectors as sides. In a scaled diagram such as this one, you can use a ruler to measure your resultant velocity. The resultant is 150km/h in the East direction. Now to find the angle, you can still measure it with a protractor. It should approximately measure up to 36.90. It can also be calculated by using Pythagoras theorem as examples shown earlier. If the vectors are drawn to scale 1cm=20km/h, then the resultant will be 150km/h. To work out the angle using a calculator, 90/120 = 0.75. Press shift button on your calculator, then press tan button. You should get the answer 36.9 on your calculator. Remember, it is the angle and should be written in degrees. 90 tan  = = 0.75 = 36.90 120 The velocity and direction written in true bearing is 150kmh-1 N53.10E The parallelogram method to add vectors acting at the same point but not in a straight line is used. The order in adding vectors does not affect the magnitude and the direction of the resultant. There are four different vectors shown in the diagram on the next page. If you add the four vector head to tail in different ways but keeping its direction, you will have the same effect for their resultant. 33  GR 11 PHYS M2 MOTION (KINEMATICS) Two possible ways of joining your vectors are shown in the diagram below. d c a c b R d a = = R d b R b = a c Figure 17 Resultant of a number of vectors The resultant of many vectors is a single vector that would have the same effect as all the single vectors added head to tail. Subtracting vectors In certain problems, it is necessary to subtract vectors. For instance, when subtracting vector B from vector A, the direction of vector B is reversed and added to A (head to tail). A - (-B) Here, vector B has its direction reversed, and when added to a vector A, it falls under the subtraction process for an unlike sign. The determination of resultant is in the same way. For example, if you want to subtract a displacement of 12m south which is B and a displacement of 18m north which is A. Then we add the vectors and obtain an answer 30m north. Other vectors can be subtracted in the same way as the displacement. A = 18m North = +18 B=12m South = -12 18m - (-12)m = 30m North 34  GR 11 PHYS M2 MOTION (KINEMATICS) 12m South - = + = 18m North 18m North 12m North 30m North Figure 18 Vector illustration on subtracting opposing direction Now check what you have just learnt by trying out the learning activity below! Learning Activity 2 20 minutes Answer the following questions by writing all your answers on the spaces provided. 1. A flat top truck is moving west at 12ms-1. A person on the back of the truck runs from A to B at 3ms-1. Calculate the velocity of the person with respect to the road. B A 2. An aeroplane is flying due North at 300m/s and a Westerly wind of 25m/s is blowing it off course. Calculate the velocity and direction of the plane. 35  GR 11 PHYS M2 MOTION (KINEMATICS) 3. A car travelling at 50km/h enters a restricted speed zone and reduces speed to 35km/h. Calculate the change in velocity. 4. A person rows a boat across the river at 4km/h. The boat is kept headed at right angles to the banks. The river is flowing at 6km/h and is 200m across. a) In what direction from the bank does the person’s boat go about the shore? b) How long does it take the person to cross the river? c) How far from the person’s landing point downstream is the starting point? d) How long would it take the person to cross the river if there were no current? 5. A pilot aims his plane due east at 100m/s while the wind is blowing north at 2m/s. What is his actual speed and direction? 36  GR 11 PHYS M2 MOTION (KINEMATICS) 6. A car travels 80m East and 60m South in 10s. a) What is the change in position? b) What distance has it travelled? c) What is the speed of the car? d) What is the car’s velocity? 7. A plane is flying at 300kmh-1 at 30o east of north. What is the northerly and easterly component of the velocity? 8. A power boat which can travel at 10kmh-1 in still water heads due south across a river flowing from the west to east at 4kmh-1. What is the boats velocity relative to the river bed? Thank you for completing learning activity 2. Now check your work. Answers are at the end of the module. 37  GR 11 PHYS M2 MOTION (KINEMATICS) 11.2.2 Graphs of Motion The equations that we studied earlier can be used when acceleration and velocity are constant. However, the graphical analysis of motion has the advantage that it can be used for both constant and variable acceleration because the concept is the same in each case. Moreover, graphs are often useful as they give a continuous picture of the motion from data which is taken at a regular and very small time interval. For example, the gradient of a velocity time graph gives the acceleration as: v a = t If the acceleration for each of the different sections of a velocity – time graph is calculated, the results can be used to plot acceleration - time graph. The time scales for the two graphs are often same. In the same way, the gradient of the displacement - time graph gives the velocity. We can also work backwards using the area under the velocity – time graph. As velocity is calculated from the displacement divided by time, displacement equals velocity multiplied by time. It is where the area under the velocity versus time graph gives us the size of the displacement. We cannot calculate without additional information, therefore, we find out where an object is at a particular time from a velocity –time graph. It is simply because the graph only gives us the displacement or change in position. To find out the object’s final position, we need to know its initial position to which the displacement can be added. The area under the acceleration-time graph has the units of acceleration multiplied by time (a x t) which gives the units of velocity ms-2 x s = ms-1. In this case, a velocity-time graph can be created by calculating the area under the acceleration-time graph. An acceleration–time graph comes from a velocity–time graph after calculating the velocity. The area under the acceleration-time graph measures these changes but give no information about the initial velocity, so you cannot calculate the final velocity from the graph. We have seen that velocity is almost the same thing as speed, but it is a vector quantity. In a similar way, displacement is almost the same as distance, but it is a vector quantity. Displacement has a size (called distance) and direction. If you move to a different part of the room, you might have a displacement of 2 metres in an Easterly direction. The direction is important. The ticker timer measurements are sometimes used to draw displacement-time graphs, although they are not as useful as velocity time– graphs. The techniques of graphing are very useful in kinematics. Drawing graphs can simplify calculation of velocity, displacement and acceleration. 38  GR 11 PHYS M2 MOTION (KINEMATICS) Distance –Time Graph A body travelling with constant speed covers equal distances in equal time. If distance –time graph is a straight line, like OL in the diagram below, it shows a velocity of 10ms-1. The slope of the graph is OL/OM = 40m/4s = 10m/s DISPLACEMENT-TIME GRAPH FOR CONSTANT SPEED 50 L 40 30 Distance (m) 20 10 O M 0 0 1 2 3 4 5 Time (s) When the speed of the body is changing, the slope of the distance-time graph varies and at any point, equals the slope of the tangent. For example, the slope of the tangent is AB/BC = 40ms-1/2s = 20m/s. The velocity at that instant corresponding to T is therefore 20ms-1. DISPLACEMENT-TIME GRAPH FOR NON CONSTANT SPEED 50 40 A 30 Distance (m) 20 T 10 0 C B 0 1 2 3 4 5 Time (s) The two graphs shown above are distance–time graph for a constant speed and a non- constant speed. We can now conclude about the slope of the two graphs. The slope of a distance -time graph gives the speed. 39  GR 11 PHYS M2 MOTION (KINEMATICS) Displacement-time graph If the car moves away at constant velocity, then the displacement will increase. If the car moves away at a higher velocity, then the line will be steeper. The velocity is shown by the slope (or gradient) of the displacement–time graph. The diagram shows a displacement– time graph for a car at rest (stationary, not moving). A DISPLACEMENT-TIME GRAPH FOR AT REST at rest displacement time B DISPLACEMENT-TIME GRAPH FOR CONSTANT VELOCITY displacement constant velocity time C DISPLACEMENT-TIME GRAPH FOR ACCELERATING displacement accelerating time 40  GR 11 PHYS M2 MOTION (KINEMATICS) D DISPLACEMENT-TIME GRAPH FOR LIFT 40 C displacement 30 B 20 lift 10 A 0 0 1 2 3 4 5 6 Time in seconds The graph shown above as well as the three on the previous page is displacement-time graph for four different bodies. We can now conclude about the slope of the four graphs. The slope of a displacement – time graph gives the velocity. You can still convert a displacement-time graph into a distance time graph. Consider the displacement time graph given below. We may extract following information from it. The initial displacement is zero. This means that a body started from rest and moves with a positive velocity of 1ms-1 for 5 seconds. For the next 5 seconds, the body was resting. It then moves with a negative velocity of -1ms-1 for the next 20 seconds. This means that the body is now travelling in the opposite direction between 10 seconds and 20 seconds.  Velocity at 4s =1ms-1  Velocity at 15s = -1ms-1 DISPLACEMENT-TIME GRAPH 6 4 Displacement (m) 2 0 0 5 10 15 20 25 -2 Time (s) -4 -6 If you convert the graph into a distance–time graph, it will be like the one shown on the next page. There is no negative value of distance travelled. Distance time graph cannot be changed to a displacement-time graph as there is not enough information available. 41  GR 11 PHYS M2 MOTION (KINEMATICS) Speed is the gradient of the distance–time graph at the time indicated. Speed at time 4s = 1ms-1. Speed at time 15s = 1ms-1 DISTANCE –TIME GRAPH 16 14 12 10 Distance (m) 8 6 4 2 0 0 5 10 15 20 25 Time (s) Velocity–Time Graph When the velocity of a body is plotted against time, the graph obtained will be a velocity – time graph. It provides a way of solving motion problems. Tape charts are crude velocity- time graphs which show the velocity in jumps rather than smoothly, as occurs in practice. A motion sensor gives a smoother plot. CONSTANT VELOCITY TIME GRAPH 25 20 A B 15 Velocity (m/s) 10 5 0 O C 0 1 2 3 4 5 6 Time (s) In the graph, AB is the velocity-time graph for a body moving with a constant velocity of 20m/s. Since distance = average velocity x time, after 5 seconds it will have moved 20m/s x 5s = 100m. It is the shaded area under the graph, the rectangle OABC. 42  GR 11 PHYS M2 MOTION (KINEMATICS) The diagram below shows another graph with constant velocity. The PQ is the velocity–time graph for a body moving with uniform acceleration. At the start of the timing, the velocity is 20m/s but increase steadily to 40m/s after 5 seconds. If the distance covered equals the area under the PQ, that is the shaded area OPQRS, then Distance = area of rectangle OPRS + area of triangle PQR Distance = OP x OS + ½ x PR x QR (area of triangle = ½ base x height) Distance = 20m/s x 5s + ½ x 5s x 20m/s = 100m+ 50m = 150m VELOCITY TIME GRAPH FOR CONSTANT ACCELERATION 40 Q 35 30 Velocity (m/s) 25 20 P R 15 10 5 0 O S 0 1 2 3 4 5 6 Time (s) 43  GR 11 PHYS M2 MOTION (KINEMATICS) The diagram shows a velocity – time graph for a car travelling at different velocities. It shows that the car is accelerating or increasing its acceleration. VELOCITY TIME GRAPH FOR INCREASING ACCELERATION 35 30 25 Velocity (m/s) 20 15 10 5 0 0 2 4 6 Time(s) What would the graph look like if the car accelerated gently? Because its velocity would increase the graph would slope upwards. In this graph, the car started from a rest where the velocity is zero and accelerated uniformly (steadily). A straight line graph means a uniform constant acceleration. The graph below shows the car decreasing its velocity. VELOCITY TIME GRAPH FOR DECREASING ACCELERATION 25 20 Velocity (m/s) 15 10 5 0 0 1 2 3 4 5 6 Time (s) Remember: If an object starts from rest, the velocity is zero. 44  GR 11 PHYS M2 MOTION (KINEMATICS) A straight line graph means uniform, constant acceleration. What would the graph look like if the car accelerates more rapidly? Since its velocity increased more rapidly, the graph would be steeper. VELOCITY TIME GRAPH FOR CONSTANT DECELERATION 3.5 3 Velocity (m/s) 2.5 2 1.5 1 0.5 0 0 1 2 3 4 5 6 Time (s) VELOCITY TIME GRAPH FOR CONSTANT NEGATIVE ACCELERATION 3 2 Velocity (m/s) 1 0 0 1 2 3 4 5 6 7 -1 -2 -3 Time (s) The slope of a velocity- time graph shows how fast a body accelerates. The steeper the slope, the acceleration will be greater. This simply means the body is moving faster in a time of one second. If the slope of the graph is less steep, then it has a gentle acceleration. Hence, the body is moving slowly in a time of one second. 45  GR 11 PHYS M2 MOTION (KINEMATICS) The velocity-time graph below will help compare the motion of two bodies moving with constant velocities. VELOCITY TIME GRAPH COMPARING THE MOTION OF TWO BODIES MOVING WITH CONSTANT VELOCITY y 60 50 Velocity (m/s) 40 30 A 20 10 B 0 x 0 1 2 3 4 5 6 Time (s) To calculate the acceleration of the two bodies, you can use the formula below. The answer, you obtain from calculating the slope should give some idea about the motion of the two bodies. Unless the bodies are accelerating, then you must plot a line of best fit or tangent as shown earlier. This will make it easy for you to calculate the slope. Regardless of where you want to work out the slope, the value for each will be same anywhere on the line graph. This is because both motions are constant. Now let us try and calculate our slope. The acceleration is shown by the slope (gradient) of the velocity-time graph). rise Slope = run y Slope = x v2 - v1 Acceleration = t2 - t1 46  GR 11 PHYS M2 MOTION (KINEMATICS) Slope A Slope B rise rise Slope = Slope = run run 40 - 20 15 - 0 Slope = Slope = 4- 2 3-0 20 15 Acceleration = Acceleration = 2 3 = 10ms-2 = 5ms-2 You will now see the acceleration is much higher for line graph A and is less for line graph B. The diagram shows a velocity-time graph for a car starting off from one set of traffic light and stopping at the next set of lights. VELOCITY- TIME GRAPH FOR A CAR velocity C D B E A time In this graph, you will see that the car accelerates from A to C, than travels at a constant velocity from C to D and then decelerates (brakes) rapidly to stop at point E. The car is accelerating most at point B and C. 47  GR 11 PHYS M2 MOTION (KINEMATICS) VELOCITY- TIME GRAPH FOR A MOTORBIKE H I velocity F G J K E time From the diagram, you will see that the motorbike is accelerating more rapidly at E and F. The motorbike will shift gear at point G and H where the speed changes. The fastest speed will be achieved between F and G. The rider will start to use the brake to stop the car at point I. At point J and K, the graph shows the rider hitting a solid brick wall. Distance travelled is shown by the area under the velocity – time graph. VELOCITY TIME GRAPH FOR CONSTANT NEGATIVE ACCELERATION 20 E F 15 D 10 C Velocity (m/s) 5 A B G H J 0 0 20 40 60 80 100 120 Time (s) -5 I -10 The diagram above shows the velocity of a stock car (insert) when starting a race crashing into another car and then reversing. From the diagram, we can see that the driver has to wait 10 seconds before the starter waves his flag to start the race. The car acceleration is greatest at point C because of the 48  GR 11 PHYS M2 MOTION (KINEMATICS) steepness of the graph. The fastest speed will be achieved between point D and E where the velocity will be 15m/s. The driver will start to stop his car by pressing the brake at point E and it will come to a stop at point G after hitting another car. The time for the car to travel this far is 60 seconds. The driver rested for 10 seconds before reversing his car at point H. When the car is reversing the velocity is negative. The car’s maximum velocity at reverse will be negative 5m/s (-5m/s) at point I. The driver took 30 seconds to reverse the car and finally came to a stop at point J. Area under the velocity-time graph The graph below shows a body that starts from rest and accelerates uniformly for 10s reaching a velocity of 20ms-1. It then travels at this velocity for 40s, before deceleration to rest in 50s. Example 1 VELOCITY TIME GRAPH 25 20 Velocity (m/s) 15 10 1 2 3 5 0 0 20 40 60 80 100 120 Time (s) To calculate the total displacement you can use two methods. a) Calculate the displacement for each part of the graph and then add them together. Displacement for segment 1 = average velocity x time = (0 + 20) x 10 /2 = 100m Displacement for segment 2 = velocity x time = 20 x 40 = 800m Displacement for segment 3 = average velocity x time (20 + 0 x 50 /2 = 500m 49  GR 11 PHYS M2 MOTION (KINEMATICS) Therefore, total displacement = 100 + 800 + 500 = 1400m You will see from the graph that two segments are triangles and the other is a rectangle. You can use area, A = ½ base x height for triangles or area, A = length x width for rectangles. b) Use the fact that the area under the velocity vs time graph is the displacement. Now the shape of the area between the graph and the ‘x’ axis is a trapezium. The formula for the area of the trapezium is. Area = half the sum of the two parallel sides multiplied by the height. Where, Area = (40 + 100)/2 x 20 = 140 x 10 = 1400m. You will see that the answers obtained either way are same but method B is much quicker. The acceleration of the velocity- time graph is the slope of the graph. rise v2 - v1 20 - 0 Segment 1. Slope = = = = 2ms-2 run t 10 Segment 2. The acceleration = 0 (since there is no change in velocity) rise v2 - v1 0 - 20 = - 0.4ms-2 Segment 3. The acceleration = Slope = = = run t 50 Since the acceleration has a negative value, we conclude that the object’s motion is slowing down or decelerating. 50  GR 11 PHYS M2 MOTION (KINEMATICS) Example 2 VELOCITY TIME GRAPH FOR A CAR TRAVELLING ALONG A STRAIGHT ROAD 15 10 Velocity (m/s) 5 1 2 3 0 0 5 10 15 4 -5 -10 Time (s) Given is a velocity–time graph for a car travelling along a straight road. Find the: a) acceleration at 1s, 3s and 6s. b) displacement after 12s. You must know that the acceleration is the gradient of the graph. Between t =1 and t = 2, the gradient of the graph is 10/2 = 5. Therefore, the acceleration is 5ms-2. Even if you calculate the gradient at t = 1s, it will still be 5ms-2. Between t = 5 and 12, the gradient is -17.5/7 = -2.5, so the acceleration is -2.5ms-2. The displacement is the area under the graph. You divide the total area into four parts and add your answer. Total area = area of triangle 1 + area of rectangle + area of triangle 2 + area of triangle 3 = 1/2 x (10 x 2) + 10 (5-2) + 1/2 x (10 x4) + 1/2 (-7.5 x 3) = 60 – 11.25 = 48.75m The area under the time axis is a negative displacement. It means that the moving object is travelling in the opposite direction. The distance covered in the above example would be, 60 + 11.25 = 71.25. 51  GR 11 PHYS M2 MOTION (KINEMATICS) The acceleration for the velocity time graph is represented in the acceleration time graph below. Acceleration – time graphs It is also possible to draw acceleration–time graphs from velocity–time graphs. The following example is based on the above velocity-time graph. ACCELERATION –TIME GRAPH 6 5 4 Acceleration (m/s2) 3 2 1 0 0 2 4 6 8 10 12 14 -1 Time (s) -2 -3 Now check what you have just learnt by trying out the learning activity below! Learning Activity 3 60 minutes Answer the following questions on the spaces provided. 1. The graph below shows a distance-time graph for a girl on a cycle ride. 60 50 F D E 40 Distance (km) B C 30 20 A 10 0 0 2 4 6 8 Time (hr) 52  GR 11 PHYS M2 MOTION (KINEMATICS) a) How far did she travel? b) How long did she take to travel? c) What was her average speed in km/h? d) How many stops did she make? e) How long did she stop for altogether? f) What was her average speed, excluding stops? g) How can you tell from the shape of the graph, when she travelled fastest? Over which stage did this happen? 53  GR 11 PHYS M2 MOTION (KINEMATICS) 2. The graph below shows the distance travelled by a car plotted against time. DISTANCE AGAINST TIME GRAPH OF A CAR 120 A 100 80 Distance (m) 60 40 20 0 0 2 4 6 8 Time (s) a) How far has the car travelled at the end of 5s? b) What is the speed of the car during the 5s? c) What happened to the car after A? d) Draw a graph showing the speed of the car during the 5s. 54  GR 11 PHYS M2 MOTION (KINEMATICS) 3. A car traveling at 20m/s slows down to 10m/s in 2 seconds. Find its a) acceleration. b) deceleration. 4. A truck is travelling at 12m/s and increases its speed to 28m/s in 4 seconds. What is its acceleration? 5. The velocity–time graph below show the motion of a bus. VELOCITY TIME GRAPH OF A BUS 25 20 Velocity (m/s) 15 10 5 0 0 20 40 60 80 Time (s) a) Find the acceleration at t= 10s and t= 30s. 55  GR 11 PHYS M2 MOTION (KINEMATICS) b) Find the velocity at time = 60s. c) Find the distance covered in first 40s. d) Find the acceleration at t= 70s. e) Draw an acceleration-time graph for the above motion. ACCELERATION TIME GRAPH 1 0.5 Acceleration (m/s2) 0 0 20 40 60 80 -0.5 -1 -1.5 -2 -2.5 Time (s) 56  GR 11 PHYS M2 MOTION (KINEMATICS) 6. The graph below shows a velocity–time graph for an object moving in a time interval of 20s. 8 6 4 Velocity (m/s) 2 0 0 5 10 15 20 25 -2 -4 Time (s) a) Find the distance travelled. b) Find the displacement. c) Work out the average velocity. 57  GR 11 PHYS M2 MOTION (KINEMATICS) d) Determine the acceleration during each part of the motion and draw an acceleration–time graph.