Notas de Física

8/7/2019 Notas de Fsica

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PHYSICS 209LECTURE NOTES

Dr. John W. NorburyProfessor of Physics

University of Wisconsin-Milwaukee

c 2003 John W. Norbury

September 24, 2003


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Contents

1 INTRODUCTION 51.1 What is Physics? . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 How to Study Physics . . . . . . . . . . . . . . . . . . . . . . 61.3 Units . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Powers of Ten . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Conversion of Units . . . . . . . . . . . . . . . . . . . . . . . 81.6 Signicant Figures . . . . . . . . . . . . . . . . . . . . . . . . 91.7 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 10

2 CALCULUS REVIEW 112.1 Derivative Equals Slope or Rate of Change . . . . . . . . . . 11

2.1.1 Slope of a Straight Line . . . . . . . . . . . . . . . . . 112.1.2 Slope of a Curve . . . . . . . . . . . . . . . . . . . . . 13

2.1.3 Some Common Derivatives . . . . . . . . . . . . . . . 162.1.4 Extremum Value of a Function . . . . . . . . . . . . . 21

2.2 Integral Equals Antiderivative or Area . . . . . . . . . . . . . 212.2.1 Integral Equals Antiderivative . . . . . . . . . . . . . . 212.2.2 Integral Equals Area Under Curve . . . . . . . . . . . 222.2.3 Summary . . . . . . . . . . . . . . . . . . . . . . . . . 252.2.4 Denite and Indenite Integrals . . . . . . . . . . . . . 26

2.3 Problems (10 questions) . . . . . . . . . . . . . . . . . . . . . 28

3 STRAIGHT LINE MOTION 293.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

3.2 Position, Distance and Displacement . . . . . . . . . . . . . . 303.3 Average Velocity and Average Speed . . . . . . . . . . . . . . 313.4 Position and Velocity Graphs . . . . . . . . . . . . . . . . . . 333.5 Instantaneous Velocity and Instantaneous Speed . . . . . . . 353.6 Acceleration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

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4 CONTENTS

3.7 Constant Acceleration Equations . . . . . . . . . . . . . . . . 37

3.7.1 Algebraic Derivation . . . . . . . . . . . . . . . . . . . 373.7.2 Summary of Constant Acceleration equations . . . . . 413.7.3 Calculus Derivation . . . . . . . . . . . . . . . . . . . 42

3.8 Free Fall . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 443.9 Historical Note . . . . . . . . . . . . . . . . . . . . . . . . . . 453.10 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 47

4 VECTORS 494.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 494.2 Trigonometry . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.3 Vector Components . . . . . . . . . . . . . . . . . . . . . . . . 554.4 Unit Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . 584.5 Vector Addition . . . . . . . . . . . . . . . . . . . . . . . . . . 604.6 Vector Multiplication . . . . . . . . . . . . . . . . . . . . . . . 62

4.6.1 Scalar Product . . . . . . . . . . . . . . . . . . . . . . 624.6.2 Vector Product . . . . . . . . . . . . . . . . . . . . . 64

4.7 Problems (7 questions) . . . . . . . . . . . . . . . . . . . . . . 65

5 2- AND 3-DIMENSIONAL MOTION 675.1 Displacement, Velocity and Acceleration . . . . . . . . . . . . 675.2 Constant Acceleration Equations . . . . . . . . . . . . . . . . 68

5.3 Pro jectile Motion . . . . . . . . . . . . . . . . . . . . . . . . . 725.4 Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 785.5 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 81

6 NEWTONS LAWS OF MOTION 836.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 836.2 Forces and the Second Law . . . . . . . . . . . . . . . . . . . 84

6.2.1 Weight . . . . . . . . . . . . . . . . . . . . . . . . . . . 846.2.2 Normal Force . . . . . . . . . . . . . . . . . . . . . . . 856.2.3 Tension . . . . . . . . . . . . . . . . . . . . . . . . . . 856.2.4 Spring . . . . . . . . . . . . . . . . . . . . . . . . . . . 856.2.5 Friction . . . . . . . . . . . . . . . . . . . . . . . . . . 85

6.3 Circular Motion . . . . . . . . . . . . . . . . . . . . . . . . . . 956.4 Historical Note . . . . . . . . . . . . . . . . . . . . . . . . . . 986.5 Problems (10 questions) . . . . . . . . . . . . . . . . . . . . . 99


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CONTENTS 5

7 WORK AND ENERGY 105

7.1 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1057.2 Simple Machines . . . . . . . . . . . . . . . . . . . . . . . . . 1077.2.1 Ramp . . . . . . . . . . . . . . . . . . . . . . . . . . . 1077.2.2 Pulley . . . . . . . . . . . . . . . . . . . . . . . . . . . 1107.2.3 Lever . . . . . . . . . . . . . . . . . . . . . . . . . . . 1127.2.4 Hydraulic Press . . . . . . . . . . . . . . . . . . . . . . 114

7.3 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 1157.4 Work-Energy Theorem . . . . . . . . . . . . . . . . . . . . . . 1177.5 Gravitational Potential Energy . . . . . . . . . . . . . . . . . 1197.6 Conservation of Energy . . . . . . . . . . . . . . . . . . . . . 1197.7 Spring Potential Energy . . . . . . . . . . . . . . . . . . . . . 1217.8 Appendix: alternative method to obtain potential energy . . 1227.9 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 124

8 MOMENTUM AND COLLISIONS 1258.1 Center of Mass . . . . . . . . . . . . . . . . . . . . . . . . . . 125

8.1.1 Many Particle Systems . . . . . . . . . . . . . . . . . . 1268.1.2 Rigid Bodies . . . . . . . . . . . . . . . . . . . . . . . 129

8.2 Newtons Second Law for a Many Particle System . . . . . . 1318.3 Momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132

8.3.1 Point Particle . . . . . . . . . . . . . . . . . . . . . . . 1328.3.2 Many Particles . . . . . . . . . . . . . . . . . . . . . . 1328.3.3 Conservation of Momentum . . . . . . . . . . . . . . . 133

8.4 Collisions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1358.4.1 Collisions in 1-dimension . . . . . . . . . . . . . . . . 1358.4.2 Collisions in 2-dimensions . . . . . . . . . . . . . . . . 138

8.5 Center of Mass Frame . . . . . . . . . . . . . . . . . . . . . . 1418.6 Problems (7 questions) . . . . . . . . . . . . . . . . . . . . . . 143

9 ROTATIONAL MOTION 1459.1 Angular Displacement, Velocity, Acceleration . . . . . . . . . 145

9.1.1 Constant Angular Acceleration Equations . . . . . . . 1469.2 Kinetic Energy . . . . . . . . . . . . . . . . . . . . . . . . . . 1489.3 Moment of Inertia . . . . . . . . . . . . . . . . . . . . . . . . 149

9.4 Torque and Newtons Second Law . . . . . . . . . . . . . . . 1539.5 Work and Kinetic Energy . . . . . . . . . . . . . . . . . . . . 1539.6 Angular Momentum . . . . . . . . . . . . . . . . . . . . . . . 156

9.6.1 Many Particle System . . . . . . . . . . . . . . . . . . 1579.6.2 Rigid Body . . . . . . . . . . . . . . . . . . . . . . . . 157


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6 CONTENTS

9.6.3 Conservation of Angular Momentum . . . . . . . . . . 157

9.7 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 16010 GRAVITY 163

10.1 Newtons Gravitational Force Law . . . . . . . . . . . . . . . 16610.2 Gravity near the Surface of Earth . . . . . . . . . . . . . . . . 167

10.2.1 Gravity Inside Earth . . . . . . . . . . . . . . . . . . . 16910.3 Potential Energy . . . . . . . . . . . . . . . . . . . . . . . . . 17010.4 Escape Speed . . . . . . . . . . . . . . . . . . . . . . . . . . . 17410.5 Kepler s Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . 17710.6 Einsteins Theory of Gravity . . . . . . . . . . . . . . . . . . 18010.7 Problems (9 questions) . . . . . . . . . . . . . . . . . . . . . . 181

11 FLUIDS 183

12 OSCILLATIONS 18512.1 Introduct ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18512.2 Simple Harmonic Motion . . . . . . . . . . . . . . . . . . . . 185

12.2.1 Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 19012.3 Pendulums . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19112.4 Navigation and Clocks . . . . . . . . . . . . . . . . . . . . . . 19712.5 Problems (7 questions) . . . . . . . . . . . . . . . . . . . . . . 198

13 WAVES 201

13.1 Introduct ion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20113.2 Wavelength, Frequency, Speed . . . . . . . . . . . . . . . . . . 20113.3 Interference, Standing Waves and Resonance . . . . . . . . . 20513.4 Sound . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20713.5 Doppler Effect . . . . . . . . . . . . . . . . . . . . . . . . . . 21013.6 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 213

14 THERMODYNAMICS 21714.1 Temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . 21714.2 Heat . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

14.2.1 Heat Capacity . . . . . . . . . . . . . . . . . . . . . . 221

14.2.2 Specic Heat . . . . . . . . . . . . . . . . . . . . . . . 22214.2.3 Molar Specic Heat . . . . . . . . . . . . . . . . . . . 22214.2.4 Heats of Transformation . . . . . . . . . . . . . . . . . 224

14.3 Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22414.4 First Law of Thermodynamics . . . . . . . . . . . . . . . . . . 226


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CONTENTS 7

14.4.1 Adiabatic Processes . . . . . . . . . . . . . . . . . . . 226

14.4.2 Constant-volume Processes . . . . . . . . . . . . . . . 22614.4.3 Cyclical Processes . . . . . . . . . . . . . . . . . . . . 22714.4.4 Free Expansion . . . . . . . . . . . . . . . . . . . . . . 227

14.5 Kinetic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . 22814.5.1 Ideal Gas . . . . . . . . . . . . . . . . . . . . . . . . . 22814.5.2 Work Done by an Ideal Gas . . . . . . . . . . . . . . . 22914.5.3 Speed, Energy and Temperature . . . . . . . . . . . . 232

14.6 Problems (8 questions) . . . . . . . . . . . . . . . . . . . . . . 234


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8 CONTENTS


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Chapter 1

INTRODUCTION

1.1 What is Physics?

A good way to dene physics is to use what philosophers call an osten-sive denition, i.e. a way of dening something by pointing out examples.Physics includes the following general topics, such as:MotionThermodynamicsElectricity and MagnetismOptics and LasersRelativityQuantum mechanicsAstronomy, Astrophysics and CosmologyNuclear Physics and Elementary particlesPhysics of SurfacesCondensed Matter PhysicsAtoms and MoleculesSolids, Liquids, GasesElectronicsAcousticsMaterials scienceGeophysics

BiophysicsChemical PhysicsMathematical Physics and Applied MathematicsComputational PhysicsEngineering Physics

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10 CHAPTER 1. INTRODUCTION

Physics is a very fundamental science which explores nature from the

scale of the tiniest particles to the behaviour of the universe and many thingsin between. Most of the other sciences such as biology, chemistry, geology,medicine rely heavily on techniques and ideas from physics. For example,many of the diagnostic instruments used in medicine (MRI, x-ray) weredeveloped by physicists. All elds of technology and engineering are verystrongly based on physics principles. The electronics and computer industryis based on physics principles. Much of the communication today occurs viaber optical cables which were developed from studies in physics. Also theWorld Wide Web was invented at the famous physics laboratory called theEuropean Center for Nuclear Research (CERN). Thus anyone who plans towork in any sort of technical area needs to know the basics of physics. Thisis what an introductory physics course is all about, namely getting to knowthe basic principles upon which most of our modern technological society isbased.

1.2 How to Study Physics

If you want to learn to ride a bicycle or play the piano, we all know thatreading a book alone will never suffice. One must practice . This inevitablymeans falling off the bicycle a few times or bungling a few tunes. The sameis true with physics. You will never learn physics only by reading a book.It is essential to practice physics by doing problems. A strong emphasis of any good physics course will be on examples and homework problems. Thisis what we call active learning .

Here are some tips that will help you succeed:

1. Read the relevant section of the book before it is covered in class.This will help you enormously in understanding what is presented inlectures.

2. Carefully study the examples in the book. The best way to do thisis to read the problem statement and cover the solution. Then spend10 minutes trying to work out the example yourself. Only then take alook at the solution. Remember its all about active learning.

3. As mentioned above practice is of utmost importance. You should doevery homework problem.

4. What if you cant gure out a homework problem? There is a lot of research showing that students learn very well by working together.


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1.3. UNITS 11

This is called peer instruction . Get together with your classmates

and help each other understand the material. If you cant work outa problem then discuss it with your classmates. Remember activelearning. If none of you can work it out then go and see your instructorand ask for help.

5. Obviously the best way to prepare for a piano exam is to practice themusic pieces that you have been learning. Similarly in preparing foryour physics exams you should work out examples and problems. Goback through the book and again read the example statements in thebook (and cover the solution) and work out the problem. Do the samewith the homework. The best way to prepare for your physics exams isto make sure you can do every example and homework problem from

scratch. The worst way to prepare is to simply read over the book andhomework. You must practice by doing . Active learning!

Lets summarize. Passive learning, such as just reading over the bookor lecture notes or problems, is not effective. Learning physics is all aboutactive learning and there is a great deal of educational research literaturethat proves this.

1.3 Units

We shall come across a wide variety of different units being used for dif-ferent physical quantities. Some that you may already be familiar with aredistance measured in feet or meters. The British unit is foot, but the inter-national system (SI) of units uses meter. In science one of the most commonstandards is to use SI units, and these will be used throughout this book.Some SI units are listed in Table 1.1.

Table 1.1 SI Units

Quantity Unit Symbol

Time second sLength meter mMass kilogram kg


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1.4 Powers of Ten

Because the study of physics involves very large systems, such as the be-havior of galaxies, and very small systems, such as the behavior of atoms,we will need to use very large and very small numbers. Instead of writing60,000 meters, i.e. 60 thousand m, we instead write it as 60 103 m or60 km. Similarly 6 one hundredths of a meter is 0.06 m which is written6102 m or 6 cm. Some common prexes are listed in Table 1.2.

Table 1.2 Prexes for large and small numbers.

Number Familiar name Prex Symbol

103 = 1 , 000 thousand kilo- k106 = 1 , 000, 000 million mega- M109 billion giga- G1012 trillion tera- T

101 = 0 .1 tenth deci- d102 = 0 .01 hundredth centi- c103 = 0 .001 thousandth milli- m106 = 0 .000001 millionth micro- 109 billionth nano- n1012 trillionth pico- p1015 femto- f

1.5 Conversion of Units

We shall often have to convert units from one to another. The method shownbelow is based on substitution . That is you just nd what one quantity isin terms of another and then substitute. The units should be treated asalgebraic quantities that can be multiplied, divided, squared etc.


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1.6. SIGNIFICANT FIGURES 13

Example Convert 20 m to km.Solution We know that km = 1000 m so that 1 m = km1000 = 10 3km. Thus 20 m = 20 103 km.

Example Convert 1 minute 2 to s2.

Solution 1 minute 2 = (60 sec) 2 = 3600 sec2

Another thing that we will come across is the little word per , as used forexample in 55 miles per hour or 55 mph. It is very important to realize thatthe word per means divided by . Thus

55 miles per hour = 55mileshour

= 55 miles hour 1

1.6 Signicant Figures

The number of signicant gures reects how accurately a certain numberhas been measured. A number such as 4.7 has 2 signicant gures. 4700has 4 signicant gures which can also be written as 4 .700103, which stillhas 4 signicant gures. However 47 has only 2 signicant gures and isre-written as 4 .7

101. It would be incorrect to write it as 4 .70

101, which

would have 3 signicant gures.Suppose we can measure the length of a table very accurately, say 5.135

m, but suppose we cannot measure the width as accurately, say 2.3 m. Thearea is the length times the width and we might write area = 11.81 m.However quoting such a large number of signicant gures would imply weknow the area to better accuracy than the width, which does not make sense.Obviously the area should be rounded off to reect the least accuracy in thewidth or length, namely 12 m.

When numbers are multiplied or divided, the number of signi-cant gures in the answer should be the same as the number of

signicant gures in the least accurate number.

Now consider addition and subtraction. Suppose we measure the lengthof a sidewalk in two stages. Suppose the rst length is measured as 1.23 mand the second length is measured as 22 m. The total length is not 23.23 m,


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because that would mean we know the total length more accurately than

one of the measured lengths. Instead the total length should be written as23 m.

When numbers are added or subtracted, the number of decimal points in the answer should be the same as the number of decimalpoints in the least accurate number.

1.7 Problems (8 questions)

1. Based on the discussion in the text, write a summary on how you aregoing to approach your study of physics.

2. Based on the discussion in the text, discuss how you should not studyphysics.

3. Convert 24 hours to seconds (s). Write your answer in scientic nota-tion using the correct number of signicant gures.

4. Convert 36 seconds to hours. Write your answer in scientic notationusing the correct number of signicant gures.

5. Convert 55 miles per hour to km/s. Write your answer in scienticnotation using the correct number of signicant gures. (Note that1 mile = 1.61 km.)

6. A car acclerates at 10 m/s 2. Convert this to km/hour 2. Write youranswer in scientic notation using the correct number of signicantgures.

7. What is 23 .178 + .01 to the correct number of signicant gures?

8. What is 23 .178 .01 to the correct number of signicant gures?


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16 CHAPTER 2. CALCULUS REVIEW

(x)

x=2

y=4

Figure 2.1 Plot of y(x) = 2 x + 1. The slope is y x = 2.

Rather than always having to verify the slope graphically, lets do itanalytically for all lines. Take x i = x as the initial x value and xf = x + xas the nal value. Obviously xf x i = x. The initial value of y is

yi y(x i ) = mx i + c= mx + c (2.3)

and the nal value is

yf y(xf ) = mx f + c= y(x + x) = m(x + x) + c (2.4)

Thus y = yf yi = m(x + x) + c mx c = m x. Therefore the slopebecomes y x

=m x x

= m (2.5)

which is a proof that y = mx + c has a slope of m.

From above we can re-write our formula (2.2) using yf = y(x + x) andyi = y(x), so that

Slope y x

=yf yixf x i

=y(x + x) y(x)

x(2.6)


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2.1. DERIVATIVE EQUALS SLOPE OR RATE OF CHANGE 17

2.1.2 Slope of a Curve

A straight line always has constant slope m. Thats why its called straight .The parabola y(x) = x2 + 1 is plotted in Fig.2.2 and obviously the slopechanges. In fact the concept of the slope of a parabola doesnt make any sense because the parabola continuously curves . However we might thinkabout little pieces of the parabola. If you look at any tiny little piece it looksstraight. These tiny little pieces are all tiny little line segments, each withtheir own slope. Notice that the slope of the tiny little line segments keepschanging . At x = 0 the slope is 0 (the tiny little line is at) whereas aroundx = 1 the slope is larger.

(x)

Figure 2.2 Plot of y(x) = x2 + 1. Some tiny little pieces areindicated, which look straight.

One of the most important ideas in calculus is the concept of the deriva-tive, which is nothing more than

Derivative Slope of tiny little line segment.In Fig.2.1 we got the slope from y and x on the large triangle in the topright hand corner. But we would get the same answer if we had used thetiny triangles in the bottom left hand corner. What characterizes these tiny

triangles is that x and y are both tiny (but their ratio, y x = 2 always).Another way of saying that x is tiny is to say

x is tiny lim x0That is the limit as x goes to zero is another way of saying x is tiny.


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Let us now evaluate some expressions involving these tiny limits.

Examples

1) lim x0

[ x + 3] = 3

2) lim x0

x = 0

3) lim x0

[( x)2 + 4] = 4

4) lim x0

( x)2 + 4 x x

= lim x0

( x + 4) = 4

5) lim x0

3 = 3

For a curve like the parabola we cant draw a big triangle, as in Fig.2.1,because the hypotenuse would be curved . But we can get the slope at a point by drawing a tiny triangle at that point. Thus lets dene the

Slope of curve ata point

lim x0

y x

=Slope of tinylittle linesegment

Derivative

So its the same denition as before in (2.6) except lim x0

is an instruction

to use a tiny triangle. Now y x =y(x+ x)y(x) x from (2.6) and the derivative

is given a fancy new symbol dydx so that

dydx lim x0

y(x + x) y(x) x

(2.7)

The symbol dy simply means

dy tiny yThat is, usually y can be big or small. If we are talking about a tiny ywe write dy instead. Similarly for x.


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Example Calculate the derivative of the straight line y(x) = 3 x.Solution y(x) = 3 x

y(x + x) = 3( x + x)

dydx

= lim x0

3(x + x) 3x x

= lim x0

3x + 3 x 3x x

= lim x0

3 x x

= lim x0

3 = 3

Thus the derivative is the slope!

Example Calculate the derivative of the straight line y(x) = 4Solution y(x) = 4

y(x + x) = 4

dydx

= lim x0

4 4 x

= 0

The line y(x) = 4 has 0 slope and therefore 0 derivative.

The derivative was dened to give us the slope of a curve at a point. The

two examples above show that it also works for a straight line. A straightline is a special case of a curve. Now we do some examples for real curves.

Example Calculate the derivative of the parabola y(x) = x2

Solution y(x) = x2

y(x + x) = ( x + x)2

= x2 + 2 x x + ( x)2

dydx

= lim x0

y(x + x) y(x) x

= lim x0

x2 + 2 x x + ( x)2 x2 x

= lim x0

(2x + x)

= 2 x


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Example Calculate the slope of the parabola y(x) = x2 at thepoints x = 2, x = 0, x = 3.Solution We already have dydx = 2 x. Thus

dydx x= 2

= 4dydx x=0

= 0

dydx x=3

= 6

which shows how the slope of a tiny little line segment varies aswe move along the parabola.

Example Calculate the slope of the curve y(x) = x2 + 1 (seeFig.2.2) at the points x = 2, x = 0, x = 3Solution y(x) = x2 + 1

y(x + x) = ( x + x)2 + 1= x2 + 2 x x + ( x)2 + 1

dydx

= lim x

0

y(x + x) y(x) x

= lim x0

x2 + 2 x x + ( x)2 + 1 (x2 + 1) x= lim

x02x + x

= 2 x

Thus the slopes are the same as in the previous example.

2.1.3 Some Common Derivatives

In a previous example we saw that the derivative of y(x) = 4 was dydx = 0,which make sense because a graph of y(x) = 4 reveals that the slope isalways 0. This is true for any constant c. Thus

dcdx

= 0 (2.8)


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We also saw in a previous example that ddx x2 = 2 x. In general we have

dxn

dx= nx n1 (2.9)

This is a very important result. We have already veried it for n = 2. Letsverify it for n = 3.

Example Check that (2.9) is correct for n = 3.

Solution Formula (2.9) gives

dx3dx

= 3 x31 = 3 x2

We wish to verify this. Take y(x) = x3.

y(x + x) = ( x + x)3

= x3 + 3 x2 x + 3 x( x)2 + ( x)3

dydx

= lim x0

y(x + x) y(x) x

= lim x0x3 + 3 x2 x + 3 x( x)2 + ( x)3

x3

x= lim

x03x2 + 3 x x + ( x)2

= 3 x2 in agreement with our result above.

A list of very useful results for derivatives is given in Tables 2.1 and 2.2.These results are proved in calculus books.


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Table 2.1 Properties of Derivatives and Derivatives of Partic-ular Functions. [nnn from Tipler, pg. AP-16, 1991].

Multiplicative constant rule:

The derivative of a constant times a function equals the constanttimes the derivative of the function:

ddx

[Cy(x)] = C dy(x)

dx

Addition rule:

The derivative of a sum of functions equals the sum of the deriva-tives of the functions:

ddx

[y(x) + z(x)] =dy(x)

dx+

dz(x)dx

Chain rule:

If y is a function of z and z is in turn a function of x, the deriva-tive of y with respect to x equals the product of the derivativeof y with respect to z and the derivative of z with respect to t:

d

dxy(z) =

dy

dz

dz

dxDerivative of a product:

The derivative of a product of functions y(x)z(x) equals the rstfunction times the derivative of the second plus the second func-tion times the derivative of the rst:

ddx

[y(x)z(x)] = y(x)dz(x)

dx+

dy(x)dx

z(x)

Reciprocal derivative:

The derivative of y with respect to x is the reciprocal of the

derivative of x with respect to y, assuming that neither derivativeis zero:dydx

=dxdy

1 if dxdy

= 0


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Table 2.2 Derivatives of Particular Functions. [from Tipler,pg. AP-16, 1991].

dC dx

= 0 where C is a constant

d(xn )dx

= nx n1

d

dxsin x = cos x

ddx

cos x = sin xd

dxtan x = sec2 x

ddx

ebx = bebx

ddx

ln bx =1bx

Example Use of multiplicative constant rule,d

dx[Cy (x)] = C

dy(x)dx

This just means, for instance, that

ddx

(3x2) = 3dx2

dx= 3 2x = 6 x

Example Use of addition rule,d

dx[y(x) + z(x)] =

dy(x)dx

+dz(x)

dx

Take for instance y(x) = x and z(x) = x2. This rule just meansd

dx(x + x2) =

dxdx

+dx2

dx= 1 + 2 x


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Now consider the chain rule,dydx

=dydz

dzdx

. A rough proof of this is to

just note that the dz cancels in the numerator and denominator. The useof the chain rule is best seen in the following example, where y is not givenas a function of x.

Example Verify the chain rule for y = z3 and z = x2.Solution We have y(z) = z3 and z(x) = x2. Thus y(x) = x6.

dydx

= 6 x5

dy

dz= 3 z2

dzdx

= 2 x

Now dydzdzdx = (3 z

2)(2x) = (3 x4)(2x) = 6 x5.Thus we see that dydx =

dydz

dzdx .

Now consider the product rule,d

dx[y(x)z(x)] = y(x)

dz(x)dx

+dy(x)

dxz(x).

The use of this arises when multiplying two functions together as follows.

Example If y(x) = x3 and z(x) = x2, verify the product rule.Solution y(x)z(x) = x5

d

dx[y(x)z(x)] =

dx5

dx= 5 x4

Now lets show that the product rule gives the same answer.

y(x)dz(x)

dx= x3

dx2

dx= x32x = 2 x4

dy(x)dx

z(x) =dx3

dxx2 = 3 x2x2 = 3 x4

y(x)dz(x)

dx+

dy(x)dx

z(x) = 2 x4 + 3 x4 = 5 x4

in agreement with our answer above.


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2.2. INTEGRAL EQUALS ANTIDERIVATIVE OR AREA 25

2.1.4 Extremum Value of a Function

A nal important use of the derivative is that it can be used to tell us whena function attains a maximum or minimum value. This occurs when thederivative or slope of the function is zero.

Example What are the ( x, y ) coordinates of the place wherethe parabola y(x) = x2 + 3 has its minimum value?

Solution The minimum value occurs where the slope is 0. Thus

0 =dydx

=d

dx(x2 + 3) = 2 x

x = 0y = x2 + 3 y = 3

Thus the minimum is at ( x, y ) = (0 , 3). You can verify this byplotting a graph.

We have completed our review of the derivative. Now lets turn to thesecond major topic.

2.2 Integral Equals Antiderivative or Area

2.2.1 Integral Equals Antiderivative

The derivative of y(x) = 3 x is dydx = 3. The derivative of y(x) = x2 is

dydx = 2 x. The derivative of y(x) = 5 x

3 is dydx = 15 x2.

Lets play a game. I tell you the answer and you tell me the question.Or I tell you the derivative dydx and you tell me the original function y(x)that it came from. Ready?

If dy

dx= 3 then y(x) = 3 x

If dydx

= 2 x then y(x) = x2

If dydx

= 15 x2 then y(x) = 5 x3


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We can generalize this to a rule.

If dydx

= xn then y(x) = 1n + 1

xn +1

Actually I have cheated. Lets look at the following functions

y(x) = 3 x + 2y(x) = 3 x + 7y(x) = 3 x + 12y(x) = 3 x + C (C is an arbitrary constant)y(x) = 3 x

All of them have the same derivative dydx = 3. Thus in our little game of re-constructing the original function y(x) from the derivative dydx there isalways an ambiguity in that y(x) could always have some constant added toit.

Thus the correct answers in our game are

If dydx

= 3 then y(x) = 3 x + constant

Actually instead of always writing constant, let me just write C .

If dy

dx= 2 x then y(x) = x2 + C

If dydx

= 15 x2 then y(x) = 5 x3 + C

If dydx

= xn then y(x) =1

n + 1xn +1 + C.

This original function y(x) that we are trying to get is given a specialname called the antiderivative or integral , but its nothing more than theoriginal function.

2.2.2 Integral Equals Area Under Curve

Lets see how to extract the integral from our original denition of derivative.The slope of a curve is y x or dydx when the increments are tiny. Notice

that y(x) is a function of x but so also is dydx . Lets call it

f (x) dydx

= y x

(2.10)


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Thus if f (x) = dydx = 2 x then y(x) = x2 + C , and similarly for the other

examples.In equation (2.10) I have written y x also becausedydx is just a tiny version

of y x .Obviously then

y = f x (2.11)

ordy = f dx (2.12)

What happens if I add up many y increments? For instance suppose youare aged 18. Then if I add up many age increments in your life, such as

Age = Age 1 + Age 2 + Age 3 + Age 4 or1 year + 3 years + 0.5 year + 5 years + 0.5 year + 5 years + 3 years

= 18 years

I get your complete age. Thus if I add up all possible increments of y thenI get back y. That is

y = y1 + y2 + y3 + y4 + We use a special symbol for this,

y =i

yi (2.13)

where yi = f i x i (2.14)

Now looking at Fig.2.3 we can see that the area of the shaded section is justf i x i . Thus yi is an area of a little shaded region. Add them all up andwe have the total area under the curve. Thus

Area undercurve f (x) =

if i x i =

i

yi x i

x i =i

yi = y (2.15)

Lets now make the little intervals yi and x i very tiny. Call them dy anddx. If I am using tiny intervals in my sum then I am going to use a new

symbol . Thus Area = fdx = dydx dx = dy = y (2.16)which is just the tiny version of (2.15). Notice that the dx cancels.


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x1 x2 xi x1 i

f 1

f i

f(x)

x

Figure 2.3 A general function f (x). The area under the shadedrectangle is approximately f i x i . The total area under the curveis therefore

if i x i . If the x i are tiny then write x i = dx

and writei

= . The area is then f (x)dx.In formula (2.16) recall the following. The derivative is f (x) dydx andy is my original function which we called the integral or antiderivative.

We now see that the integral or antiderivative or original function can beinterpreted as the area under the derivative curve f (x) dydx .By the way

f dx reads integral of f with respect to x.


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2.2.3 Summary

Let us briey summarize what we have so far. If we have

f (x) =dydx

then this implies

y = f dx + cFor example we obtain the following derivatives

y(x) = x2 dydx

= 2 x f (x)y(x) = x2 + 4

dydx

= 2 x f (x)and we get back the original functions by integrating

dydx

= 2 x y(x) = f dx + c = x2 + cwhere c = 0 for the rst example and c = 4 for the second case.

Our derivations are summarized as follows.

Let f (x) =dydx

= y x

y = f xor dy = f dx

Any function y is a sum of tiny increments, as in

y =i

yi = dy=

if i x i = f dx

= Area under curve f (x)

= Antiderivative

y = f dx + c


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Example What is x dx ?Solution The derivative function is f (x) = dydx = x. Thereforethe original function must be 12 x

2 + c. Thus

x dx = 12x2 + c

2.2.4 Denite and Indenite Integrals

The integral x dx is supposed to give us the area under the curve x, butour answer in the above example ( 12 x2 + c) doesnt look much like an area.We would expect the area to be a number.Example What is the area under the curve f (x) = 4 betweenx1 = 1 and x2 = 6?

Solution This is easy because f (x) = 4 is just a horizontalstraight line as shown in Fig.2.4. The area is obviously 4 5 = 20.

(x)

1 6

Figure 2.4 Plot of f (x) = 4. The area under the curve betweenx1 = 1 and x2 = 6 is obviously 4 5 = 20.


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Consider

4dx = 4 x + c. This is called an indenite integral or an-

tiderivative . The integral which gives us the area is actually a thing calledthe denite integral written

x 2

x 14dx [4x + c]x 2x 1 (4x2 + c) (4x1 + c)

= [4x]x 2x 1 = 4 x2 4x1 (2.17)Lets explain this. The formula 4x+ c by itself does not give the area directly.For an area we must always specify x1 and x2 (see Fig.2.4) so that we knowwhat area we are talking about. In the previous example we got 4 5 = 20from 4x2 4x1 = (4 6) (4 1) = 24 4 = 20, which is the same as(2.17). Thus (2.17) must be the correct formula for area! Notice here that it doesnt matter whether we include the constant c because it cancels out when we calculate area.

Thus 4dx = 4 x + c is the antiderivative or indenite integral and itgives a general formula for the area but not the value of the area itself. Toevaluate the value of the area we need to specify the edges x1 and x2 of thearea under consideration as we did in (2.17). Using (2.17) to work out theprevious example we would write

6

14dx = [4x + c]61 = [(4 6) + c][(4 1) + c]

= 24 + c 4 c= 24 4 = 20 (2.18)

or leaving out the constant c we get

6

14dx = [4x]61 = (4 6) (4 1)

= 24 4 = 20

Example Evaluate the area under the curve f (x) = 3 x2 be-tween x1 = 3 and x2 = 5.Solution

5

33x2dx = [x3 + c]53 = (125 + c) (27 + c) = 98

or leaving out the constant c we get

5

33x2dx = [x3]53 = 125 27 = 98


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1. Calculate the derivative of y(x) = 5 x + 2.

2. Calculate the slope of the curve y(x) = 3 x2 + 1 at the points x = 1,x = 0 and x = 2.3. Calculate the derivative of x4 using the formula

ddx

xn = nx n1

Verify your answer by calculating the derivative from

dydx = lim x0

y(x + x)

y(x)

x

4. Prove that ddx (3x2) = 3 ddx x

2.

5. Prove that ddx (x + x2) = dxdx +

dx 2dx .

6. The chain rule for derivatives is

dydx

=dydz

dzdx

Verify that this is true by taking y = z3 and z = x2 and calculating

the left and right hand sides of the chain rule showing they are equal.7. The product rule for derivatives is

ddx

(yz) = ydzdx

+ zdydx

Verify that this is true by taking y = x and z = x2 and calculatingthe left and right hand sides of the chain rule showing they are equal.

8. Where do the extremum values of y(x) = x2 4 occur? Verify youranswer by plotting a graph.

9. Evaluate x2

dx and 3x3

dx.10. What is the area under the curve f (x) = x between x1 = 0 and x2 = 3?

Work out your answer A) graphically and B) with the integral.


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Chapter 3

STRAIGHT LINE MOTION

3.1 Introduction

When you drive you car and go on a journey there are several things youare interested in. Typically these are the distance travelled and the speed with which you travel. Often you want to know how long a journey will takeif you drive at a certain speed over a certain distance. Also you are ofteninterested in the acceleration of your car, especially for a very short journeysuch as a little speed race with you and your friend. You want to be ableto accelerate quickly so that you reach your top speed more quickly. In thischapter we will spend a lot of time studying the concepts of distance, speed

and acceleration.

Experiment

Drop a ball from different heights. It falls straight down and itgoes faster at the bottom if released from higher distances.

In the experiment an object is dropped from a certain height, it starts off with zero speed and ends up hitting the ground with a large speed. If you

think about it, thats a pretty amazing phenomenom. Why did the speedof the ball increase? You might say gravity. But whats that? The speedof the ball increased, and therefore gravity provided an acceleration . Buthow? Why? When? We shall address these deep questions in this and laterchapters.

33


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34 CHAPTER 3. STRAIGHT LINE MOTION

3.2 Position, Distance and Displacement

In 1-dimension, positions are measured along the x-axis with respect to someorigin . It is up to us to dene where to put the origin , because the x-axis is just something we invented to put on top of, say a real landscape.

The following example explains what is meant by the term position ,which is given the symbol x. The example also shows how position changesdepending on where the origin is located.

Example Chicago is 100 miles South of Milwaukee and thereis a town called Glendale which is 10 miles North of Milwaukee.A) If we dene the origin of the x-axis to be at Glendale whatis the position of someone in Chicago, Milwaukee and Glendale?B) If we dene the origin of x-axis to be at Milwaukee, what isthe position of someone in Chicago, Milwaukee and Glendale?Solution A) For someone in Chicago, x=110 miles.For someone in Milwaukee, x = 10 miles.For someone in Glendale, x = 0 miles.B) For someone in Chicago, x = 100 miles.For someone in Milwaukee, x = 0 miles.For someone in Glendale, x =

10 miles. This is a negative

position.

Displacement is dened as a change in position . Specically,

x xf x i (3.1)Note: We always write anything anything f anything i where anything f is the nal value and anything i is the initial value. This applies to suchthings as position, speed, time etc.

Distance is best understood simply as what the odometer on your carreads. The odometer does not read displacement (except if displacment and

distance are the same, as is the case for a one way straight line journey).You can see that if x i is bigger than xf then the displacement can benegative. Distance will be the magnitude of the displacement. For example,if the displacement is 100 m then the distance is also 100 m. But if thedisplacement is 100 m then the distance is still 100 m.


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3.3. AVERAGE VELOCITY AND AVERAGE SPEED 35

Example What is the displacement for someone driving fromMilwaukee to Chicago? What is the distance?

Solution With the origin at Milwaukee, then the initial positionis x i = 0 miles and the nal position is xf = 100 miles, so that x = xf xi = 100 miles. You get the same answer with theorigin dened at Gendale. Try it.The distance is also 100 miles.

Example What is the displacement for someone driving fromMilwaukee to Chicago and back? What is the distance?

Solution With the origin at Milwaukee, then the initial positionis x i = 0 miles and the nal position is also xf = 0 miles, so that x = xf xi = 0 miles. Thus there is no displacement if thebeginning and end points are the same. You get the same answerwith the origin dened at Gendale. Try it.

The distance is 200 miles. This is what the odometer in your carwould read.

3.3 Average Velocity and Average Speed

Average velocity is dened as the ratio of displacement divided by the corre-sponding time interval.

vx x t

=xf x it f t i

(3.2)

whereas average speed is just the total distance divided by the time interval,

v total distance

t=

d t

(3.3)


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Example What is the average velocity and averge speed forsomeone driving from Milwaukee to Chicago who takes 2 hoursfor the journey?

Solution x = 100 miles and t = 2 hours , giving

vx =100 miles2 hours

= 50mileshour 50miles per hour 50mph

The average speed is the same as average velocity in this casebecause the total distance is the same as the displacement. Thusv = 50 mph.

Example What is the average velocity and averge speed forsomeone driving from Milwaukee to Chicago and back to Mil-waukee who takes 4 hours for the journey?

Solution x = 0 miles and t = 2 hours, giving vx = 0.

However the total distance is 200 miles completed in 4 hoursgiving v = 200 miles4 hours = 50 mph again.


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3.4. POSITION AND VELOCITY GRAPHS 37

3.4 Position and Velocity Graphs

A very important thing to understand is how to read graphs of position andtime and graphs of velocity and time, and how to interpret such graphs.

It is useful to see how the average velocity is obtained from a position-time graph. In Fig. 3.1 an arbitrary position time graph is shown. Eventhough the motion is quite complicated, it is an easy matter to obtain theaverage velocity. We simply substitute the initial and nal times and pos-tions into (3.2). It does not matter how complicated the motion is betweenx i and xf .

t

x

xf

xi

tf

ti

Figure 3.1 Arbitrary Position - time graph.

Let us now study the position-time ( x, t ) and velocity-time ( vx , t ) graphsfor an object standing still and an object moving at constant speed. Thiscan be realised in the following simple demonstration.


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ExperimentA) Take a billiard ball and let it sit at rest.B) Take a billiard ball and let it roll in a straight line on a smoothhorizontal table. (We want the table to be smooth, so that theball does not slow down.)The position-time graphs are shown in Fig. 3.2. For the ballat rest the position x does not change and is therefore just astraight horizontal line. For the ball moving at constant velocity,the position keeps increasing and so the position-time graph isan inclined straight line.

Now the (v, t ) graph is the slope of the (x, t ) graph. Thus forthe ball at rest, the slope of the ( x, t ) is zero, so that the ( v, t )is always at zero. For the ball moving at constant speed the(x, t ) graph has a constant slope, so that the ( v, t ) graph is justa straight horizontal line. This is displayed in Fig. 3.2.

t

x

t

v

t

t

x

t

v

t

(A) (B)

Figure 3.2 Position - time and Velocity - time graphs forA) object standing still and B) object moving at constant speed.


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3.5. INSTANTANEOUS VELOCITY AND INSTANTANEOUS SPEED 39

3.5 Instantaneous Velocity and Instantaneous Speed

When you drive to Chicago with an average velocity of 50 mph you probablydont drive at this velocity the whole way. Sometimes you might pass a truckand drive at 70 mph and when you get stuck in the traffic jams you mightonly drive at 20 mph.

Now when the police use their radar gun and clock you at 70 mph, youmight legitimately protest to the officer that your average velocity for thewhole trip was only 50 mph and therefore you dont deserve a speeding ticket.However, as we all know police officers dont care about average velocity oraverage speed. They only care about your speed at the instant that youpass them. Thus lets introduce the concept of instantaneous velocity andinstantaneous speed .

What is an instant ? It is nothing more than an extremely short timeinterval. The way to describe this mathematically is to say that an instantis when the time interval t approaches zero, or the limit of t as t 0(approaches zero). We denote such a tiny time interval as dt instead of t.The corresponding distance that we travel over that tiny time interval willalso be tiny and we denote that as dx instead of x.

Thus instantaneous velocity or just velocity is dened as

vx lim t0 x t

=dxdt

(3.4)

This is the derivative of x with respect to t.

The instantaneous speed , or just speed , is dened as simply the magnitudeof the instantaneous velocity or magnitude of velocity .

The units of distance and displacement are m and the units of time ares. Therefore the units for velocity or speed are m/sec.

3.6 Acceleration

We saw that velocity tells us how quickly position changes. Acceleration tells us how much velocity changes. Average acceleration is dened as

ax =vxf vxi

t f t i=

vx

t(3.5)

and the instantaneous acceleration , or just acceleration , is dened as

ax =dvxdt

(3.6)


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Given that the units of velocity are m/sec, it follows that the units of

acceleration are m/sec2

. If the velocity is written in miles per hour then theacceleration is miles per hour 2.Now because vx = dxdt we can write ax =

ddt vx =

ddt

dxdt which is often

written instead as ddtdxdt d

2 xdt 2 , that is the second derivative of position

with respect to time. Another way to write acceleration isusing the chainrule as ax = dvxdt =

dvxdx

dxdt = vx

dvxdx . Thus the acceleration can be written in

the alternative forms

ax =dvxdt

=d2xdt2

= vxdvxdx

(3.7)

You can check that the units are the same throughout.

Example When driving your car, what is your average accel-eration if you are able to reach 20 mph from rest in 5 seconds?

Solution

vxf = 20 mph vxi = 0t f = 5 seconds t i = 0

ax =20 mph 0

5 sec 0=

20 miles per hour5 seconds

= 4miles

hour seconds= 4 mph per sec

= 4miles

hour 13600 hour= 14 , 400 miles per hour 2

Now lets return to our previous Experiment and plot the acceleration-

time graphs corresponding to Fig. 3.2. The acceleration is simply the slopeof the velocity-time graph. Both velocity-time graphs have zero slope andso the acceleration-time graphs are always zero. This makes sense. Anobject at rest or constant speed in a striaght line does not accelerate. Thisis plotted in Fig. 3.3.


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3.7. CONSTANT ACCELERATION EQUATIONS 41

t

a

t(A)

t

a

t(B)

Figure 3.3 Acceleration-time graphs for motion depicted pre-viously in Fig. 3.2.

3.7 Constant Acceleration Equations

Velocity describes changing position and acceleration describes changing ve-locity. A quantity called jerk describes changing acceleration. However, veryoften the acceleration is constant , and we dont consider jerk. When drivingyour car the acceleration is usually constant when you speed up or slowdown or put on the brakes. (When you slow down or put on the brakes theacceleration is constant but negative and is called deceleration.) When youdrop an object and it falls to the ground it also has a constant acceleration.

When the acceleration is constant, then we can derive ve very handyequations that will tell us everything about the motion. We will now showhow to derive these equations using only algebra. When this is nished weshow the derivations using calculus.

3.7.1 Algebraic Derivation

We are going to use the following values:

t i 0t f t

and acceleration a is a constant so that axf = axi ax . Thus now t = t f t i = t 0 = t x = xf x i vx = vxf vxi ax = axf axi = ax ax = 0


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( a must be zero because we are only considering constant a.)

Also, because acceleration is constant then average acceleration is alwaysthe same as instantaneous acceleration

ax = ax

Now use the denition of average acceleration

ax = ax = vx t

=vxf vxi

t 0=

vxf vxit

ax t = vxf vxior

vxf = vxi + ax t (3.8)which is the rst of our constant acceleration equations. If you plot this ona (v, t ) graph, then it is a straight line of slope a, for a = constant. In thatcase the average velocity is (see Example below)

vx =12

(vxf + vxi )

From the denition of average velocity

vx = x t

=xf x i

t

xf x it = 12(vxf + vxi )

=12

(vxi + ax t + vxi )

giving

xf x i = vxi t +12

ax t2 (3.9)

which is the second of our constant acceleration equations. To get the otherthree constant acceleration equations, we will just combine the rst two asthe following examples show.


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Example When the acceleration is constant, show that

vx =12

(vxf + vxi )

Solution If the acceleration is constant then the equation vxf =vxi + ax t shows that a ( vx , t ) graph is a straight line of slope a.Such a graph is plotted in Figure 3.4.

t

vx

vxf

vxi

t

Figure 3.4 (vx , t ) graph for constant acceleration.

The area under the graph gives the position x, which is justthe area of the rectangle plus the area of the triangle. Thus

x = vxi t +12

(vxf vxi ) t=

12

(vxi + vxf ) t

giving

vx = x t= 1

2(vxf + vxi )


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Example Prove thatv2xf = v

2xi + 2 ax (x xi ) (3.10)

Solution Obviously t has been eliminated. From (3.8)

t =vxf vxi

axSubstituting into (3.9) gives

xf x i = vxivxf vxi

ax+

12

axvxf vxi

ax

2

ax (xf x i ) = vxi vxf v2xi +

12(v

2xf 2vxf vxi + v

2xi )

=12

(v2xf v2xi ) v

2xf = v

2xi + 2 ax (xf x i )

Example Prove that

xf x i =12

(vxi + vxf )t (3.11)

Solution Obviously ax has been eliminated. From (3.8)

ax = vxf vxit

Substituting into (3.9) gives

xf x i = vxi t +12

vxf vxit

t2

= vxi t +12

(vxf t vxi t)=

12

(vxi + vxf )t

The nal equation is

xf x i = vxf t 12

ax t2 (3.12)

This derivation is left to the Problems.


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3.7.2 Summary of Constant Acceleration equations

The 5 constant acceleration equations are:

vxf = vxi + ax tv2xf = v


xf x i =vxi + vxf

2t

= vxi t +12

ax t2

= vxf t 12

ax t2


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3.7.3 Calculus Derivation

The constant acceleration equations can be derived from integral calculusas follows.

Example Prove that vxf = vxi + ax t using calculus.

Solution For constant acceleration ax is not a function of po-sition x or time t. That is ax = ax (x) and ax = ax (t). Now

ax =dvxdt

and integrating both sides gives

t f

t iax dt = dvxdt dt

Given that ax is constant, we can take it outside the integralgiving

ax t f

t idt =

vxf

vxidvx

ax (t f t i ) = vxf vxiax (t

0) = vxf

vxi

vxf = vxi + ax t

Example Prove that xf x i = vxi t + 12 ax t2 using calculus.Solution We have

vx =dxdt

Intedgrate both sides with respect to t giving

vx dt = dxdt dtHowever now vx changes and so it cannot be taken outside theintegral. In fact the formula telling us how vx changes was


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vxf (t) = vxi + ax t and so we substitute this into the previous

expression to get

t f

t i(vxi + at )dt =

x f

x idx

vxi t +12

at 2t f

t i= xf x i

= vxi (t f t i ) +12

a(t f t i )2

= vxi (t 0) +12

ax (t 0)2

= vxi t +12

ax t2

which gives

xf x i = vxi t +12

ax t2

Example Prove that v2xf = v2xi + 2 ax (xf x i ) using calculus.Solution Here we use the chain rule,

ax =dvxdt

=dvxdx

dxdt

= vxdvxdx

Integrate both sides,

x f

x iax dx = vx dvxdx dxThe acceleration is constant and can be taken outside the inte-

gral,

ax x f

x idx =

vxf

vxivx dvx

ax (xf x i ) =12

v2xvxf

vxi

=12

v2xf v2xito nally give

v2xf = v2xi + 2 ax (xf x i )

One can now get the other equations using algebra.


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3.8 Free Fall

One of the most common instances of constant acceleration occurs whenan object is dropped near the surface of the Earth. An extraordinary fact,originally discovered by Galileo when dropping object from the tower of Pisa, is that all objects fall to the ground with the same acceleration if airresistance is neglected. We can easily demonstrate this with the followingexpreiment.

Experiment

A) Take two identical cans and ll one with water or dirt. Drop

them from the same height. They hit the ground at the sametime!

B) Drop a paper cup lled with water which has a hole in thebottom. Water leaks out if the cup is held stationary. Waterdoes not leak out while the cup is dropping!

We often think that lighter objects, such as a feather, fall more slowly thanheavier objects. But this is just because a light object such as a featherexperiences a lot of air resistance. In the experiment above air resistance isthe same for the two falling cans.

A famous demonstration done by Apollo astronauts on the Moon was todrop a feather and hammer at the same time. They hit the ground at thesame time because there is no air on the Moon.

If we neglect air resistance, then near the surface of Earth, all fallingobjects have same acceleration given by

a = g = 9 .8 m/sec 2

When we study gravitation in more detail we will be able to explain wherethis number comes from and also why all objects fall with the same accel-eration near the Earths surface.


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3.9. HISTORICAL NOTE 49

3.9 Historical Note

The constant acceleration equations were rst discovered by Galileo Galilei(1564 - 1642). Galileo is widely regarded as the father of modern sciencebecause he was really the rst person who went out and actually did ex-preiments to arrive at facts about nature, rather than relying solely onphilosophical argument. Galileo wrote two famous books entitled Dialoguesconcerning Two New Sciences [Macmillan, New York, 1933; QC 123.G13]and Dialogue concerning the Two Chief World Systems [QB 41.G1356].

In Two New Sciences we nd the following [Pg. 173]:

THEOREM I, PROPOSITION I : The time in which any spaceis traversed by a body starting from rest and uniformly accel-

erated is equal to the time in which that same space would betraversed by the same body moving at a unifrom speed whosevalue is the mean of the highest speed and the speed just beforeacceleration began.

In other words this is Galileos statement of our equation

x x i =12

(vxi + v)t (3.13)

We also nd [Pg. 174]:

THEOREM II, PROPOSITION II : The spaces described by afalling body from rest with a uniformly accelerated motion areto each other as the squares of the time intervals employed intraversing these distances.

This is Galileos statement of

x x i = vxi t +12

at 2 = vxf t 12

at 2 (3.14)

Galileo was able to test this equation with the simple device shown inFigure 3.5. This is basically a ball rolling down an inclined plane. Ob- jects moving down inclined planes are studied in all introductory physicscourses. Galileo is responsible for this! By the way, Galileo also inventedthe astronomical telescope.


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moveable fret wires

Figure 3.5 Galileos apparatus for verifying the constant ac-celeration equations.

[from From Quarks to the Cosmos Leon M. Lederman andDavid N. Schramm (Scientic American Library, New York, 1989)]


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3.10. PROBLEMS (8 QUESTIONS) 51


1. The following functions give the position as a function of time:

i) x = Aii) x = Bt

iii) x = Ct 2

iv) x = D cos tv) x = E sin t

where A,B,C,D,E, are constants.

A) What are the units for A,B,C,D,E, ?

B) Work out expressions for the velocity and acceleration. Indicate forwhat functions the acceleration is constant .

C) Sketch graphs of x,v,a as a function of time.

2. The gures below show position-time graphs. Sketch the correspond-ing velocity-time and acceleration-time graphs.

t

x

t

x

t

x

3. Suppose you drop an object from a height H above the ground.

A) Derive a formula for the speed with which the object hits the ground.(This is the speed the instant before the object touches the ground.)

B) Check your formula by making sure that the units on the right handside are the same as those on the left hand side.

C) If the height is 1.0 m, what is the numerical value of the nal speed?


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4. If you start your car from rest and accelerate to 30 mph in 10 seconds,

what is your acceleration in mph per sec and in miles per hour2

?5. If you throw a ball up vertically at speed V , with what speed does it

return to the ground? Prove your answer using the constant accelera-tion equations, and neglect air resistance.

6. Show that xf x i = vxf t 12 ax t2 follows from the other constantacceleration equations. Use only algebra.7. A car is travelling at constant speed v1 and passes a second car moving

at speed v2. The instant it passes, the driver of the second car decidesto try to catch up to the rst car, by stepping on the gas pedal andmoving at acceleration a. The rst car travels at constant speed v1and does not accelerate. Derive a formula for how long it takes tocatch up. Your formula should only involve t, v1, v2 and a.

8. A car is stopped at a red traffic light. When the light turns greenthe car starts accelerating with a constant acceleration of ac. At theinstant the light turns green, a truck travelling at constant speed vtpasses the car at the traffic light. Assuming the car keeps accelerating,then after some time the car will eventually cath up and pass the truck.Derive a formula for how long this takes.


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Chapter 4

VECTORS

4.1 Introduction

When we considered 1-dimensional motion in the last chapter we only hadtwo directions to worry about, namely motion to the Right or motion tothe Left and we indicated direction with a + or sign. We found thatthe following quantities had a direction (i.e. could take a + or sign):displacement, velocity and acceleration . Quantities that dont have a signwere distance, speed and magnitude of acceleration.

Now in 2 and 3 dimensions we need more than a + or sign. Thatswhere vectors come in.Vectors are quantities with both magnitude and direction.Scalars are quantities with magnitude only.

Examples of Vectors are: displacement, velocity, acceleration,force, momentum, electric eld

Examples of Scalars are: distance, speed, magnitude of acceler-ation, time, temperature

Vectors are usually either written as boldface quantities such as A or asquantities with a little arrow over the top as in A. Usually textbooks usethe A notation, but when writing things by hand or on the blackboard it iseasier to use the A notation, becuase it is difficult to write bold face whenwriting by hand. Throughout this book we use the A notation.

Before delving into vectors consider the following problem.

53


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54 CHAPTER 4. VECTORS

Example Joe and Mary are rowing a boat across a river whichis 40 m wide. They row in a direction perpendicular to the bank.However the river is owing downstream and by the time theyreach the other side, they end up 30 m downstream from theirstarting point. Over what total distance did the boat travel?

Solution Obviously the way to do this is with the triangle inFig. 4.1, and we deduce that the distance is 50 m.

30 m

40 m

50 m

Figure 4.1 Graphical solution to river problem.


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4.1. INTRODUCTION 55

Another way to think about the previous problem is with vectors , which

are little arrows whose orientation species direction and whose length spec-ies magnitude. The displacement along the river is represented as

Figure 4.2 Displacement along the river.

with a length of 30 m, denoted as A and the displacement across the river,denoted B ,

Figure 4.3 Displacement across the river.

with length of 40 m. To re-construct the previous triangle, the vectors are

added head-to-tail as in Fig. 4.4.

Figure 4.4 Vector addition solution to the river problem.

The resultant vector, denoted C , is obtained by lling in the triangle. Math-ematically we write C = A + B .


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4.2. TRIGONOMETRY 57

The side opposite the right angle is always called the Hypotenuse. Consider

one of the other angles, say .

Hypotenuse

Adjacent

Opposite

Figure 4.6 Right-angled triangle showing sides Opposite and Adjacent tothe angle .

The side adjacent to is called Adjacent and the side opposite is calledOpposite. Now consider the other angle . The Opposite and Adjacent sidesare switched because the angle is different.

Hypotenuse

Opposite

Adjacent

Figure 4.7 Right-angled triangle showing sides Opposite and Adjacent to

the angle .

Lets label Hypotenuse as H , Opposite as O and Adjacent as A. Pythago-ras theorem states

H 2 = A2 + O2


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This is true no matter how the Opposite and Adjacent sides are labelled, i.e.

if Opposite and Adjacent are interchanged, it doesnt matter for Pythagorastheorem.Often we are interested in dividing one side by another. Some possible

combinations are OH ,AH ,

OA . These special ratios are given special names.

OH

is called Sine. AH is called Cosine.OA is called Tangent. Remember them by

writing SOH, CAH, TOA. The names are usually abbreviated to sin, cos,tan.

Example Using the previous triangle for the river problem,write down sin , cos , tan , sin ,cos , tan .

Solution

sin =OH

=40 m50 m

=45

= 0 .8

cos =AH

=30 m50 m

=35

= 0 .6

tan =OA

=40 m30 m

=43

= 1 .33

sin =OH

=30 m50 m

=35

= 0 .6

cos =AH

=40 m50 m

=45

= 0 .8

tan =OA =

30 m40 m =

34 = 0 .75

30 m

40 m

50 m

Figure 4.8 Triangle for river problem.


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4.3. VECTOR COMPONENTS 59

Now whenever the sin of an angle is 0.8 the angle is always 53.1. Thus

= 53 .1. Again whenever tan of an angle is 0.75 the angle is always 36.9.So if we have calculated any of the ratios, sin, cos or tan then we alwaysknow what the corresponding angle is.

4.3 Vector Components

An arbitrary vector has both x and y components. These are like shadowson the x and y areas, as shown in Figure 4.9.

x

y

Ax

Ay A

Figure 4.9 Components, Ax and Ay , of vector A .

The components are denoted Ax and Ay and are obtained by dropping aperpendicular line from the vector to the x and y axes. Thats why weconsider trigonometry and right-angled triangles !


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A physical understanding of components can be obtained.

Experiment

Pull a cart with a rope at some angle to the ground, as shownin Fig. 4.11. Vary the angle and notice how the acceleration of the cart varies, even thought the pulling force is kept constant.

In this experiment the cart will move with a certain acceleration, deter-mined not by the force F , but by the component F x in the x direction. If you change the angle, the acceleration of the cart will change.


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4.3. VECTOR COMPONENTS 61

F

F x

Figure 4.10 Pulling a cart with a force F .

Lets re-draw Figure 4.10, writing A instead of F as follows:

A

Ax

Ay

Figure 4.11 Components and angles for Fig. 4.10.


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Lets denote the magnitude or length of A simply as A. A better notation

is |A |, but its quicker to just write A. However sometimes we will also use|A |. Pythagoras theorem givesA2 = A2x + A

2y

and alsotan =

AyAx

andtan =

AxAy

(Also sin =AyA , cos =

AxA , sin =

AxA , cos =

AyA .)

Thus if we have the components, Ax and Ay we can always get themagnitude and direction of the vector, namely A and (or ). Similarly if we start with A and (or ) we can always nd Ax and Ay .

4.4 Unit Vectors

A vector is completely specied by writing down magnitude and direction (i.e. A and ) or x and y components (Ax and Ay).

Theres another very useful and compact way to write vectors and that isby using unit vectors . The unit vector i is dened to always have a length of 1 and to always lie in the positive x direction, as in Fig. 4.12. (The symboli is used to denote these unit vectors, when writing them by hand.)

x

y

i

Figure 4.12 Unit vector i.


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4.4. UNIT VECTORS 63

Similarly the unit vector j is dened to always have a length of 1 also but

to lie entirely in the positive y direction.

x

y

j

Figure 4.13 Unit vector j .

The unit vector k lies in the positive z direction.

x

y

k

z

Figure 4.14 Unit vector k .

Thus any arbitrary vector A is now written as

A = Ax i + Ay j + Az k


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4.5 Vector Addition

Finally we will now see the use of components and unit vectors. Rememberhow we discussed adding vectors graphically using a ruler and protractor. Abetter method is with the use of components, because then we can get ouranswers by pure calculation.

In Fig. 4.16 we have shown two vectors A and B added to form C , butwe have also indicated all the components.

A x

C x

A y

B x

B yC y

BC

A

x

y

Figure 4.15 Adding vectors by components.

By carefully looking at the gure you can see that

C x = Ax + Bx

C y = Ay + By

This is a very important result.


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4.5. VECTOR ADDITION 65

Now lets back-track for a minute. When we write

C = A + B

you should say, Wait a minute! What does the + sign mean? We are usedto adding numbers such as 5 = 3 + 2, but in the above equation A , B andC are not numbers. They are these strange arrow-like objects called vectorswhich are added by putting head-to-tail. We should really write

C = A B

whereis a new type of addition, totally unlike adding numbers. HoweverAx , Bx , Ay , By , C x , C y are ordinary numbers and the + sign we used

above does denote ordinary addition. Thus C = A B actually meansC x = Ax + Bx and C y = Ay + By . The statement C = A B is reallyshorthand for two ordinary addition statements. Whenever anyone writes

something like D = F + E it actually means two things, namely Dx = F x + E xand Dy = F y + E y .

All of this is much more obvious with the use of unit vectors. WriteA = Ax i + Ay j and B = Bx i + By j and C = C x i + C y j. Now

C = A + B

is simply

C x i + C y j = Ax i + Ay j + Bx i + By j= ( Ax + Bx )i + ( Ay + By) j

and equating coefficients of i and j gives

C x = Ax + Bx

andC y = Ay + By


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Example Do the original river problem using components.Solution

A = 30 i B = 40 jC = A + B

C x i + C y j = Ax i + Ay j + Bx i + By jAy = 0 Bx = 0

C x i + C y j = 30 i + 40 jC x = 30 C y = 40

or C x = Ax + Bx = 30 + 0 = 30C y = Ay + By = 0 + 40 = 40C 2 = C 2x + C

2y = 30

2 + 40 2 = 900 + 1600 = 2500

C = 50

4.6 Vector Multiplication

4.6.1 Scalar Product

We know how to add vectors. Now lets learn how to multiply them.When we add vectors we always get a new vector, namely C = A +B . When we multiply vectors we get either a scalar or vector. There aretwo types of vector multiplication called scalar product and vector product.These are often also called dot product and cross product.

The scalar product is dened as

A B AB cos (4.1)where A and B are the magnitude of A and B respectively and is theangle between A and B . The whole quantity A B = AB cos is a scalar,i.e. it has magnitude only.

Based on our denition (4.1) we can work out the scalar products of allof the unit vectors.


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4.6. VECTOR MULTIPLICATION 67

Example Evaluate i iSolution i i = ii cos but i is the magnitude of i which is 1, and the angle is 0. Thus

i i = 1

Example Evaluate i jSolution i j = ij cos90 = 0

Thus we have i i = j j = k k = 1 and i j = i k = j k = j i = k i =k j = 0.Now any vector can be written in terms of unit vectors as A = Ax i +Ay j + Az k and B = Bx i + By j + Bz k . Thus the scalar product of any twoarbitrary vectors is

A B = AB cos = ( Ax i + Ay j + Az k ) (Bx i + By j + Bz k )= Ax Bx + AyBy + Az Bz

Thus we have a new formula for scalar product, namely

A B = Ax Bx + AyBy + Az Bz (4.2)which has been derived from the original denition (4.1) using unit vectors.

Whats the good of all this? Well for one thing its now easy to gureout the angle between vectors, as the next example shows.

Example What is the angle between A = i + j and B = i j?Solution We have

A B = AxBx + AyBy= 1

1 = 0

= AB cos

= 90o


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4.6.2 Vector Product

In making up the denition of vector product we have to dene its magnitudeand direction. The symbol for vector product is A B . Given that the resultis a vector lets write C A B . The magnitude is dened as

C = AB sin

and the direction is dened to follow the right hand rule. ( C = thumb, A= forenger, B = middle nger.)

Example Evaluate i jSolution |i j| = ij sin90 = 1The direction is in direction k

Thus i j = k

Example Evaluate k kSolution |k k | = kk sin 0 = 0Thus k k = 0

Therefore

i j = k j k = i k i = j j i = k k j = i i k = j

andi i = j j = k k = 0

Thus the vector product of any two arbitrary vectors is

A B = ( Ax i + Ay j + Az k ) (Bx i + By j + Bz k )which gives a new formula for vector product, namely

A B = ( AyBz Az By)i + ( Az Bx Ax Bz ) j+( Ax By AyBx )k


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4.7. PROBLEMS (7 QUESTIONS) 69


1. Calculate the angle between the vectors A = i + 2 j and B = j k .Use A) the scalar product and then B) the vector product to obtainyour answer, which should be the same in both cases.

2. Evaluate ( A + 2 B ) C where A = i + 2 j and B = j k and C = i j.3. Two vectors are dened as A = j + k and B = i + j. Evaluate:

A) A + B

B) A BC) A BD) A B

4. Using the right hand rule, gure out the direction of the followingvector products.A) i kB) i (k )C) j iD) ( j) (k )E) ( j) iF) (k ) (i)

5. Write A = Ax i + Ay j + Az k and similarly for B and C . Work out a

formula for A (B C ) in terms of the components of A , B and C .6. An airplane pilot wishes to y 100 mph North. However a wind is

blowing at 30 mph towards the West. What speed and direction shouldthe pilot y the plane? Give the direction as the number of degreesNorth of East. Work out the problem using unit vectors.

7. A pilot is aiming her plane in a direction 45 o South of West at a speedof 200 mph. A wind is blowing at 50 mph in a direction 30 o Eastof North. What will be the resulting speed and direction of the planewith respect to the ground? Use unit vectors to work out this problem.


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Chapter 5

2- AND 3-DIMENSIONALMOTION

5.1 Displacement, Velocity and Acceleration

In this chapter we will go over everything we did in Chapter 3 concerningmotion, except that now the entire discussion will involve vectors.

In Chapter 3 we used the coordinate x alone to denote position. Howeverfor 3-dimensions position is generally described with the position vector

r = xi + y j + zk .

Now in Chapter 3, displacement was dened as a change in position, namelydisplacement = x = xf x i . In 3-dimensions, displacement is dened asthe change in position vector,

displacement = r = r f r i= xi + y j + zk= ( xf x i)i + ( yf yi ) j + ( zf zi )k

Thus displacement is a vector .In 1-dimension, the average velocity was dened as displacement divided

by time interval or vx x t =x f x it f t i . Similarly, in 3-dimensions averagevelocity is dened as

v r t

=r f r it f t i

= xi + y j + zk

t

71


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72 CHAPTER 5. 2- AND 3-DIMENSIONAL MOTION

= x t

i + y t

j + z t

k

= vx i + vy j + vz k

For 1-dimension, the instantaneous velocity, or just velocity, was dened asvx dxdt . In 3-dimensions we dene velocity as

v drdt

=ddt

(xi + y j + zk )

=dxdt

i +dydt

j +dzdt

k

= vx i + vy j + vz k

Thus velocity is a vector . Note the following.

The instantaneous velocity of a particle is always tangent to thepath of the particle.

Of course thats because the velocity is the derivative of the positionvector r , and we saw that the derivative is just the slope.

Again we follow the denitions made for 1-dimension. In 3-dimensions,the average acceleration is dened as

a v t

=v f v it f t i

and acceleration (instantaneous acceleration) is dened asa =

dvdt

Thus acceleration is also a vector .

5.2 Constant Acceleration Equations

In 1-dimension, our basic denitions were

vx = x t

vx =dxdt

ax = v t

ax =dvdt


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We found that if the acceleration is constant, then from these equations we

can prove thatvxf = vxi + ax tv2xf = v


xf x i =vxi + vxf

2t

= vxi t +12

ax t2

= vxf t 12

ax t2

which are known as the 5 constant acceleration equations.

In 3-dimensions we hadv

r t

orvx i + vy j + vz k =

x t

i + y t

j + z t

k

orvx =

x t

, vy + y t

, vz = z t

These 3 equations are the meaning of the rst vector equation v r t .Similarlyv

drdt

orvx =

dxdt

, vy =dydt

, vz =dzdt

Similarly

a v t

orax =

vx t

, ay = vy t

, az = vz t

anda

dvdt

orax =

dvxdt

, ay =dvydt

, az =dvzdt

So we see that in 3-dimensions the equations are the same as in 1-dimension except that we have 3 sets of them; one for each dimension. Thus


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if the 3-dimensional acceleration vector a is now constant, then ax , ay and

az must all be constant. Thus we will have 3 sets of constant accelerationequations, namely

vxf = vxi + ax tv2xf = v


xf x i =vxi + vxf

2t

= vxi t +12

ax t2

= vxf t 12

ax t2

and

vyf = vyi + aytv2yf = v

2yi + 2 ay(yf yi )

yf yi =vyi + vyf

2t

= vyi t +12

ayt2

= vyf t 12

ayt2

and

vzf = vzi + az tv2zf = v

2zi + 2 az (zf zi)

zf zi =vzi + vzf

2t

= vzi t +12

az t2

= vzf t 12

az t2

These 3 sets of constant acceleration equations are easy to remember. Theyare the same as the old ones in 1-dimension except now they have subscriptsy and z in addition to x.


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5.3 Projectile Motion

Most motion in 3-dimensions actually only occurs in 2-dimensions. Theclassic example is kicking a football off the ground. It follows a 2-dimensionalcurve, as shown in Fig. 5.1. Thus we can ignore all motion in the z directionand just analyze the x and y directions. Also we shall ignore air resistance.

v i

vxi

vyi

range, R

Figure 5.1 Projectile Motion.


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5.3. PROJECTILE MOTION 77

Example A football is kicked off the ground with an initialvelocity of v i at an angle to the ground. Write down the xconstant acceleration equation in simplied form. (Ignore airresistance)

Solution The x direction is easiest to deal with, because there isno acceleration in the x direction after the ball has been kicked,i.e. ax = 0. Thus the constant acceleration equations in the xdirection become

vxf = vxiv2xf = v

2xi

xf x i = vxi + vxf

2t = vxi t = vxf t

xf x i = vxi t= vxf t (5.1)

The rst equation ( vxf = vxi ) makes perfect sense because if ax = 0 then the speed in the x direction is constant , whichmeans vxf = vxi . The second equation just says the same thing.If vxf = vxi then of course also v2xf = v2xi . In the third equationwe also use vxf = vxi to get

vxi + vxf 2 =

vxi + vxi2 = vxi or

vxi + vxf 2 =

vxf + vxf 2 = vxf . The fourth and fth equations are also consistent

with vxf = vxi , and simply say that distance = speed timewhen the acceleration is 0.Now, what is vxi in terms of vi |v i | and ? Well, from Fig. 5.1we see that vxi = vi cos and vyi = vi sin . Thus (5.1) becomes

xf x i = vi cos t


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Example What is the form of the y-direction constant acceler-ation equations from the previous example?

Solution Can we also simplify the constant acceleration equa-tions for the y direction? No. In the y direction the accelerationis constant ay = g but not zero. Thus the y direction equationsdont simplify at all, except that we know that the value of ay isg or 9.8 m/sec 2.Also we can write vyi = vi sin . Thus the equations for the ydirection are

vyf = vi sin gtv2yf = ( vi sin )2 2g(yf yi )yf yi =

vi sin + vy2

t

yf yi = vi sin t 12

gt2

An important thing to notice is that t never gets an x, y or z subscript.This is because t is the same for all 3 components, i.e. t = tx = ty = tz .

LECTURE DEMONSTRATIONS1) Drop an object: it accerates in y direction.

Air track: no acceleration in x direction.

2) Push 2 objects off table at same time. One falls in vertical path andthe other on parabolic trajectory but both hit ground at same time.

3) Monkey shoot.


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Example The total horizontal distance (called the Range) thata football will travel when kicked, depends upon the initial speedand angle that it leaves the ground. Derive a formula for theRange, and show that the maximum Range occurs for = 45 .(Ignore air resistance and the spin of the football.)

Solution The Range, R is just

R = xf x i = vxi t= vi cos t

Given vi and we could calculate the range if we had t. We getthis the y direction equation. From the previous example we had

yf yi = vi sin t 12

gt2

But for this example, we have yf yi = 0. Thus0 = vi sin t

12

gt2

0 = vi sin 12

gt

t = 2vi sin g

Substituting into our Range formula above gives

R = vi cos t

=2v2i sin cos

g

=v2i sin2

g

using the formula sin2 = 2sin cos . Now R will be largest

when sin2 is largest which occurs when 2 = 90o. Thus = 45

o.


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COMPUTER SIMULATION (Interactive Physics): Air Drop.

H=200 m

R=400 m

origin

Figure 5.2 Air Drop.

Example A rescue plane wants to drop supplies to isolatedmountain climbers on a rocky ridge a distance H below. Theplane is travelling horizontally at a speed of vxi . The planereleases supplies a horizontal distance of R in advance of themountain climbers. Derive a formula in terms of H, vxi , R and

g, for the vertical velocity (up or down) that the supplies shouldbe given so they land exactly at the climbers position. If H=200m, vxi = 250 km/hr and R=400 m, calculate a numerical valuefor this speed. (See Figure 5.2.)


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Solution Lets put the origin at the plane. See Fig. 5.2. The

initial speed of supplies when released is vxi = +250 km/hourxf x i = R 0 = R

ay = gyf yi = 0 H = H (note the minus sign !)

We want to nd the initial vertical velocity of the supplies,namely vyi . We can get this from

yf yi = vyi t +12

ayt2 = H = vyi t

12

gt2

or vyi = H t +12

gt

and we get t from the x direction, namely

xf x i = vxi t = R

t =Rvxi

giving

vyi = H vxiR +12

gRvxi

which is the formula we seek. Lets now put in numbers:

= 200m 25

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