FOR SCIENTISTS AND ENGINEERS physics a strategic approach THIRD EDITION randall d. knight © 2013 Pearson Education, Inc. Chapter 2 Lecture

© 2013 Pearson Education, Inc. Chapter Goal: To learn how to solve problems about motion in a straight line. Chapter 2 Kinematics in One Dimension Slide 2-2

© 2013 Pearson Education, Inc.  Kinematics is the name for the mathematical description of motion.  This chapter deals with motion along a straight line, i.e., runners, rockets, skiers.  The motion of an object is described by its position, velocity, and acceleration.  In one dimension, these quantities are represented by x, v x, and a x.  You learned to show these on motion diagrams in Chapter 1. Chapter 2 Preview Slide 2-3

© 2013 Pearson Education, Inc. Chapter 2 Preview Slide 2-4

© 2013 Pearson Education, Inc. Chapter 2 Preview Slide 2-5

© 2013 Pearson Education, Inc. Chapter 2 Reading Quiz Slide 2-6

© 2013 Pearson Education, Inc. The slope at a point on a position-versus-time graph of an object is A.The object’s speed at that point. B.The object’s average velocity at that point. C.The object’s instantaneous velocity at that point. D.The object’s acceleration at that point. E.The distance traveled by the object to that point. Reading Question 2.1 Slide 2-7

© 2013 Pearson Education, Inc. The slope at a point on a position-versus-time graph of an object is A.The object’s speed at that point. B.The object’s average velocity at that point. C.The object’s instantaneous velocity at that point. D.The object’s acceleration at that point. E.The distance traveled by the object to that point. Reading Question 2.1 Slide 2-8

© 2013 Pearson Education, Inc. The area under a velocity-versus-time graph of an object is A.The object’s speed at that point. B.The object’s acceleration at that point. C.The distance traveled by the object. D.The displacement of the object. E.This topic was not covered in this chapter. Reading Question 2.2 Slide 2-9

© 2013 Pearson Education, Inc. The area under a velocity-versus-time graph of an object is A.The object’s speed at that point. B.The object’s acceleration at that point. C.The distance traveled by the object. D.The displacement of the object. E.This topic was not covered in this chapter. Reading Question 2.2 Slide 2-10

© 2013 Pearson Education, Inc. The slope at a point on a velocity-versus-time graph of an object is A.The object’s speed at that point. B.The object’s instantaneous acceleration at that point. C.The distance traveled by the object. D.The displacement of the object. E.The object’s instantaneous velocity at that point. Reading Question 2.3 Slide 2-11

© 2013 Pearson Education, Inc. The slope at a point on a velocity-versus-time graph of an object is A.The object’s speed at that point. B.The object’s instantaneous acceleration at that point. C.The distance traveled by the object. D.The displacement of the object. E.The object’s instantaneous velocity at that point. Reading Question 2.3 Slide 2-12

© 2013 Pearson Education, Inc. Suppose we define the y-axis to point vertically upward. When an object is in free fall, it has acceleration in the y-direction A. a y   g, where g   9.80 m/s 2. B. a y  g, where g   9.80 m/s 2. C. Which is negative and increases in magnitude as it falls. D. Which is negative and decreases in magnitude as it falls. E. Which depends on the mass of the object. Reading Question 2.4 Slide 2-13

© 2013 Pearson Education, Inc. Suppose we define the y-axis to point vertically upward. When an object is in free fall, it has acceleration in the y-direction A. a y   g, where g   9.80 m/s 2. B. a y  g, where g   9.80 m/s 2. C. Which is negative and increases in magnitude as it falls. D. Which is negative and decreases in magnitude as it falls. E. Which depends on the mass of the object. Reading Question 2.4 Slide 2-14

© 2013 Pearson Education, Inc. At the turning point of an object, A.The instantaneous velocity is zero. B.The acceleration is zero. C.Both A and B are true. D.Neither A nor B is true. E.This topic was not covered in this chapter. Reading Question 2.5 Slide 2-15

© 2013 Pearson Education, Inc. At the turning point of an object, A.The instantaneous velocity is zero. B.The acceleration is zero. C.Both A and B are true. D.Neither A nor B is true. E.This topic was not covered in this chapter. Reading Question 2.5 Slide 2-16

© 2013 Pearson Education, Inc. A 1-pound block and a 100-pound block are placed side by side at the top of a frictionless hill. Each is given a very light tap to begin their race to the bottom of the hill. In the absence of air resistance A.The 1-pound block wins the race. B.The 100-pound block wins the race. C.The two blocks end in a tie. D.There’s not enough information to determine which block wins the race. Reading Question 2.6 Slide 2-17

© 2013 Pearson Education, Inc. A 1-pound block and a 100-pound block are placed side by side at the top of a frictionless hill. Each is given a very light tap to begin their race to the bottom of the hill. In the absence of air resistance A.The 1-pound block wins the race. B.The 100-pound block wins the race. C.The two blocks end in a tie. D.There’s not enough information to determine which block wins the race. Reading Question 2.6 Slide 2-18

© 2013 Pearson Education, Inc. Chapter 2 Content, Examples, and QuickCheck Questions Slide 2-19

© 2013 Pearson Education, Inc.  If you drive your car at a perfectly steady 60 mph, this means you change your position by 60 miles for every time interval of 1 hour.  Uniform motion is when equal displacements occur during any successive equal-time intervals.  Uniform motion is always along a straight line. Uniform Motion Slide 2-20 Riding steadily over level ground is a good example of uniform motion.

© 2013 Pearson Education, Inc.  An object’s motion is uniform if and only if its position-versus-time graph is a straight line.  The average velocity is the slope of the position- versus-time graph.  The SI units of velocity are m/s. Uniform Motion Slide 2-21

© 2013 Pearson Education, Inc. Tactics: Interpreting Position-versus-Time Graphs Slide 2-22

© 2013 Pearson Education, Inc. Example 2.1 Skating with Constant Velocity Slide 2-23

© 2013 Pearson Education, Inc. Example 2.1 Skating with Constant Velocity Slide 2-24

© 2013 Pearson Education, Inc. Example 2.1 Skating with Constant Velocity Slide 2-25

© 2013 Pearson Education, Inc. Example 2.1 Skating with Constant Velocity Slide 2-26

© 2013 Pearson Education, Inc.  Consider an object in uniform motion along the s-axis, as shown in the graph.  The object’s initial position is s i at time t i.  At a later time t f the object’s final position is s f.  The change in time is  t  t f  t i.  The final position can be found as: The Mathematics of Uniform Motion Slide 2-27

© 2013 Pearson Education, Inc.  The distance an object travels is a scalar quantity, independent of direction.  The displacement of an object is a vector quantity, equal to the final position minus the initial position.  An object’s speed v is scalar quantity, independent of direction.  Speed is how fast an object is going; it is always positive.  Velocity is a vector quantity that includes direction.  In one dimension the direction of velocity is specified by the  or  sign. Scalars and Vectors Slide 2-28

© 2013 Pearson Education, Inc. An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are A.50 cm and 50 cm. B.30 cm and 50 cm. C.50 cm and 30 cm. D.50 cm and –50 cm. E.50 cm and –30 cm. QuickCheck 2.1 Slide 2-29

© 2013 Pearson Education, Inc. An ant zig-zags back and forth on a picnic table as shown. The ant’s distance traveled and displacement are A.50 cm and 50 cm. B.30 cm and 50 cm. C.50 cm and 30 cm. D.50 cm and –50 cm. E.50 cm and –30 cm. QuickCheck 2.1 Slide 2-30

© 2013 Pearson Education, Inc.  An object that is speeding up or slowing down is not in uniform motion.  In this case, the position-versus-time graph is not a straight line.  We can determine the average speed v avg between any two times separated by time interval  t by finding the slope of the straight-line connection between the two points.  The instantaneous velocity is the object’s velocity at a single instant of time t.  The average velocity v avg   s/  t becomes a better and better approximation to the instantaneous velocity as  t gets smaller and smaller. Instantaneous Velocity Slide 2-31

© 2013 Pearson Education, Inc. Instantaneous Velocity Slide 2-32 Motion diagrams and position graphs of an accelerating rocket.

© 2013 Pearson Education, Inc.  As ∆t continues to get smaller, the average velocity v avg  ∆s/∆t reaches a constant or limiting value.  The instantaneous velocity at time t is the average velocity during a time interval ∆t centered on t, as ∆t approaches zero.  In calculus, this is called the derivative of s with respect to t.  Graphically, ∆s/∆t is the slope of a straight line.  In the limit ∆t  0, the straight line is tangent to the curve.  The instantaneous velocity at time t is the slope of the line that is tangent to the position-versus-time graph at time t. Instantaneous Velocity Slide 2-33

© 2013 Pearson Education, Inc. The slope at a point on a position-versus-time graph of an object is A.The object’s speed at that point. B.The object’s velocity at that point. C.The object’s acceleration at that point. D.The distance traveled by the object to that point. E.I really have no idea. QuickCheck 2.2 Slide 2-34

© 2013 Pearson Education, Inc. The slope at a point on a position-versus-time graph of an object is A.The object’s speed at that point. B.The object’s velocity at that point. C.The object’s acceleration at that point. D.The distance traveled by the object to that point. E.I really have no idea. QuickCheck 2.2 Slide 2-35

© 2013 Pearson Education, Inc. Example 2.4 Finding Velocity from Position Graphically Slide 2-36

© 2013 Pearson Education, Inc. Example 2.4 Finding Velocity from Position Graphically Slide 2-37

© 2013 Pearson Education, Inc. Example 2.4 Finding Velocity from Position Graphically Slide 2-38

© 2013 Pearson Education, Inc. Example 2.4 Finding Velocity from Position Graphically Slide 2-39

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? QuickCheck 2.3 Slide 2-40

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which position-versus-time graph matches this motion diagram? QuickCheck 2.3 Slide 2-41

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? QuickCheck 2.4 Slide 2-42 E. None of the above.

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? QuickCheck 2.4 Slide 2-43 E. None of the above.

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? QuickCheck 2.5 Slide 2-44

© 2013 Pearson Education, Inc. Here is a motion diagram of a car moving along a straight road: Which velocity-versus-time graph matches this motion diagram? QuickCheck 2.5 Slide 2-45

© 2013 Pearson Education, Inc.  ds/dt is called the derivative of s with respect to t.  ds/dt is the slope of the line that is tangent to the position-versus-time graph.  Consider a function u that depends on time as u  ct n, where c and n are constants:  The derivative of a constant is zero:  The derivative of a sum is the sum of the derivatives. If u and w are two separate functions of time, then: A Little Calculus: Derivatives Slide 2-46

© 2013 Pearson Education, Inc. Suppose the position of a particle as a function of time is s = 2t 2 m where t is in s. What is the particle’s velocity? Derivative Example Slide 2-47  Velocity is the derivative of s with respect to t:  The figure shows the particle’s position and velocity graphs.  The value of the velocity graph at any instant of time is the slope of the position graph at that same time.

© 2013 Pearson Education, Inc. Here is a position graph of an object: At t = 1.5 s, the object’s velocity is A.40 m/s. B.20 m/s. C.10 m/s. D.–10 m/s. E.None of the above. QuickCheck 2.6 Slide 2-48

© 2013 Pearson Education, Inc. Here is a position graph of an object: At t = 1.5 s, the object’s velocity is A.40 m/s. B.20 m/s. C.10 m/s. D.–10 m/s. E.None of the above. QuickCheck 2.6 Slide 2-49

© 2013 Pearson Education, Inc. Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A.40 m/s. B.20 m/s. C.10 m/s. D.–10 m/s. E.None of the above. QuickCheck 2.7 Slide 2-50

© 2013 Pearson Education, Inc. Here is a position graph of an object: At t = 3.0 s, the object’s velocity is A.40 m/s. B.20 m/s. C.10 m/s. D.–10 m/s. E.None of the above. QuickCheck 2.7 Slide 2-51

© 2013 Pearson Education, Inc. When do objects 1 and 2 have the same velocity? A.At some instant before time t 0. B.At time t 0. C.At some instant after time t 0. D.Both A and B. E.Never. QuickCheck 2.8 Slide 2-52

© 2013 Pearson Education, Inc. When do objects 1 and 2 have the same velocity? A.At some instant before time t 0. B.At time t 0. C.At some instant after time t 0. D.Both A and B. E.Never. QuickCheck 2.8 Slide 2-53 Same slope at this time

© 2013 Pearson Education, Inc.  Suppose we know an object’s position to be s i at an initial time t i.  We also know the velocity as a function of time between t i and some later time t f.  Even if the velocity is not constant, we can divide the motion into N steps in which it is approximately constant, and compute the final position as:  The curlicue symbol is called an integral.  The expression on the right is read, “the integral of v s dt from t i to t f.” Finding Position from Velocity Slide 2-54

© 2013 Pearson Education, Inc.  The integral may be interpreted graphically as the total area enclosed between the t-axis and the velocity curve.  The total displacement ∆s is called the “area under the curve.” Finding Position From Velocity Slide 2-55

© 2013 Pearson Education, Inc. Here is the velocity graph of an object that is at the origin (x  0 m) at t  0 s. At t  4.0 s, the object’s position is A.20 m. B.16 m. C.12 m. D.8 m. E.4 m. QuickCheck 2.9 Slide 2-56

© 2013 Pearson Education, Inc. Here is the velocity graph of an object that is at the origin (x  0 m) at t  0 s. At t  4.0 s, the object’s position is A.20 m. B.16 m. C.12 m. D.8 m. E.4 m. QuickCheck 2.9 Slide 2-57 Displacement  area under the curve

© 2013 Pearson Education, Inc. Example 2.6 The Displacement During a Drag Race Slide 2-58

© 2013 Pearson Education, Inc. Example 2.6 The Displacement During a Drag Race Slide 2-59

© 2013 Pearson Education, Inc.  Taking the derivative of a function is equivalent to finding the slope of a graph of the function.  Similarly, evaluating an integral is equivalent to finding the area under a graph of the function.  Consider a function u that depends on time as u  ct n, where c and n are constants:  The vertical bar in the third step means the integral evaluated at t f minus the integral evaluated at t i.  The integral of a sum is the sum of the integrals. If u and w are two separate functions of time, then: A Little More Calculus: Integrals Slide 2-60

© 2013 Pearson Education, Inc. Which velocity-versus-time graph goes with this position graph? QuickCheck 2.10 Slide 2-61

© 2013 Pearson Education, Inc. QuickCheck 2.10 Slide 2-62 Which velocity-versus-time graph goes with this position graph?

© 2013 Pearson Education, Inc.  Imagine a competition between a Volkswagen Beetle and a Porsche to see which can achieve a velocity of 30 m/s in the shortest time.  The table shows the velocity of each car, and the figure shows the velocity-versus-time graphs.  Both cars achieved every velocity between 0 and 30 m/s, so neither is faster.  But for the Porsche, the rate at which the velocity changed was: Acceleration Slide 2-63

© 2013 Pearson Education, Inc.  The SI units of acceleration are (m/s)/s, or m/s 2.  It is the rate of change of velocity and measures how quickly or slowly an object’s velocity changes.  The average acceleration during a time interval ∆t is:  Graphically, a avg is the slope of a straight-line velocity- versus-time graph.  If acceleration is constant, the acceleration a s is the same as a avg.  Acceleration, like velocity, is a vector quantity and has both magnitude and direction. Motion with Constant Acceleration Slide 2-64

© 2013 Pearson Education, Inc. A cart slows down while moving away from the origin. What do the position and velocity graphs look like? QuickCheck 2.11 Slide 2-65

© 2013 Pearson Education, Inc. A cart slows down while moving away from the origin. What do the position and velocity graphs look like? QuickCheck 2.11 Slide 2-66

© 2013 Pearson Education, Inc. A cart speeds up toward the origin. What do the position and velocity graphs look like? QuickCheck 2.12 Slide 2-67

© 2013 Pearson Education, Inc. A cart speeds up toward the origin. What do the position and velocity graphs look like? QuickCheck 2.12 Slide 2-68

© 2013 Pearson Education, Inc. Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration a x is A.Positive. B.Negative. C.Zero. QuickCheck 2.13 Slide 2-69

© 2013 Pearson Education, Inc. Here is a motion diagram of a car speeding up on a straight road: The sign of the acceleration a x is A.Positive. B.Negative. C.Zero. QuickCheck 2.13 Slide 2-70 Speeding up means v x and a x have the same sign.

© 2013 Pearson Education, Inc. Example 2.10 Running the Court Slide 2-71

© 2013 Pearson Education, Inc. Example 2.10 Running the Court Slide 2-72

© 2013 Pearson Education, Inc. Example 2.10 Running the Court Slide 2-73

© 2013 Pearson Education, Inc. Example 2.10 Running the Court Slide 2-74

© 2013 Pearson Education, Inc. A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.14 Slide 2-75

© 2013 Pearson Education, Inc. A cart speeds up while moving away from the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.14 Slide 2-76

© 2013 Pearson Education, Inc. A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.15 Slide 2-77

© 2013 Pearson Education, Inc. A cart slows down while moving away from the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.15 Slide 2-78

© 2013 Pearson Education, Inc. A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.16 Slide 2-79

© 2013 Pearson Education, Inc. A cart speeds up while moving toward the origin. What do the velocity and acceleration graphs look like? QuickCheck 2.16 Slide 2-80

© 2013 Pearson Education, Inc.  Suppose we know an object’s velocity to be v is at an initial time t i.  We also know the object has a constant acceleration of a s over the time interval ∆t  t f − t i.  We can then find the object’s velocity at the later time t f as: The Kinematic Equations of Constant Acceleration Slide 2-81

© 2013 Pearson Education, Inc.  Suppose we know an object’s position to be s i at an initial time t i.  It’s constant acceleration a s is shown in graph (a).  The velocity-versus-time graph is shown in graph (b).  The final position s f is s i plus the area under the curve of v s between t i and t f : The Kinematic Equations of Constant Acceleration Slide 2-82

© 2013 Pearson Education, Inc.  Suppose we know an object’s velocity to be v is at an initial position s i.  We also know the object has a constant acceleration of a s while it travels a total displacement of ∆s  s f − s i.  We can then find the object’s velocity at the final position s f : The Kinematic Equations of Constant Acceleration Slide 2-83

© 2013 Pearson Education, Inc. The Kinematic Equations of Constant Acceleration Slide 2-84

© 2013 Pearson Education, Inc. The Kinematic Equations of Constant Acceleration Slide 2-85 Motion with constant velocity and constant acceleration. These graphs assume s i = 0, v is > 0, and (for constant acceleration) a s > 0.

© 2013 Pearson Education, Inc. The Kinematic Equations of Constant Acceleration Slide 2-86

© 2013 Pearson Education, Inc. Example 2.12 Friday Night Football Slide 2-87

© 2013 Pearson Education, Inc. Example 2.12 Friday Night Football Slide 2-88

© 2013 Pearson Education, Inc. Example 2.12 Friday Night Football Slide 2-89

© 2013 Pearson Education, Inc. Example 2.12 Friday Night Football Slide 2-90

© 2013 Pearson Education, Inc. Example 2.12 Friday Night Football Slide 2-91

© 2013 Pearson Education, Inc. Which velocity-versus-time graph goes with this acceleration graph? QuickCheck 2.17 Slide 2-92

© 2013 Pearson Education, Inc. Which velocity-versus-time graph goes with this acceleration graph? QuickCheck 2.17 Slide 2-93

© 2013 Pearson Education, Inc.  The motion of an object moving under the influence of gravity only, and no other forces, is called free fall.  Two objects dropped from the same height will, if air resistance can be neglected, hit the ground at the same time and with the same speed.  Consequently, any two objects in free fall, regardless of their mass, have the same acceleration: Free Fall Slide 2-94 In the absence of air resistance, any two objects fall at the same rate and hit the ground at the same time. The apple and feather seen here are falling in a vacuum.

© 2013 Pearson Education, Inc.  Figure (a) shows the motion diagram of an object that was released from rest and falls freely.  Figure (b) shows the object’s velocity graph.  The velocity graph is a straight line with a slope:  Where g is a positive number which is equal to 9.80 m/s 2 on the surface of the earth.  Other planets have different values of g. Free Fall Slide 2-95

© 2013 Pearson Education, Inc. A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration a y is A.Positive. B.Negative. C.Zero. QuickCheck 2.18 Slide 2-96

© 2013 Pearson Education, Inc. A ball is tossed straight up in the air. At its very highest point, the ball’s instantaneous acceleration a y is A.Positive. B.Negative. C.Zero. QuickCheck 2.18 Slide 2-97

© 2013 Pearson Education, Inc. Example 2.14 Finding the Height of a Leap Slide 2-98

© 2013 Pearson Education, Inc. Example 2.14 Finding the Height of a Leap Slide 2-99

© 2013 Pearson Education, Inc. Example 2.14 Finding the Height of a Leap Slide 2-100

© 2013 Pearson Education, Inc. Example 2.14 Finding the Height of a Leap Slide 2-101

© 2013 Pearson Education, Inc.  Figure (a) shows the motion diagram of an object sliding down a straight, frictionless inclined plane.  Figure (b) shows the the free-fall acceleration the object would have if the incline suddenly vanished.  This vector can be broken into two pieces: and.  The surface somehow “blocks”, so the one-dimensional acceleration along the incline is  The correct sign depends on the direction the ramp is tilted. Motion on an Inclined Plane Slide 2-102

© 2013 Pearson Education, Inc. A ball rolls up the ramp, then back down. Which is the correct acceleration graph? QuickCheck 2.19 Slide 2-103

© 2013 Pearson Education, Inc. A ball rolls up the ramp, then back down. Which is the correct acceleration graph? QuickCheck 2.19 Slide 2-104

© 2013 Pearson Education, Inc. Tactics: Interpreting Graphical Representations of Motion Slide 2-105

© 2013 Pearson Education, Inc. Example 2.16 From Track to Graphs Slide 2-106

© 2013 Pearson Education, Inc. Example 2.16 From Track to Graphs Slide 2-107

© 2013 Pearson Education, Inc. Example 2.16 From Track to Graphs Slide 2-108

© 2013 Pearson Education, Inc. Example 2.16 From Track to Graphs Slide 2-109

© 2013 Pearson Education, Inc. Example 2.17 From Graphs to Track Slide 2-110

© 2013 Pearson Education, Inc. Example 2.17 From Graphs to Track Slide 2-111

© 2013 Pearson Education, Inc.  Figure (a) shows a realistic velocity-versus-time graph for a car leaving a stop sign.  The graph is not a straight line, so this is not motion with a constant acceleration.  Figure (b) shows the car’s acceleration graph.  The instantaneous acceleration a s is the slope of the line that is tangent to the velocity-versus-time curve at time t. Instantaneous Acceleration Slide 2-112

© 2013 Pearson Education, Inc.  Suppose we know an object’s velocity to be v is at an initial time t i.  We also know the acceleration as a function of time between t i and some later time t f.  Even if the acceleration is not constant, we can divide the motion into N steps of length ∆t in which it is approximately constant.  In the limit ∆t  0 we can compute the final velocity as:  The integral may be interpreted graphically a s the area under the acceleration curve as between t i and t f. Instantaneous Acceleration Slide 2-113

© 2013 Pearson Education, Inc. Example 2.19 Finding Velocity from Acceleration Slide 2-114

© 2013 Pearson Education, Inc. Example 2.19 Finding Velocity from Acceleration Slide 2-115

© 2013 Pearson Education, Inc. Chapter 2 Summary Slides Slide 2-116

© 2013 Pearson Education, Inc. General Principles Slide 2-117

© 2013 Pearson Education, Inc. General Principles Slide 2-118

© 2013 Pearson Education, Inc. Important Concepts Slide 2-119

© 2013 Pearson Education, Inc. Important Concepts Slide 2-120