
Motion in one direction
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1. If the uniform speed of a car 300 meters in a half minute, then its speed equals .......................... meter/sec

300
30
150
10
2. If car (A) moves at speed 40 km/h and car (B) moves at speed 50 km/h in the same direction, the speed of car (B) relative to a passenger in car (A) is ............................

90
10
50
40
1. If the uniform speed of a car 300 meters in a half minute, then its speed equals .......................... meter/sec

2. If car (A) moves at speed 40 km/h and car (B) moves at speed 50 km/h in the same direction, the speed of car (B) relative to a passenger in car (A) is ............................

3. A person takes a time of 10 minutes to transfer from home to work moving at an average speed of 3 m/sec. which of the following equals the distance between home and work? ........................
4.The distance that a moving object covers within a unit of time is ..........................

5. The total distances divided by the total time taken to cover this distance is .........................

Physics motion in one direction
Graphic representation of motion in one direction, speed, types of speed, acceleration, regular speed, irregular speed, average speed, wind velocity, relative speed, scalar quantity, vector quantity, distance, displacement, velocity, relative speed
Graphic representation of motion in one direction
PASSIVE TRANSPORT Cell membranes help organisms maintain homeostasis by controlling what substances may enter or leave cells. Some substances can cross the cell membrane without any input of energy by the cell in a process known as passive transport. DIFFUSION The simplest type of passive transport is diffusion. Diffusion is the movement of molecules from an area of higher concentration to an area of lower concentration. This difference in the concentration of molecules across a distance is called a concentration gradient. Consider what happens when you add a sugar cube to a beaker of water. As shown in Figure 5-1, the sugar cube sinks to the bottom of the beaker. This sinking makes the concentration of sugar mole- cules greater at the bottom of the beaker than at the top. As the cube dissolves, the sugar molecules begin to diffuse slowly through the water, moving towards the lower concentration at the top. Diffusion is driven entirely by the molecules’ kinetic energy. Molecules are in constant motion because they have kinetic energy. Molecules move randomly, traveling in a straight line until they hit an object, such as another molecule. When they hit some- thing, they bounce off and move in a new direction, traveling in another straight line. If no object blocks their movement, they con- tinue on their path. Thus, molecules tend to move from areas where they are more concentrated to areas where they are less concentrated, or “down” their concentration gradient. In the absence of other influences, diffusion will eventually cause the molecules to be in equilibrium—the concentration of molecules will be the same throughout the space the molecules occupy. Returning to the example in Figure 5-1, if the beaker of water is left undisturbed, at some point the concentration of sugar molecules will be the same throughout the beaker. The sugar concentration will then be at equilibrium. SECTION 1 OBJECTIVES ● Explain how an equilibrium is established as a result of diffusion. ● Distinguish between diffusion and osmosis. ● Explain how substances cross the cell membrane through facilitated diffusion. ● Explain how ion channels assist the diffusion of ions across the cell membrane. VOCABULARY passive transport diffusion concentration gradient equilibrium osmosis hypotonic hypertonic isotonic contractile vacuole turgor pressure plasmolysis cytolysis facilitated diffusion carrier protein ion channel Sugar Water 1 2 3 FIGURE 5-1 Sugar molecules, initially in a high concentration at the bottom of a beaker, , will move about randomly through diffusion, , and eventually reach equilibrium, . At equilibrium the sugar concentration will be the same throughout the beaker. Diffusion occurs naturally because of the kinetic energy the molecules possess. 3 2 1 Copyright © by Holt, Rinehart and Winston. All rights reserved. 98 CHAPTER 5 It is important to understand that even at equilibrium the ran- dom movement of molecules continues. But because there is an equal concentration of molecules everywhere, molecules are just as likely to move in one direction as in any other. The random movements of many molecules in many directions balance one another, and equilibrium is maintained. Diffusion Across Membranes Cell membranes allow some molecules to pass through, but not others. If a molecule can pass through a cell membrane, it will diffuse from an area of higher concentration on one side of the membrane to an area of lower concentration on the other side. Diffusion across a membrane is also called simple diffusion, and only allows certain molecules to pass through the membrane. The simple diffusion of a molecule across a cell membrane depends on the size and type of molecule and on the chemical nature of the membrane. A membrane can be made, in part, of a phospho- lipid bilayer, and certain proteins can form pores in the membrane. Molecules that can dissolve in lipids may pass directly through the membrane by diffusion. For example, because of their nonpolar nature, both carbon dioxide and oxygen dissolve in lipids. Molecules that are very small but not soluble in lipids may diffuse across the membrane by moving through the pores in the membrane.
Chapter 7 - Review Data and Decision Making *Glow bus due at midnight, name and student number: answer questions using content in class People have created wonderful things for centuries, and management Management can be traced as far back as 500 bc when the ancient Sumerians used written records to improve government and business activities Why is it important to lean from the past Not to repeat our mistakes Classical management approaches Scientific management Administrative Principles Bureaucratic organisation Behavioural Management Approaches Follett’s Organizations as communities The Hawthorne studies Maslow’s theory of human needs Mcgregor’s Theory x and Theory Y Argyris Personality and organisation Modern Management foundations Organises as systems Contingency thinking Quality management Quantitative and analysis and tools Evidence-based management Contributions Frederick Taylor - Father of Scientific management He noticed that workers often did their jobs with wasted motions and without a constant approach. His resulted in inefficiency and low performance He believed the problem could be fixed if workers were taught to do their jobs in the best ways and ten were helped and guided by supervisors Four guiding principles of scientific management Rules of motion, standardized work and proper working conditions Select workers with the right abilities Train workers and give them incentives Support workers by planning and smoothing the way as they do their work Frank and Lillian Gilbreth Pioneered use of motitono studies as a management tool In one famous case, the gilbreaths cut down the number of motions used by bricklayers adn tripled their productivity Contributions from scientific management Make results-based compensation a performance incentive Carefully design jobs with efficient work methods Carefully select workers with the ability to perform the job Trian workers to execute activities to the best of their abilities Train supervisors to support workers so they can perform jobs to the best of their abilities Classical Management Adiminstative principle (Henro Fayol) 1919, after a career in French industry, Henri F published “adminisration Industrielle et Generale” (General and industrial management) in which we out like his views on the management of organiztion and workers Rules and duties in management Foresight - to complete a plan of action for the future Organization - To provide and mobilize resources to implement the plan Common- to lead, select and evaluate workers to get the best work toward the plan Coordination- to fit diverse efforts together and ensure information is shared and problems solved Control- to make sure things happen according to plan and to take necessary corrective action Classical management Bureacratic organiztion (Max Weber) Max weber (Bureaucrativ organization) - late 19th century German political economist who had a major impact in the fields of management and sociology Bureaucratic Organization An ideal, intentionally rational adn very efficient form of organization Based on the principles of logic, order and legitimate authority Characteristics of BO Clear division of labour Clear hierarchy of authority Formal rules and procedure Impersonality Careers based on merit What are some disadvantages of bureaucracy Takes a long time for problems to become solved bec there are procedures and there is a chain of people in command Having the power Rules have to follow Excessive paperwork or “red tape” Slowness in handling problems Rigidity in the face of shifting needs Resistance to change Employee apathy Behavioural Management Approaches (focus on understanding the elements that affect human behaviour in organisations) Follett’s Organizations as communites Mary park follett contributed to the transition from classical thinking inot behavioural management Groups and human cooperation Groups allow individuales too combine their talents for a greater good Organizations are cooperating “communites” of managers adn workers Managers job is to help people copperate and achive an integration of goals and intrests Forward-looking managment insight: Making every emploee an owner creates a sense of collective responsibility Prescursor of employrr ownership, profit sharing and gain sharing Buniess problems invovle a varity of inter realted factors Prescursor of systems thinking Private profits realtive to public good Precursor of managerial ethics and social respinsibility Hawthorne studies Took place at western electric chicago plan, a tran led by Harvards Elton Mayo set out to learn how econmic incentives and workplace conditions affected workers output Maing objective Intial study examined how ecomoin incentives adn physical conditions affected worker output (productivity) No consistent relationship found During experientmetn they had 2 groups The expertiant groups (impoved wokring ocnditions ) The control group ( no changes to original working conidtions) No consitant relationship found, perfomance in both groups increased even after removing incentives Social setting and human relations Concluded New “social setting” led workers to do good job Good “Human relations” = higher productivity The contect - The Great Depression (1929-1940) Employee attitudes and groups processes Osme thinsf satisifed some workers but not others People resticited output to adhere to groups norms (Avoid layoffs) Lessons from he hawthrone stufirs Social and human concerns are keys to prductivity Hawthrone effect - People who are singled out for special attention perform as expected Maslow’s Theory of human needs Human needs The work of psychologist Abraham Maslow in the area if human “needs,” also has had a major impact in the behavioual apporach to management Maslow’s hierarchy of human needs Self actualization needs Higherst level: need foe self fulfillment to grow and use abilites to fullest and most creative extent Esteem needs Needs fro esteem in eyes of others need for respect, prestige, recognition; need for self esteem, personal sense of competence, mastery Social needs Need for love, affection, sense of belongingness in ones relationship either other people Safett needs Need for security, protection and stability in teh events of day to day life Physiological needs Most basic of all human needs: need for biological maintence; food, water and phydical well being Principles Defict principle: A satidifed need is not a motivator of behaviour Progress principles: A need becomes a motivator once the preceding lower-level need is satisfied Both principles cease to operate at self actulilzation level McGregor’s Theories Thepry x assumes that workers; Dislike work Lack ambition Are irresponsible Resist change Prefer to be led Theoyry y assumes that workers are Willing to work Willing to accept responsibility Capable of self control Capable of self direction Imaginative and creative According to McGregor, Managers create: Self fulfilling prophecies Implications of Theory x and y Theory x managers: Create situations where workers become dependent, passive and reluctant Theory y managers create situations where workers respond with initiative and high performance Central to notions of empowerment and self management Argyris’s theory of adult personality Classical management principles and practices inhibit worker maturation and are inconsistent with the mature adult personality Management practices should accommodate the mature personality: Increasing task responsibility Increasing task variety Using participative decision making Modern Management Foundation Quantitative analysis and Tools Analytics: the use of large data bases and mathematics to solve problems and make informed decision using systematic analysis Organization as systems System Collection of interrelated parts that function together to achieve a common purpose Subsystem A smaller component of a larger system Open systems Organisations that interact with their environment Contingency thinking Tires to maths managerial responses with problem (situation) No “one best way” to manage The “appropriate way to to manage depends on the situations Quality management Qality anc competitive advantafe are linked Total quality managment (TQM) Comprehensive approach to contiou impovment on teh entire organization ISO certification Gloval quality management standards Refine and upgrade quality to meet ISO requirments Evidednce Based Managment Making management decision on “hard facts” about what really works
What is Electric Force? Electric force is just one of many types of forces in the world of physics. Forces are how and why things move, and can be explained by Newton's Laws of Motion. On the smallest scale, electric force is the resulting interaction between two charged particles. These charges can be either positive or negative. Larger objects can be charged by having an abundance of either of these particles, and therefore can create an electric force on a larger scale. Electric force is the reason why hair will sometimes stand up on its own and is also why we have electricity, allowing us to live in the modern world with lights and technology. Even out in nature electric force is present, as electric force causes lightning to strike. Electric force is fundamental to our everyday way of living. Reviewing Newton's Laws of Motion Newton's Laws of motion are the basic principles or ground rules that are applied all across physics. They describe how objects move and can be used to describe the interaction of charges. They are the following: An object in motion will stay in motion unless an external force is applied The force exerted on an object is equal to the mass times the acceleration of the object. ( ) Every force has an equal and opposite force Newton's laws explain how and why charged particles move. Since there is a force involved (e.g. electric force), particles will move around, which is explained by the first law. The second law describes how acceleration of charges can be calculated once the electric force is known. The third law explains how attractive and repulsive forces between charged objects are equal and opposite. Electric Force Examples and Types of Charge As previously mentioned, there are only two types of charges; positive and negative. Two like charges will repel (or move away from) each other, and two opposite charges will attract (or move towards) each other. In other words, two positive or two negative charges will repel, while a positive and a negative charge will attract. Opposite charges will attract while like charges will repel. Attraction versus Repelling Forces Notice how the forces acting upon each other are equal and opposite, as Newton's third law states. Both charges are exerting forces onto each other. Charges in Atoms An atom is made up of three types of particles; protons, neutrons, and electrons. Protons have a positive charge, neutrons have no charge, and electrons have a negative charge. There are no positive or negative charges smaller than protons and electrons. Objects on a larger scale result in an overall positive or negative charged due to an uneven distribution of protons to electrons. An atom consisting of more protons than electrons would be considered positive, and an atom with more electrons than protons would be considered negative. Protons are held close to the nucleus and are tightly bound to an atom, so it's difficult for protons to escape an atom. Electrons, on the other hand, are much further away from the nucleus of an atom. This makes it much easier for them to be removed from an atom. Electrons can leave or join atoms, making them positive or negative depending on the amount of protons. Similarly, for the bigger picture, overall materials and objects with more electrons than protons would be considered negative, and vice versa. Electric Force Examples Hair standing up: When hair is brushed, the hairbrush can strip electrons from hair strands, resulting in the hair being positively charged. This addition of electrons to the hairbrush in turn makes the hairbrush negatively charged. Since the hair is now positively charged, and like forces repel, hair strands will move away from each other, resulting in the hair standing up. Current electricity: All of our everyday technology is powered through current electricity, which is the consistent flow of electrons through conductive materials. This flow is caused by the electric force, as the electrons flow from a negative source to a positive source. Lightning: During a storm, it is common for an abundance of electrons to build up on the bottom of a cloud, making that part of the cloud negatively charged. Positive charges in the ground start to gather on the surface or even on tall objects such as trees as they are attracted towards the negatively charged undersides of clouds. Lightning strikes as a result of these charges becoming extremely built up. Lightning is caused by electric force Lightning Electric Force Equation: Coulomb's Law The magnitude of the electric force, or the amount of force in which objects repel or attract, depends on the distance between the two charged objects and the amount of charge each object carries. The electric force is stronger the closer together the two charges are, and weaker as the two charges move apart. Electric force is also stronger with more charge, and weaker with less charge. This effect on electric force is predictable, and is known as Coulomb's Law. It can be calculated using a mathematical equation, and the resulting magnitude of electric force is measured in Newtons. Coulomb's Law Electric force can be calculated using the following equation known as Coulomb's Law: In this equation, F is the electric force measured in newtons, K is a constant known as the electrostatic constant, and are charges one and two measured in coulombs, and is the radial distance in meters between the two charges. Since the distance is squared and on the denominator, the electric force drops off exponentially as charges move away from each other. This means that the Electric force is inversely proportional to distance. As charges move away from each other, the electric force between them gets smaller and smaller, until the force is negligible. The amount of charges are in the numerator of this equation, making the magnitude of the force larger with more charge. This means that the force is directly proportional to the amount of charge. When the charges are smaller, the amount of force will be smaller. When there is a lot of charge, the force will be much greater. When calculating the electric force using Coulomb's law, the resulting answer only gives the magnitude of the force and not the direction. In order to know the direction, you must know the types of charges. Once again, like forces repel, and unlike forces attract. It helps to draw a visual representation, or a free-body diagram, of the charges and forces acting upon them in order to understand the resulting force direction. Electric Field versus Electric Force An electric field is a direct result of an electric force. Its pure definition is electric force per unit charge, and can be thought of as a mapping of the force vectors. An electric field is present anytime there is an electric force. Therefore, when there are two or more charged particles, there is a surrounding electric field. The direction of the electric field is the direction a positive charge would flow if it were placed within the field. The electric field moves out from a positive charge and goes into a negative charge. Particles with unlike charges move towards each other, and their corresponding electric field lines move out from the positive charge and into the negative charge. The strength of the force at any given point can be seen through the spacing of the electric field lines. The electric force is strongest where the electric field lines are closest together, and weaker as these lines move apart. Like Coulomb's law expresses, electric field lines show how the electric force is strongest with a minimum distance between the two charges. Unlike charges will result in a repelling force, and the resulting electric field is a visual representation of this effect. Electric fields of two positive charges have the electric field moving out away from both of them. As with two negative charges, the field lines move in towards each negative. Lesson Summary An electric force is created when there are two or more charged particles or objects. These charges can be either positive or negative. Like charges will attract (move towards each other) while unlike charges will repel (move away from each other). As Newton's third law suggests, the forces acting upon each other are both equal and opposite. Electrons and protons within an atom are the two smallest types of charges there are. Electrons carry a negative charge while protons carry a positive charge. Electrons can be easily removed or added to atoms, making the overall charge positive or negative. Objects with more electrons than protons are negatively charged. Electric force is strengthened with increased charge and a shorter distance between the charges. This effect is known as Coulomb's law and can be calculated with the Coulomb's law equation. The magnitude of the force is measured in Newtons, and the direction can be determined by knowing whether the charges are attracting or repelling each other. An electric field is present wherever there is an electric force. The direction of this electric field is the direction a positive charge would flow if it where to be dropped in the field, which is from the positive to the negative.
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
## Fill in the Blank Fill in the blank with the correct words: 1. Work is the use of _____________ to move an object. _____________ 2. Only the part of the applied force that acts in the same _____________ as motion does work. _____________ 3. Work transfers _____________ from one form to another. _____________ 4. The unit of work is the _____________ (J). _____________ 5. The formula for work is _____________ . _____________ [Word Bank]: force, direction, energy, joule, $W = F \times d$ ## Multiple Choice Questions Choose the correct answer from the choices for each question: 1. Which statement best describes when work is done on an object? * A. A force acts on the object while it moves in the same direction as the force. * B. A force acts on the object while it does not move. * C. The object moves but no force is applied. * D. A force acts opposite to the motion and the object does not move. 2. If a force of 10 N moves an object 3 m in the same direction as the force, how much work is done? * A. 3 J * B. 10 J * C. 30 J * D. 13 J 3. Which unit correctly measures work? * A. Newton (N) * B. Meter (m) * C. Joule (J) * D. Watt (W) 4. If a force acts perpendicular to the motion of an object, how much work does that component do? * A. Maximum work * B. Zero work * C. Negative work * D. Equal to force times distance 5. Using the example from the passage, the work done lifting a 30 N dog 2 m is: * A. 15 J * B. 60 J * C. 32 J * D. 90 J ## Open-Ended Questions Answer the following questions in complete sentences: 1. Explain in your own words why only the component of force in the direction of motion does work. 2. A student pushes a box with a 25 N horizontal force and the box moves 4 m horizontally. Calculate the work done and show your steps using the formula from the passage. 3. Describe one real-life example (not in the passage) where work transfers energy from one form to another. Identify the force, the distance, and the energy transfer.