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Useful negative words with their meaning that starts form letter 'D' set1
Quiz┬аby Sanjeev Kumar Singhal
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Why and How Managers Plan Importance of planning The planing process Benefits of planning Planning and time management Types of PLans used by managers Long term and short term plans Strageic and tactical plans Operational plans Planning Tools and Techiqunes Forecasting Contrigency planning Scenario planning Benchmaking Use of staff planners Implementing Plans to Achive Results Goal setting Goal management Goal alignment Participation and involvement Planning Def: The process of setting objectives and determining how best to accomplish them Planning at Eaton Corporation тАЬMaking the hard decision before events force them upon you, an anticipating the future needs of the market before the demand asset itself Objectives and goals Identifity the specific results or desired outcomes that one intends to achieve Plan Def: A statement of action steps to be taken in order to accomplish the objectives (goals) Steps in the planning process: Define your objectives Determine where you stand vis-a-vis objectives Develpo premises reagrdsing future conditions Analyze alternatives and make a plan Implement the plan and evaluate results What are the benefits of planning Improves focus and flexibility Imporves action orteitation Imporves coordination and control Imporves time management Time Managment Personal time management tips Do say тАЬnoтАЭ to request that distract you form what you should be doing Dont get bogged down inn details that can be addressed later Do screen telephone calls, emails and meeting request Dont let drop in visitors, text messaging use up your time Do prioritize your important and urgent work Dont become calendar bound by letting other control your schedule Do follow priorities; do most important and urgent work first Some 77% of mangers in one survey said that digital age has increased th number of decisions they have to make 43% said there was less time available to make these decisions Types of plans used by Managers What is teh time horizon Long term vs Short term Long term Look three or more years into teh future Short term plans Typically cover one year or less However: the increasing environmental complexity and dynamism of recent years has severely tested the concept of тАЬlong-termтАЭ planning Plans are subject to frequent revisions Most executives would likely agree that these complexities adn uncertainties challenge how er actually go about planning and how far ahead we can really plan At the very least we can conclude that there is a lot less permanency to long term plans today and that tey are subject to frequent revision Managment reaeracher Eillot Jaques believes tha people vary in their capability to think with different time horizons Types of Plans used by Managers (3 of 5) Strategic plans Set broad, comprehensive and linger term action directions for teh entire organization or major division Vision Clarifies purpose of the organization and what it hopes to be on the future Typical plans Specify how the organizations resources are used to implement strategy Tactical plans in business often take the form of functional plans Functional plans Incidate how different component within the organiztion will help accompnlish the overall strategy Production plans Finacial plans Facilites Plans Logisitc plans Marketing plans Human Resource Plans Operation plans Describe short-term activities to implement strategic plans Policies: Are standing plans that communicate guidelines for decisions Ex: Policies on office romances: The media is quick to report when a top executive or public figures runs into trouble over an office affair. Are there ant policies on office romances? Employer polices on office raltioshiis vary. One survey find teh following: 24% prohibit relationships among employees in the same department 13% prohibit relationships among employees who have the smae supervisor 80% prohibit relationships between supervisors and subordinates 5% have no restrictions on office romances Procedures: Are rules that describe actions to be taken in specific situations Budgets: are single use plans that commit resources to projects or activities Zero based budgets: allocate resources as if each budget were brand new There is no guarantee that any past funding will be renwer. All propsales, old and new, must compete for available funds at teh start of each new budget cycle Forcasting Attempts to predict the future Qualitaive forecasting uses expert opinions Quantitative forecasting uses mathematical models and statiscal aanylsis of historical data dna surveys Contingency planning Identify alternative course of action to take when things go wrong Anticipate changing conditions Contain trigger points to indicate when to activate plan (or a specific course of action) Scenario planning A long term version of contingency planning Identifying alternative future scenarios Plans made for each future scenario Increases organizations flexibility and preparation for future shocks Benchmarking Use of external and internal comparisons to better evaluate current performance Adopting best practices: things people adn organization do that lead to superior performance Staff Planners Experts who assist in all steps of the planning process They help bring focus and expertise to a wide variety of planning tasks Important: Communication between staff planers landline managers is essential for teh success of teh planning process Goal Setting - Always set SMART goal The solution: Goal Aligment Between Team Leader and Team Member Jonintly plan: Set objectives, set standards, choose actions Individually acy: Perform tasks (member), provide support (leader) Jointly control: Review results, discuss implications, renew cycle x4 Collective effort and commitment Participatroy planning Includes in all planning steps that people who will be affected by the plans adn askedd to help implement them Unloacks motivational potential of goal setting Management by objective (MBO) promotes participation Participation increases understanding and acceptance of plan and commitment to success Participatory planning - Number of people involved in teh decision making process Amazon is intensely focused on what it does. It believes in creating tight single-threaded teams, also known as тАЬ2 pizza team.тАЭ Data and Decision Making What are some of the important competencies managers must have today? Delegate Marketing and technology Manager must have Technological competency Ability to understand new technologies and to use them to their best advantage Information competency Ability to locate, gather, organize and display information for decision-making and problem solving Analytical competency Ability to evaluate and analyze information to make actual decisions and solve real problems What is the difference between Data and Information Data Raw facts and observation Information Data made useful and meaningful for decision-making Important concepts Big data Exists in huge quantities and is difficult to process without sophisticated mathematical and analytical techniques Data production today Bernard Marr is an internationally best-selling author. He helps organizations improve their business performance, use data more intelligently Data mining The process of analyzing data to produce useful information for decision-makers Management Analytics The systematic evaluation and analysis of data to make informed decision Information drives management Bad Data Refers to information that can be erroneous, misleading, and without general formatting The challenge: Can er use the data that is available in the тАЬBig DataтАЭ Needs to be valid Can not trust everything out there Being ethical Look at the trends Data is structured and unstructured Data BIg Data = Structured + Unstructured Information Drive Management decision making What are the characteristics of useful information Easy to access If its credible Accurate Characteristics of useful information: Timely High quality Complete Relevant Understandable What about bad data It's not credible Miss information If it is not structured/ organized Bias based on opinions Confusing If its updated Bad data Refers to information that can be erroneous miss What are some examples of Management information system Business intelligence -BI Information systems to extract and report data in organized ways that are useful to decision-makers Executive dashboards Visually update and display key performance metrics (or Key Performance Indicators -KPIs) and information on a real-time basis Information needs in organization External Environment Information exchanges with the external environment Gather intelligence information Provide public information Information needs within the organizations (internal Enviroement) Information exchange within the organization Facilitate decision making Facilitate problem-solving Managers as information processors Continually gather, share and receive information Now as much electronic as it is face-to-face Always on, always connected How many people telecommute at least once a week 70% of people globally work remotely at least once a week, Work at home after covid 19 our forecast Our best estimate it that 25-30% of the workforce will be working form home multiple days a week by the end of 2021 As of 2023, 12.7% of full time employees work from home, while 28.2% work a hybrid model Managers as problem solvers Problem-solving The process of identifying a discrepancy between actual and desired performance and taking action to resolve it Ishikawa Fishbone diagram To identify the cause of problems Decision A choice among possible alternative courses of action Performance threat Something is wrong or has the potential to go wrong Performance opportunity The situation offers the chance for a better future if the right steps are taken Problem-solving approaches or style - from textbook Problem avoiders Inactive in information gathering and solving problems Problem seekers Proactive in anticipation of problems and opportunities and taking appropriate action to gain an advantage Problem solvers Reactive in gathering information and solving problem Managers - can approach problems in a systematic or intuitive manner Systematic thinking approaches problem in rational, step-by-step and analytical fashion Intuitive thinking approaches problems in a flexible and spontaneous fashion Multidimensional thinking- applies both intuitive and systematic thinking Managers face structured and unstructured problems Structure problems Are ones that are familiar, straight forward, and clear with respect to information needs Program decisions apply solutions that are readily available from past experiences to solve structured problems Know how to solve them Familiar Know what we are dealing with Unstructured problems Are ones that are full of ambiguities and information deficiencies Nonprogrammed decisions apply a specific solution to meet the demands of a unique problem Commonly faced by higher-level management Crisis decision making A crisis involves an unexpected problem that can lead to disaster if not resolved quickly and appropriately Ruled for crisis management Figure out what is going on Remember that speed matters Remember that slow counts, too Respect the danger of the unfamiliar Value the skeptic Be ready to тАЬfight fire with fireтАЭ Managers make decisions with various amounts of information Certain environment Offers complete information on possible action alternatives and their consequences Risk environment Lacks complete information but offers probabilities of the likely outcomes for possible action alternatives Uncertain environment Lacks so much information that it is difficult to assign probabilities to the likely outcomes of alternative Ex: Certain and uncertain environments: The worldwide Governance Indicators for over 200 countries, comparing distinct environments (Canada-Brazil) Step 1-Identify and define the problem Focuses on information gathering information processing and deliberation Decision objectives should be established What are some common mistakes in definding problems? Common mistakes in defining problems Defining the problem too broadly or too narrowly Focusing on symptoms instead of causes Choosing the wrong problem to deal with Step 2- Generate and Evaluate Alternative Courses of Action Potential solutions are formulated and more information is gathered, data are analyzed, the advantages and disadvantages of alternative solutions are identified Common mistakes: Abandoning the search for alternatives too quickly Step 3- Decide on a preferred course of Action Two different approaches Behavioural model leads to satisficing decisions Classical model les to optimising decisions Behavioural Model Rationality is bounded because: There are limits our thinks capacity Available information (incomplete) Time constraints Step 4-Implement the decision Involves taking action to make sure the solution decided upon becomes a reality Managers need to have the willingness and ability to implement action plans Problems: Lack of participation error should be avoided Step 5 - Evaluate Results Involves comparing actual and desired results The positive and negative consequences of the chosen course of action should be examined If actual results fall short desire results, the manager returns to earlier steps in the decision-making process At all steps, check ethical reasoning Ask these spotlight questions Utility Does teh decision satisfy all constituents or stakeholders Rights Does the description respect the rights and duties of everyone? Justice Is the decision consistent with the canons of justice Caring Is the decision consistent with my responsibilities to care? Issues in decision-making How do errors happen? Heuristics: are strategies for simplifying decision-making Availability Bias: Bases a decision on recent information or events Representativeness bias: Bases a decision on similarity to other situations Anchoring and Adjustment Bias: Bases a decision on incremental adjustment from a prior decision point Framing error: Tring to solve a problem in the context perceived, positive or negative Confirmation Error: Focusing on information that confirms a decision already made Escalating commitment: Continuing a course of action even though it is not working Creative Decision making Creativity is the generation of a novel idea or unique approach that solves a problem or crafts an opportunity Big C: Creativity occurs when extraordinary things are done by exceptional people Little C: Creativity occurs when average people come up with unique ways to deal with daily events and situations The three types of situational creativity drivers Chapter review What are objectives and goals? The specific results or desired outcomes What are the 5 characteristics of great (SMART) goals? Forecasting - Attempts Qualitative forecasting uses options Quantitative forecasting uses mathematical models and statistical analysis of historical data and surveys Scenarios-OracleтАЩs crystal ball combines qualitative and quantitative methodsCovalent Molecules and Compounds Just as an atom is the simplest unit that has the fundamental chemical properties of an element, a molecule is the simplest unit that has the fundamental chemical properties of a covalent compound. Some pure elements exist as covalent molecules. Hydrogen, nitrogen, oxygen, and the halogens occur naturally as the diatomic (тАЬtwo atomsтАЭ) molecules H2, N2, O2, F2, Cl2, Br2, and I2 (part (a) in Figure 3.1.1). Similarly, a few pure elements exist as polyatomic (тАЬmany atomsтАЭ) molecules, such as elemental phosphorus and sulfur, which occur as P4 and S8 (part (b) in Figure 3.1.1). Each covalent compound is represented by a molecular formula, which gives the atomic symbol for each component element, in a prescribed order, accompanied by a subscript indicating the number of atoms of that element in the molecule. The subscript is written only if the number of atoms is greater than 1. For example, water, with two hydrogen atoms and one oxygen atom per molecule, is written as H2O. Similarly, carbon dioxide, which contains one carbon atom and two oxygen atoms in each molecule, is written as CO2. Covalent compounds that predominantly contain carbon and hydrogen are called organic compounds. The convention for representing the formulas of organic compounds is to write carbon first, followed by hydrogen and then any other elements in alphabetical order (e.g., CH4O is methyl alcohol, a fuel). Compounds that consist primarily of elements other than carbon and hydrogen are called inorganic compounds; they include both covalent and ionic compounds. In inorganic compounds, the component elements are listed beginning with the one farthest to the left in the periodic table, as in CO2 or SF6. Those in the same group are listed beginning with the lower element and working up, as in ClF. By convention, however, when an inorganic compound contains both hydrogen and an element from groups 13тАУ15, hydrogen is usually listed last in the formula. Examples are ammonia (NH3) and silane (SiH4). Compounds such as water, whose compositions were established long before this convention was adopted, are always written with hydrogen first: Water is always written as H2O, not OH2. The conventions for inorganic acids, such as hydrochloric acid (HCl) and sulfuric acid (H2SO4), are described elswhere. Note! For organic compounds: write C first, then H, and then the other elements in alphabetical order. For molecular inorganic compounds: start with the element at far left in the periodic table; list elements in same group beginning with the lower element and working up. Write the molecular formula of each compound. a. The phosphorus-sulfur compound that is responsible for the ignition of so-called strike anywhere matches has 4 phosphorus atoms and 3 sulfur atoms per molecule. b. Ethyl alcohol, the alcohol of alcoholic beverages, has 1 oxygen atom, 2 carbon atoms, and 6 hydrogen atoms per molecule. c. Freon-11, once widely used in automobile air conditioners and implicated in damage to the ozone layer, has 1 carbon atom, 3 chlorine atoms, and 1 fluorine atom per molecule. Solution: a. тАв A The molecule has 4 phosphorus atoms and 3 sulfur atoms. Because the compound does not contain mostly carbon and hydrogen, it is inorganic. тАв B Phosphorus is in group 15, and sulfur is in group 16. Because phosphorus is to the left of sulfur, it is written first. тАв C Writing the number of each kind of atom as a right-hand subscript gives P4S3 as the molecular formula. b. тАв A Ethyl alcohol contains predominantly carbon and hydrogen, so it is an organic compound. тАв B The formula for an organic compound is written with the number of carbon atoms first, the number of hydrogen atoms next, and the other atoms in alphabetical order: CHO. тАв C Adding subscripts gives the molecular formula C2H6O. c. тАв A Freon-11 contains carbon, chlorine, and fluorine. It can be viewed as either an inorganic compound or an organic compound (in which fluorine has replaced hydrogen). The formula for Freon-11 can therefore be written using either of the two conventions. тАв B According to the convention for inorganic compounds, carbon is written first because it is farther left in the periodic table. Fluorine and chlorine are in the same group, so they are listed beginning with the lower element and working up: CClF. Adding subscripts gives the molecular formula CCl3F. тАв C We obtain the same formula for Freon-11 using the convention for organic compounds. The number of carbon atoms is written first, followed by the number of hydrogen atoms (zero) and then the other elements in alphabetical order, also giving CCl3F. Write the molecular formula for each compound. a. Nitrous oxide, also called тАЬlaughing gas,тАЭ has 2 nitrogen atoms and 1 oxygen atom per molecule. Nitrous oxide is used as a mild anesthetic for minor surgery and as the propellant in cans of whipped cream. b. Sucrose, also known as cane sugar, has 12 carbon atoms, 11 oxygen atoms, and 22 hydrogen atoms. c. Sulfur hexafluoride, a gas used to pressurize тАЬunpressurizedтАЭ tennis balls and as a coolant in nuclear reactors, has 6 fluorine atoms and 1 sulfur atom per molecule. Answer: a. N2O b. C12H22O11 c. SF6. Ionic Compounds The substances described in the preceding discussion are composed of molecules that are electrically neutral; that is, the number of positively-charged protons in the nucleus is equal to the number of negatively-charged electrons. In contrast, ions are atoms or assemblies of atoms that have a net electrical charge. Ions that contain fewer electrons than protons have a net positive charge and are called cations. Conversely, ions that contain more electrons than protons have a net negative charge and are called anions. Ionic compounds contain both cations and anions in a ratio that results in no net electrical charge. Note! Ionic compounds contain both cations and anions in a ratio that results in zero electrical charge.An ionic compound that contains only two elements, one present as a cation and one as an anion, is called a binary ionic compound. One example is MgCl2, a coagulant used in the preparation of tofu from soybeans. For binary ionic compounds, the subscripts in the empirical formula can also be obtained by crossing charges: use the absolute value of the charge on one ion as the subscript for the other ion. This method is shown schematically as follows: Crossing charges. One method for obtaining subscripts in the empirical formula is by crossing charges. When crossing charges, it is sometimes necessary to reduce the subscripts to their simplest ratio to write the empirical formula. Consider, for example, the compound formed by Mg2+ and O2тИТ. Using the absolute values of the charges on the ions as subscripts gives the formula Mg2O2:Polyatomic Ions Polyatomic ions are groups of atoms that bear net electrical charges, although the atoms in a polyatomic ion are held together by the same covalent bonds that hold atoms together in molecules. Just as there are many more kinds of molecules than simple elements, there are many more kinds of polyatomic ions than monatomic ions. Two examples of polyatomic cations are the ammonium (NH4+) and the methylammonium (CH3NH3+) ions. P. The method used to predict the empirical formulas for ionic compounds that contain monatomic ions can also be used for compounds that contain polyatomic ions. The overall charge on the cations must balance the overall charge on the anions in the formula unit. Thus, K+ and NO3тИТ ions combine in a 1:1 ratio to form KNO3 (potassium nitrate or saltpeter), a major ingredient in black gunpowder. Similarly, Ca2+ and SO42тИТ form CaSO4 (calcium sulfate), which combines with varying amounts of water to form gypsum and plaster of Paris. The polyatomic ions NH4+ and NO3тИТ form NH4NO3 (ammonium nitrate), a widely used fertilizer and, in the wrong hands, an explosive. One example of a compound in which the ions have charges of different magnitudes is calcium phosphate, which is composed of Ca2+ and PO43тИТ ions; it is a major component of bones. The compound is electrically neutral because the ions combine in a ratio of three Ca2+ ions [3(+2) = +6] for every two ions [2(тИТ3) = тИТ6], giving an empirical formula of Ca3(PO4)2; the parentheses around PO4 in the empirical formula indicate that it is a polyatomic ion. Writing the formula for calcium phosphate as Ca3P2O8 gives the correct number of each atom in the formula unit, but it obscures the fact that the compound contains readily identifiable PO43тИТ ions.Summary тАв There are two fundamentally different kinds of chemical bonds (covalent and ionic) that cause substances to have very different properties. тАв The composition of a compound is represented by an empirical or molecular formula, each consisting of at least one formula unit.Contributors The atoms in chemical compounds are held together by attractive electrostatic interactions known as chemical bonds. Ionic compounds contain positively and negatively charged ions in a ratio that results in an overall charge of zero. The ions are held together in a regular spatial arrangement by electrostatic forces. Most covalent compounds consist of molecules, groups of atoms in which one or more pairs of electrons are shared by at least two atoms to form a covalent bond. The atoms in molecules are held together by the electrostatic attraction between the positively charged nuclei of the bonded atoms and the negatively charged electrons shared by the nuclei. The molecular formula of a covalent compound gives the types and numbers of atoms present. Compounds that contain predominantly carbon and hydrogen are called organic compounds, whereas compounds that consist primarily of elements other than carbon and hydrogen are inorganic compounds. Diatomic molecules contain two atoms, and polyatomic molecules contain more than two. A structural formula indicates the composition and approximate structure and shape of a molecule. Single bonds, double bonds, and triple bonds are covalent bonds in which one, two, and three pairs of electrons, respectively, are shared between two bonded atoms. Atoms or groups of atoms that possess a net electrical charge are called ions; they can have either a positive charge (cations) or a negative charge (anions). Ions can consist of one atom (monatomic ions) or several (polyatomic ions). The charges on monatomic ions of most main group elements can be predicted from the location of the element in the periodic table. Ionic compounds usually form hard crystalline solids with high melting points. Covalent molecular compounds, in contrast, consist of discrete molecules held together by weak intermolecular forces and can be gases, liquids, or solids at room temperature and pressure. An empirical formula gives the relative numbers of atoms of the elements in a compound, reduced to the lowest whole numbers. The formula unit is the absolute grouping represented by the empirical formula of a compound, either ionic or covalent. Empirical formulas are particularly useful for describing the composition of ionic compounds, which do not contain readily identifiable molecules. Some ionic compounds occur as hydrates, which contain specific ratios of loosely bound water molecules called waters of hydration.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: тАв Free-falling objects do not encounter air resistance. тАв 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 тАв 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 тАв a тАв 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 тАв (-8.00 m/s2) тАв d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) тАв d (16.0 m/s2) тАв 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 тАв t + ┬╜ тАв a тАв 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) тАв (4.1 s) + ┬╜ тАв (6.00 m/s2) тАв (4.10 s)2 d = (0 m) + ┬╜ тАв (6.00 m/s2) тАв (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: тАв 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. тАв 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. тАв 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. тАв 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 тАв t + ┬╜ тАв a тАв 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) тАв (t) + ┬╜ тАв (-9.8 m/s2) тАв (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) тАв (t)2 -8.52 m = (-4.9 m/s2) тАв (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 тАв a тАв 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 тАв(-9.8m/s2) тАвd 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) тАвd (-19.6 m/s2) тАв d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) тАв 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.Useful vocab for writing exam Useful English phrasesUseful Vocabulary Spanish 2 Useful words or phrases with synonymsUseful expressions in English class n┬░1