Loading...
Lesson 5: Our Teacher, Ms Nora Questions
Quiz by Mrs.Hina Masood
Customize this quiz to suit your class
Instantly translate to 100+ languages
Tag the questions with any skills you have. Your dashboard will track each student's mastery of each skill.
Give this quiz to my class
Lesson 5: Our Teacher, Ms NoraCan you make a multiple choice of test questions regarding this information given which is Curriculum from Different Points of View There are many definitions of curriculum. Because of this, the concept of curriculum is sometimes characterized as fragmentary, elusive and confusing. However, the numerous definitions indicate dynamism that connotes diverse interpretations of what curriculum is all about. The definitions are influenced by models of thought, pedagogies, political as well as cultural experiences. Let us study some of these definitions. 1. Traditional Points of View of Curriculum In early years of the 20th century, the traditional concepts held of the “curriculum is that it is a body of subjects or subject matter prepaid by the teachers for the student’s to learn”. It was synonymous to the “course of study” and “syllabus” Robert M. Hutchins views curriculum as “permanent studies” where the rules of grammar, reading, rhetoric and logic and mathematics for basic education are emphasized. Basic education should emphasize the 3 Rs and college education should be grounded on liberal education. On the other, Arthur Bestor as an essentialist, believes that the mission of the school should be intellectual training, hence curriculum should focus on the fundamental intellectual disciplines of grammar, literature and writing. It should also include mathematics, science, history and foreign language. The definition leads us to the view of Joseph Schwab that discipline is the sole source of curriculum. Thus in our education system, curriculum is divided into chunks of knowledge we call subject areas in basic education such as English, Mathematics, Science, Social Studies and others. In college, discipline may include humanities, sciences, language and many more. To Phoenix, curriculum should consist entirely of knowledge which comes from various disciplines. Academic discipline became the view of what curriculum is after the cold war and the race to space. Joseph Schwab, a leading curriculum theorist coined the term discipline as a ruling doctrine for curriculum development. Curriculum should consist only of knowledge which comes from disciplines which is the sole source. Thus curriculum can be viewed as a field of study. It is made up of its foundations (philosophical, historical, psychological and social foundations); domains of knowledge as well as its research theories and principles. Curriculum is taken as scholarly and theoretical. It is concerned with broad historical, philosophical and social issues and academics. Most of the traditional ideas view curriculum as written documents or a plan of action in accomplishing goals. 2. Progressive Points of View of Curriculum On the other hand, to a progressivist, a listing of school subjects, syllabi, course of study, and a list of courses or specific discipline do not make a curriculum. These can only be called curriculum if the written materials are actualized by the learner. Broadly speaking, curriculum is defined as the total learning experiences of the individual. This definition is anchored on John Dewey’s definition of experience and education. He believed that reflective thinking is a means that unifies curricular elements. Thought is not derived from action but tested by application. Caswell and Campbell viewed curriculum as “all experiences children have under the guidance of teachers”. This definition is shared by Smith, Stanley and Shores when they defined “curriculum as a sequence of potential experiences set up in the schools for the purpose of disciplining children and youth in group ways of thinking and acting”. Marsh and Willis on the other hand view curriculum as all the “experiences in the classroom which are planned and enacted by the teacher, and also learned by the students”. Points of View on Curriculum Development From the various definitions and concepts presented, it is clear that curriculum is a dynamic process. Development connotes changes which are systematic. A change for the better means any alteration, modification or improvement of existing condition. To produce positive changes, development should be purposeful, planned and progressive. This is how curriculum evolves. Let us look at the two models of curriculum development and concepts of Ralph Tyler and Hilda Taba. Ralph Tyler Model: Four Basic Principles. This is also popularly known as Tyler’s Rationale. He posited four fundamental questions or principles in examining any curriculum in schools. These four fundamental principles are as follows: 1. What educational purposes should the school seek to attain? 2. What educational experiences can be provided that are likely to attain these purposes? 3. How can these educational experiences be effectively organized? 4. How can we determine whether these purposes are being attained or not? In summary, Tyler’s Model show that in curriculum development, the following considerations should be made: (1) Purpose of the school, (2) Educational experiences related to the purposes, (3) Organization of the experiences, and (4) Evaluation of the experiences. On the other hand, Hilda Taba improved on Tyler’s Rationale by making a linear model. She believed that teachers who teach or implement the curriculum should participate in developing it. Her advocacy was commonly called the grassroots approach. She presented seven major steps to her model where teachers could have a major input. These steps are as follows: 1. Diagnosis of learner’s needs and expectations of the larger society. 2. Formulation of learning objectives. 3. Selection of learning content. 4. Organization of learning content. 5. Selection of learning experiences. 6. Organization of learning activities. 7. Determination of what to evaluate and the means of doing it. Thus as you look into curriculum models, the three interacting processes in curriculum development are planning, implementing and evaluating. Types of Curriculum Operating in Schools From the various concepts given, Allan Glatthorn(2000) describes seven types of curriculum operating in the schools. These are (1) Recommended curriculum- proposed by scholars and professional organizations. (2) Written Curriculum- appears in school, district, division or country documents. (3) Taught Curriculum- what teacher’s implement or deliver in the classrooms and schools. (4) Supported Curriculum- resources-textbooks, computers, audio- visual materials which support and help in the implementation of the curriculum. (5) Assessed Curriculum- that which is tested and evaluated. (6) Learned Curriculum- which the students actually learn and what is measured and (7) Hidden Curriculum- the unintended curriculum. 1. Recommended Curriculum- Most of the school curricula are recommended. The curriculum may come from a national agency like the Department of Education, Commission on Higher Education (CHED), Department of Science and Technology (DOST) or any professional organization who has stake in education. For example the Philippine Association for Teacher Education (PAFTE) or the Biology Teacher Association (BIOTA) may recommend a curriculum to be implemented in the elementary or secondary education. 2. Written Curriculum- This includes documents, course of study or syllabi handed down to the schools, districts, division, departments or colleges for implementation. Most of the written curricula are made by curriculum experts with participation of teachers. These were pilot-tested or tried out in sample schools or population. Example of this is the Basic Education Curriculum (BEC). Another example is the written lesson plan of each classroom teacher made up of objectives and planned activities of the teacher. 3. Taught Curriculum- The different planned activities which are put into action in the classroom compose the taught curriculum. These are varied activities that are implemented in order to arrive at the objectives or purposes of the written curriculum. These are used by the learners with the guidance of teachers. Taught curriculum varies according to the learning styles of students and the teaching styles of teachers. 4. Supported Curriculum- In order to have a successful teaching, other than the teacher, there must be materials which should support or help in the implementation of a written curriculum. These refer to the support curriculum that includes material resources such as textbooks, computers, audio-visual materials, laboratory equipment, playgrounds, zoos and other facilities. Support curriculum should enable each learner to achieve real and lifelong learning. 5. Assessed Curriculum- This refers to a tested or evaluated curriculum. At the duration and end of the teaching episodes, series of evaluations are being done by the teachers to determine the extent of teaching or to tell if the students are progressing. This refers to the assessed curriculum. Assessment tools like pencil-and-paper tests, authentic instruments like portfolio are being utilized. 6. Learned Curriculum- This refers the learning outcomes achieved by the students. Learning outcomes are indicated by the results of the tests and changes in behavior which can either be cognitive, affective or psychomotor. 7. Hidden Curriculum- This is the unintended curriculum which is not deliberately planned but may modify behavior or influenced learning outcomes. There are lots of hidden curricula that transpire in the schools. Peer influence, school environment, physical condition, teacher-learner interaction, mood of the teachers and many other factors made up the hidden curriculum.What is an earthquake? Would you be surprised to learn that several million earthquakes happen every year? Seriously. Most are so small in magnitude or size that we cannot even feel them. In fact, only 20 earthquakes are efficiently reported each year in the United States Geological Survey. Wow! That is a huge difference! The Earth has four major layers. Inner core, outer core, mantle, and crust. Think of the crust and top of the mantle like the skin of the earth. This skin is made up of different pieces of rock called tectonic plates. There are about 15 major slabs that join together, kind of like a puzzle. The edges around the tectonic plates are called plate boundaries. These massive pieces of rock slide back and forth under the Earth's surface, bumping up against each other and creating a lot of tension. This tension and movement create faults, which are basically huge cracks in the rock. When the faults get stuck, they build up pressure. And when they get unstuck, you guessed it, an earthquake. So basically, an earthquake is caused by the shifting and sliding of tectonic plates on the Earth's upper mantle and crust. There are three ways that tectonic plates shift or slide. They are subduction, lateral sliding, and spreading. Subduction happens when plates crash into each other. This can cause one plate to slide under another and be destroyed. Or the edges of the plate may rise up and form mountains. Lateral sliding means that the plates slide alongside each other, which can create lots of friction. And like you might have guessed, spreading happens when plates move apart from each other. When they do, melted rock between the plates rises and cools, forming new crust. Here's an interesting fact. Nearly 90% of all earthquakes begin in the Pacific Ocean, in an area called the Ring of Fire. It's called the Ring of Fire because along with earthquakes, it's filled with many active volcanoes. More than 450! Earthquakes can be powerful enough to change the surface of the earth and can do a lot of damage. And sometimes earthquakes can even cause other natural disasters, like avalanches, landslides, and tsunamis. Pretty wild, right? The epicenter is the location of an earthquake on the Earth's surface. The closer you are to the epicenter, the more of the earthquake you will feel. Earthquakes lose intensity as they travel away from the epicenter. Scientists measure the intensity of an earthquake using a special device called a seismograph. Seismometers detect and measure the vibrations given off by an earthquake. Magnitude is the number given to record the size of an earthquake. For example, a magnitude 5.5 is considered moderate. Above 8.0 is considered a major earthquake and we see one every year or two. Earthquakes measured at 2.5 or less are usually not felt, but can be recorded. And believe it or not, there are millions that happen each year. You can make a model of a seismograph at home, and we are going to show you how. It's activity time! You can print off directions for this one on our website at learnbright.org. You'll need a cardboard box, string, a plastic cup, a marker, small heavy objects, a long strip of paper, and a friend because this is an activity for at least two people. Now comes the fun part. One friend shakes the box, alternating between hard and soft and slow and fast, while the other friend is pulling the strip of paper through the bottom. Watch the marker as it records the movement. This is exactly what a seismograph does during an earthquake. So, in a way, we have not only created our own seismograph, but our own earthquake as well. Now, we can analyze the data just like scientists. Can you tell how hard the box was shaking based on the line? Can you tell when it was barely shaking at all? You are on your way to becoming a seismologist. A seismologist is a person that studies earthquakes. It's pretty cool to watch the process, but it's even more exciting to do it yourself. You can head on over to our website to get detailed instructions for this activity. Just download the lesson plan and as always have fun! Hope you had fun learning with us! Visit us at learnbright.org for thousands of Hope you had fun learning with us! Visit us at learnbright.org for thousands of free resources and turnkey solutions for teachers and homeschoolers.Lesson 5: Our new neighbor bruno AudioLesson 5: Our new neighbor bruno QuestionsLesson 5 Honoring Our ParentsIntroduction 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. Lesson 1: Why is the interior of the Earth hot? The interior of Earth is very hot (the temperature of the core reaches more than 5,000 degrees Celsius) for two main reasons: . The heat from when the planet formed, . The heat from the decay of radioactive elements. The Earth was formed by the process of accretion. After the creation of our solar system, meteorites gravitationally attracted each other and formed bigger objects, which attracted bigger masses, until our planets reach their current size. This process accumulated a lot of heat; when two objects collide, heat is generated. That is why your hands will get hot when you clap them for too long, or a nail gets very hot when you hammer it for a long time. This heat has not dissipated totally and represents about 10% of the total heat inside the Earth. The main source of heat is the decay of radioactive elements. Radioactive decay is a natural process; unstable elements like 238U (Uranium) or 40K (Potassium) stabilize with time and produce what we call daughter products: 206P (Lead) for Uranium and 40Ar (Argon) for Potassium. This process produces heat, which represents about 90% of the total heat inside the Earth. Lesson 2: How Magma Forms Magma is a molten and semi-molten rock mixture found under the surface of the Earth. This mixture is usually made up the of four parts: hot liquid base, called a melt; minerals crystallized by the melt; solid rocks incorporated into the melt from the surrounding confines; and dissolved gases. When magma is ejected by a volcano or other vent, the material is called lava. Magma that has cooled into a solid is