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Sewing side pocket
Quiz by Jennifer Dela Rosa
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​Side pocket is a pocket in or at one side of a garment.
​Jean’s pocket features ornate flap closures.
Side pocket is a pocket in or at one side of a garment.
Jean’s pocket features ornate flap closures.
Trouser pockets start at the waist and follow a diagonal line to the sideseam.
A pocket that has a curved shape is called a scoop pocket.
Western pocket is a form of patch pocket that features ornate flap closures.
The trouser side pocket is also called as slash pocket.
In preparing a side pocket bag, we must place the facing onto the right side.
The upper facing is bigger than the lower facing.
The lower facing is bigger than the upper facing.
Pinning helps to secure the fabric from uneven sew. Â
Pessimistic (adj.) seeing the worst side of things; no hope tactful (adj.) skilled in handling difficult situations or people; polite apparatus (n.) a set of equipment or tools or a machine that is used for a particular purpose daunting (adj.) making you feel slightly frightened or worried about your ability to achieve something condescend (v.) to behave in a way which shows that you consider yourself to be better or more intelligent than other people Land preparation •Carrots can be planted in beds. •The beds should be prepared before planting. (a)digging: this is done using a hoe or a spade so as to loosen the soil. •During digging , remove roots and stems of weeds. (b) manuring: add 2kg of well rotten manure per square metre. •Work the manure deeply into the soil so that the roots of the carrots do not come into directly contact with manure. •Improperly incorporated manure causes development of carrots that have side roots. •this condition is called forking. Planting •Carrots are planted in beds where they will mature. •Water the bed a day before planting. •Mark the rows 30cm apart. •Carrots seeds are very small, therefore due to their size sow the seeds thinly along the rows. •Rub the seeds between your fingers as you sow. •Cover the seeds with fine sand. •The bed can be covered with grass so that the soil remain moist. •Water the bed soon after sowing using a watering can fitted with a fine rose. Caring for carrots (a) Thinning: is when you remove some of the baby plants in a row to free up space for the rest so that they can grow to full size. • Thinning in carrots is usually done two weeks after germination. • Thin out to leave a spacing of 4 -6 cm from each plant. (b) Watering: If carrots are not watered enough they will not grow fast. • Therefore carrots need a steady supply of water to keep the soil moist during the growing period. (c) Weeding: weeds are removed by hand, especially those in the rows. (d) Top dressing: carrots are usually not top dressed. •Fertilizer can only be applied after one has noticed leaves turning yellow. (e) Pest and disease control: aphids and nematodes attack carrots. •Dimethoate is used to control aphids. Harvesting carrots •Carrots are harvested 2 to 3 months after planting. •Bigger ones are harvested first inorder to leave more room for the young seedlings to develop. •Water the bed first the lift the carrots by hand. •A garden fork can also be used to loosen the soil, then pull up by hand. •The carrots can then be tied into bunches of 5 – 7 carrotsWHAT IS SCIENCE? - is a way in which answers related to NATURAL events are proposed. - a way in which people can learn and UNDERSTAND events in the NATURAL WORLD - based on OBSERVABLE EVENTS - a study of the NATURAL WORLD - a method of DISCOVERY and UNDERSTANDING by using a PROBLEM-SOLVING process called the?? - A systematic body of knowledge based on observation and experimentation. FOUR COMMON CHARACTERISTICS OF SCIENCE: 1. It focuses on the NATURAL WORLD. 2. Goes through experiment. 3. Relies on evidence. 4. Passes through the scientific community. WHAT IS TECHNOLOGY? Brian Arthur (2009) defined technology as: 1. a means to fulfill a human purpose 2. assemblage of practices and components 3. a collection of devices and engineering practices available to a culture. SOCIETY ST (Science Technology) would not exist without society. WHAT IS STS? Science and Technology and Society (STS) is the study of how society, politics and culture affect scientific research and technological innovation and how these, in turn affects society, politics and culture. EVENTS IN THE HISTORY OF SCIENCE AND TECHNOLOGY THAT TRANSFORMED THE SOCIETY (IN THE WORLD) ANCIENT PERIOD 3500 BC. - 500 AD EUROPE - use of fire by Homo Erectus CA 750,000 - Stone Headed Spears CA 45,000 - Wooden bow and arrow CA 20,000 - The Minoans build palaces in Crete CA 2,000 THE AMERICAS - The Folsom people living on eastern side of the Rocky Mountain developed sophisticated tools CA 8,000. - Pottery is made in South America CA 6,000 - Olmec sculpture carves figurines and giant human heads. CA 1200 ASIA AND OCEANA - Earliest known clay pots are made in Japan CA 11,000. - Bronze is first made in Thailand CA 4000 - A lunar calendar is developed in China CA 2950 - Chinese doctors begin using acupuncture CA 2500 - The Hindu calendar of 360 days was introduced in India CA 1000 AFRICA AND MIDDLE EAST - Homo erectus uses stone tools CA 1000000 - CA 15000 in Africa, bone harpoons are used for fishing. - Clay tokens are used for record keeping in Mesopotamia CA 7500 - Mesopotamian mathematicians discover the Pythagorean Theorem MEDIEVAL PERIOD CA 500 -1500 - Dark ages because few written records and evidences remained - Scholastic tradition was established by Charlemagne - Vertical windmills, spectacles, mechanical clock, water mills, gothic style were invented - Johannes Gutenberg invented the printing press RENAISSANCE PERIOD 14TH – 17TH CENTURY - Rebirth of revival - Printing with movable type allowed Bible, secular books made in large amount - Nicolas Copernicus presented a heliocentric theory - Galileo Galilei invented telescope INDUSTRIAL REVOLUTION 18TH CENTURY - Skilled workers were set aside because of the machines - Iron production, steam engine and textile flourished - Scottish James Watt improved steam engine Robert Fulton (steam boat) - The following were invented: Light bulb, telephone, first steam powered locomotive 19TH CENTURY - Age of machine and tools - Herman Helmholtz (law of conservation of energy) - James Clark Maxwell (light as electro-magnetic wave) - Henry Becquerel (radioactivity) - Marie and Pierre Curie (radium) - Hans Christian Oersted (electric current near the magnet) - Michael Faraday (magnet produces electricity) - Atomic Theory proposed by John Dalton - Electron discovered by JJ. Thomson - Telegraph developed by Samuel Morse 20TH CENTURY - Communication, transportation, military research were developed - Personal computer was created - Intel developed microprocessor - Apple was introduced by Steve Jobs and Steve Wozniak - Internet was created (ARPANET) - Henry Ford's mass production of cars - Artificial Intelligence was invented SCIENCE, TECHNOLOGY AND SOCIETY (PHILIPPINE HISTORY) Stone Age - Archeological findings show that modern man from Asian mainland first came over land on across narrow channels to live in Batangas and Palawan about 48,000 B.C. - Subsequently they formed settlement in Sulu, Davao, Zamboanga, Samar, Negros, Batangas, Laguna, Rizal, Bulacan and Cagayan. Inventions - They made simple tools and weapons of stone flakes and later developed method of sawing and polishing stones around 40,000 B.C. - By around 3,000 B.C. they were producing adzes ornaments of seashells and pottery. Pottery flourished for the next 2,000 years until they imported Chinese porcelain. Soon they learned to produce copper, bronze, iron, and gold metal tools and ornaments. Iron Age - The Iron Age lasted from the third century B.C. to 11th century A.D. During this period Filipinos were engaged in extraction smelting and refining of iron from ores, until the importation of cast iron from Sarawak and later from China. INVENTIONS AND DISCOVERIES - They learn to weave cotton, make glass ornaments, and cultivate lowland rice and dike fields of terraced fields utilizing spring water in mountain regions. - They also learned to build boats for trading purposes. - Spanish chronicles noted refined plank built warships called caracoa suited for interisland trade raids 10TH CENTURY A.D. - Filipinos from the Butuan were trading with Champa (Vietnam) and those from Ma-I (Mindoro) with China as noted in Chinese records containing several references to the Philippines. These archaeological findings indicated that regular trade relations between the Philippines, China and Vietnam had been well established from the 10th century to the 15th century A.D. TRADING - The People of Ma-I and San-Hsu (Palawan) traded bee wax, cotton, pearls, coconut heart mats, tortoise shell and medicinal betel nuts, panie cloth for porcelain, leads fishnets sinker, colored glass beads, iron pots, iron needles and tin. SOME PRESPANISH FILIPINO SCIENCE AND TECHNOLOGY - Curative values of plants extract use as medicine - Alphabet (Alibata) - Counting Methods - Weights - Measuring system (isang gatang) - Calendar based on the periods of moon - Banaue Rice Terraces SPANISH REGIME  Religion the Catholic Church - The latter part of the 16th Century Development of schools: - Colegio de San Ildefonso-Cebu-1595 - Colegio de San Ignacio-Manila-1595 - Colegio De Nuestra Senora del Rosario-Manila 1597 - Colegio De San Jose-Manila-1601  Colegio De San Ildefonso De Cebu - In 1863 the colonial authorities issued a royal degree to reform the existing educational system. In 1871 the school of medicine and pharmacy were opened to UST, after 15 years it had granted the degree Of Licenciado En Medicina to 62 graduates.  Medicine - Development of hospitals San Juan Lazaro hospital the oldest in the far east was founded in 1578.  Roads and Bridges Among other Spanish contributions: - Arithmetic - Algebra - Geometry - Trigonometry - Physics - Hydrography - Meteorology - Navigation - Pilotage American Period and Post Commonwealth Era - BUREAU OF GOVERNMENT LABORATORIES (1901) - BUREAU OF SCIENCE (1905) - INSTITUTE OF SCIENCE (1946) RA 2067 OTHERWISE KNOWN AS THE “SCIENCE ACT OF 1958”. - This was enacted to integrate, coordinate, and intensify scientific and technological research and development and to foster invention including allocation of funds and other purposes. NATIONAL RESEARCH COUNCIL WAS ESTABLISHED ON DECEMBER 8, 1933. - Its Mandate (Nrcp) Promotes And Supports Fundamental Or Basic Research For The Continuing Total Improvement Of The Research Capability Of Individual Scientists Or Group Of Scientists; Provides Advice On Problems And Issues Of National Interest; Promotes Scientific And Technological Culture To All Sectors Of Society; And Fosters Linkages With Local And International Scientific Organizations For Enhanced Cooperation In The Development And Sharing Of Information NATIONAL RESEARCH COUNCIL WAS ESTABLISHED IN DECEMBER 8, 1933. - Its Mandate (NRCP) promotes and supports fundamental or basic research for the continuing total improvement of the research capability of individual scientists or group of scientists; provides advice on problems and issues of national interest; promotes scientific and technological culture to all sectors of society; and fosters linkages with local and international scientific organizations for enhanced cooperation in the development and sharing of information. It was during the American Period when Science was inclined towards: - Agriculture - Food Processing - Forestry - Medicine - Pharmacy - NursingIntroduction 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.SewingSewing Machine 8Sewing MachineSewing Tools