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The Big Football Match
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Match the word to its synonym level B1 CEFR. Use the vocabulary exactly adverb precisely except that aside from exist verb to be real existing adjective real, current Example: Flying cars are not practical with existing technology. existence noun reality Example: The existence of black holes has been confirmed by indirect observation. extraordinary adjective unusual feature noun important part of something Example: The Ramon Crater is a unique feature of the Negev Desert. feedback noun reaction figure noun shape Example: I can’t tell if that figure in the shadows is a man or a woman. figure out verb understand Example: I just can’t figure out how the magician did that amazing trick. financial adjective related to money Example: Her family is having financial problems so they can’t travel overseas this year. finance verb pay for Example: If I can’t get a loan from the bank, I won’t be able to finance a new apartment. finance noun money Example: An expert in finance predicts a global recession. finding/findings noun discoveries; results of a study Example: According to the findings of the police investigation, this is the gun which fired the fatal bullet. flexibility noun willingness to change flexible adjective adjusts easily Example: I’d prefer to meet on Monday morning but I can be flexible depending upon your schedule. flood noun a lot of water flood verb to cover with too much water flu noun type of sickness focus on/upon verb pay attention to Example: You should focus on your schoolwork if you want to improve your grades. focus noun attention People with attention deficit disorder lose focus easily. frequency noun how often frequent adjective very often Example: Hanah is a frequent customer and everyone at the store knows her. fresh adjective new Example: We need some fresh ideas if we’re going to solve this problem. frighten verb scare from preposition position, starting point gain verb make an increase, profit, earn Example: I have nothing to gain by choosing sides so I shall remain neutral. gain noun profit, amount earned generate verb create, make Example: Chat GPT can generate text written in any style you choose. guidance noun help, advice hopeful adjective optimistic, having a positive outlook Example: The farmers are hopeful that we will have rain this winter. hopefully adjective with luck ideal adjective best, most preferable Example: Nuclear power may not be an ideal solution to global warming, but it’s certainly worth considering. illness noun sickness, disease illustrate verb draw pictures illustration noun picture, image Example: Children’s storybooks have colorful illustrations. image noun picture, especially on film or television Example: The mother of the pop singer cried when she first saw her daughter’s image on television. in preposition within, inside, into in terms of regarding Example: That company makes a great product but they’re lacking in terms of customer service. in actual fact in truth Example: The mayor says the city is a safe place to live, but in actual fact the violent crime rate is very high. in connection with about Example: Police arrested four men in connection with the robbery. in that case if that is true Example: Billy Bob: “Traffic could be heavy tomorrow.” Peggy Sue: “In that case, we better leave early.” in the meantime while, during Example: The new computers won’t arrive until next week, but we can keep using the old ones in the meantime. initial adjective first Example: Her initial reaction to that song was negative, but over time she’s come to like it. initially adverb at first instruction noun teaching, order Example: Most new electronic devices come with a set of instructions. intelligence noun smartness Example: Since you have a degree from a good university, I assume you have sufficient intelligence to understand this problem. intelligent adjective smart Example: Joe isn’t very intelligent, but he is a kind person with a warm heart. interest noun attraction Example: Yossi has little interest in politics, whereas his wife goes to all the protests and demonstrations. interest verb to attract Example: Sports don’t really interest me, but my brother is a big basketball fan. introduce verb to show something new Example: Today in class I will introduce the basic concepts of literary analysis. invest verb to put money into something in order to earn money Example: Joe invested in cryptocurrency and lost a lot of money. investor noun one who puts money into something in order to earn money Example: Venture capitalists are investors who put money into risky start-up businesses. investment noun putting money into something in order to earn money Example: Buying real estate in Israel is a very safe investment because the value never goes down. investigate verb research, study Example: The police collected evidence to investigate the murder. investigation noun study Example: The police don’t have a suspect for the murder as the investigation isn’t finished yet. investigator noun detective Example: Detective Schmendrick is the lead investigator for the murder case. just about almost Example: I’m just about done here so I’ll be there shortly. keep on doing verb continue Example: You’re crazy if you keep on doing the same thing and expect different results. kind of type of Example: What kind of dog is that, a poodle? knowledge noun awareness Example: John failed the test due to lack of knowledge of the material. lack verb not having, missing Example: John failed the test due to lack of knowledge of the material. landscape noun the view of the land likely adjective, adverb probably Example: When we learn from our mistakes, we’re not likely to forget. limited adjective restricted Example: We should go to the store today because the sale is for a limited time only. limitation noun restriction little adjective small, not a lot Example: She always tells the truth. I have little reason to doubt her. look at verb see Example: People used to read newspapers on the train. Nowadays they just look at their phones. low adverb to a small amount or level Example: I have to charge my phone because the battery is running low. material noun documents, information Example: We have a lot of material to cover before the end of the semester. meaning noun significance mean verb to have significance or purpose means noun form of, by the use of Example: They communicate by means of radio. measure noun step Example: The teacher took measures to prevent cheating during the test mention verb to say, point out Example: The coach said the team played very well today but didn’t mention any player specifically. miss verb (1) fail to catch (2) wishing to see somebody Examples: (1) The football player kicked the ball but missed the goal. (2) Wow, it’s good to see you! I’ve missed you so much! misunderstand verb understand incorrectly Example: I’m afraid I misunderstood the instructions. Could you repeat them please? more or less approximately, somewhat, to a varying degree Example: This is more or less a religious neighborhood, though there are a few secular families. must modal verb have to naturally adverb as expected, normally nature noun (1) open air (2) character Examples: (1) We like to go hiking in nature reserves. (2) Pit bulls are aggressive by nature.
Uzzle, the Football Star Uzzle loves football. He plays it at the playground. He even dreams about it at night. One day, he and Jupe see a football helmet in a store window. "If I had that helmet, I would be a star," he says. Uzzle looks at the price. "It costs too much money," he says. "Maybe there is a helmet like that at a thrift store," says Jupe. "Great idea!" says Uzzle. Uzzle and Jupe go to the thrift store. There are cowboy hats. There are baseball caps. But there are no football helmets. They go to another thrift store. There are hard hats. There are top hats. But there are no football helmets. They go to the last thrift store. Uzzle sees a football helmet! Uzzle tries it on. The helmet slips over his eyes. "Does it fit?" asks Jupe. "It is too big, but monster heads grow fast," says Uzzle. Uzzle wears his too-big football helmet. He bumps into walls. He trips on the curb. He tips the helmet to look in the mirror. "I look like a star," he says. He goes to the playground. Jupe is there. "You can't play football in a too-big helmet," she says. "Yes, I can," says Uzzle. But Uzzle can't pass, punt, or kick. Instead, he misses, trips, and falls. Just then, Lurk comes to play football. Uzzle tips his helmet. He sees that Lurk's head is much bigger than his head. "That's a cool helmet," Lurk says. Uzzle thinks. "I wish I had a helmet like that," Lurk says. Uzzle thinks some more. Finally, Uzzle takes off the too-big helmet. "It doesn't fit me, but maybe it will fit you," Uzzle says. Lurk slips it on. "It fits!" he says. "Thank you, Uzzle!" Then Uzzle, Jupe, and Lurk play football. Uzzle can pass agaín. He can punt and kick again. He can even score! "You're a star even without that too-big helmet," says Jupe. "I am?" says Uzzle. "You are," says Lurk in his helmet that is just right!
For instance, carbon dioxide from the air or soil sometimes combines with water in a process called carbonation. This produces a weak acid, called carbonic acid, that can dissolve rock. Carbonic acid is especially effective at dissolving limestone. When carbonic acid seeps through limestone underground, it can open up huge cracks or hollow out vast networks of caves. Carlsbad Caverns National Park, in the U.S. state of New Mexico, includes more than 119 limestone caves created by weathering and erosion. The largest is called the Big Room.. With an area of about 33,210 square meters (357,469 square feet), the Big Room is the size of six football fields. Another type of chemical weathering works on rocks that contain iron. These rocks turn to rust in a process called oxidation. Rust is a compound created by the interaction of oxygen and iron in the presence of water. As rust expands, it weakens rock and helps break it apart. Another familiar form of chemical weathering is hydrolysis. In the process of hydrolysis, a new solution (a mixture of two or more substances) is formed as chemicals in rock interact with water. In many rocks, for example, sodium minerals interact with water to form a saltwater solution. Hydration and hydrolysis contribute to flared slopes, another dramatic example of a landscape formed by weathering and erosion. Flared slopes are sometimes nicknamed "wave rocks." Their c-shape is largely concave rock formations a result of subsurface weathering, in which hydration and hydrolysis wear away rocks beneath the landscape's surface
Weathering describes the breaking down or dissolving of rocks and minerals on the surface of the Earth. Water, ice, acids, salts, plants, animals, and changes in temperature are all agents of weathering. Once a rock has been broken down, a process called erosion transports the bits of rock and mineral away. No rock on Earth is hard enough to resist the forces of weathering and erosion. Together, these processes carved landmarks such as the Grand Canyon, in the U.S. state of Arizona. This massive canyon is 446 kilometers (277 miles) long, as much as 29 kilometers (18 miles) wide, and 1,600 meters (1 mile) deep. Weathering and erosion constantly change the rocky landscape of Earth. Weathering wears away exposed surfaces over time. The length of exposure often contributes to how vulnerable a rock is to weathering. Rocks, such as lavas, that are quickly buried beneath other rocks are less vulnerable to weathering and erosion than rocks that are exposed to agents such as wind and water, As it smoothes rough, sharp rock surfaces, weathering is often the first step in the production of soils. Tiny bits of weathered minerals mix with plants, animal remains, fungi, bacteria, and other organisms. A single type of weathered rock often produces infertile soil, while weathered materials from a collection of rocks is richer in mineral diversity and contributes to more fertile soil. Soils types associated with a mixture of weathered rock include glacial till, loess, and alluvial sediments. Weathering is often divided into the processes of mechanical weathering and chemical weathering. Biological weathering, in whichliving or once-living organisms contribute to weathering, can be a part of both processes. Mechanical Weathering Mechanical weathering, also called physical weathering and disaggregation, causes rocks to crumble. Water, in either liquid or solid form, is often a key agent of mechanical weathering. For instance, liquid water can seep into cracks and crevices in rock. If temperatures drop low enough, the water will freeze. When water freezes, it expands. The ice then works as a wedge. It slowly widens the cracks and splits the rock. When ice melts, liquid water performs the act of erosion by carrying away the tiny rock fragments lost in the split. This specific process (the freeze-thaw cycle) is called frost weathering or cryofracturing. Figure 4.3 Frost Wedging Temperature changes can also contribute to mechanical weathering in a process called thermal stress. Changes in temperature cause rock to expand (with heat) and contract (with cold). As this happens over and over again. the structure of the rock weakens. Over time, it crumbles. Rocky desert landscapes are particularly vulnerable to thermal stress. The outer layer of desert rocks undergo repeated stress as the temperature changes from day Eventually, Lo outer night. layersflake off in thin sheets, a process called exfoliation. Exfoliation contributes to the formation of bornhardts, one of the most dramatic features in landscapes formed by weathering and erosion. Bornhardts are tall, domed, isolated rocks often found areas. in tropical Sugarloaf Mountain, an iconic landmark in Rio de Janeiro, Brazil, is bornhardt. a Salt also works to weather rock in a process called haloclasty. Saltwater sometimes gets into the cracks and pores of rock. If the saltwater evaporates, salt crystals are left behind. As the crystals grow, they put pressure on the rock, slowly breaking it apart. Plants and animals can be agents of mechanical weathering. The seed of a tree may sprout in soil that has collected in a cracked rock. As the roots grow, they widen the cracks, eventually breaking the rock into pieces. Over time, trees can break apart even large rocks. Even small plants, such as mosses, can enlarge tiny cracks as they grow. Animals that tunnel underground, such as moles and prairie dogs, also work to break apart rock and soil. Other animals dig and trample rock aboveground, causing rock to slowly crumble. Chemical Weathering Chemical weathering changes the molecular structure of rocks and soil.For instance, carbon dioxide from the air or soil sometimes combines with water in a process called carbonation. This produces a weak acid, called carbonic acid, that can dissolve rock. Carbonic acid is especially effective at dissolving limestone. When carbonic acid seeps through limestone underground, it can open up huge cracks or hollow out vast networks of caves. Carlsbad Caverns National Park, in the U.S. state of New Mexico, includes more than 119 limestone caves created by weathering and erosion. The largest is called the Big Room.. With an area of about 33,210 square meters (357,469 square feet), the Big Room is the size of six football fields. Another type of chemical weathering works on rocks that contain iron. These rocks turn to rust in a process called oxidation. Rust is a compound created by the interaction of oxygen and iron in the presence of water. As rust expands, it weakens rock and helps break it apart. Another familiar form of chemical weathering is hydrolysis. In the process of hydrolysis, a new solution (a mixture of two or more substances) is formed as chemicals in rock interact with water. In many rocks, for example, sodium minerals interact with water to form a saltwater solution. Hydration and hydrolysis contribute to flared slopes, another dramatic example of a landscape formed by weathering and erosion. Flared slopes are sometimes nicknamed "wave rocks." Their c-shape is largely concave rock formations a result of subsurface weathering, in which hydration and hydrolysis wear away rocks beneath the landscape's surfaceWeathering and People Weathering is a natural process, but human activities can speed it up. For example, certain kinds of air pollution increase the rate of weathering Burning coal, natural and petroleum releases chemicals such as nitrogen oxide and gas, sulfur dioxide into the atmosphere. When these chemicals combine with sunlight and moisture, they change into acids. They then fall back to Earth as acid rain. Acid rain rapidly weathers limestone, marble, and other kinds of stone. The effects of acid rain can often be seen on gravestones, making names and other inscriptions impossible to read. Acid rain has also damaged many historic buildings and monuments. For example, at 71 meters (233 feet) tall, the Leshan Giant Buddha at Mount Emei, China is the world's largest statue of the Buddha. It was carved 1,300 years ago and sat unharmed for centuries. An innovative drainage system mitigates the natural process of erosion But in recent years, acid rain has turned the statue's nose black and made some of its hair crumble and fall.
Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.
Jackie Robinson Introduction. African Americans play a big part in professional sports today. For many years, however, black athletes weren't allowed to play with white athletes. Jackie Robinson helped change all that. The Early Years. Jack Roosevelt Robinson was born into a poor Georgia family in 1919. In college, he was a star on his school's football, track, basketball, and baseball teams. His family had little money, however. He left college in 1941 to help support his family and did not finish. Taking a Stand. In December 1941, the United States entered a war. Like many young men, Robinson had to serve in the war. One day, he and a group of soldiers got on an army bus. Robinson poses in his U.S. Army uniform. The bus was segregated. White soldiers sat in the front and black soldiers in the back. Yet Robinson knew he was as good a soldier as the white men. He would not move to the back when he was told to. Robinson was arrested, but he had only stood up for what was right. He was let go. You're Hired! After his time in the army, Robinson played baseball. In 1945, however, baseball was segregated, too. White and black athletes played in separate leagues. Robinson felt that there should not be separate baseball leagues based on skin color. So did Branch Rickey, the man who ran the Brooklyn Dodgers. Rickey wanted the Dodgers to be the first white team to include a black player. Rickey knew that this player would not only have to be a great athlete. He would also have to face prejudice because he was African American. Rickey hired Robinson. He had one condition, though. Robinson could only fight prejudice one way-by playing great baseball. Number 42 Takes the Field Wearing number 42, Robinson took the field on April 15, 1947. A crowd of twenty-six thousand people watched as he walked to the plate. Insults rang out from the other team's dugout, but Robinson just played ball. Equal rights won that day. So did the Dodgers. As for Robinson, he went on to have a great career. In 1962, he became the first African American to get into the Baseball Hall of Fame. Beyond Baseball. After Robinson stopped playing baseball in 1957, he went into business. He also continued to work for equal rights for all people. He died in 1972. Today in the United States, more people of color play in the world of sports than ever before. We all have Number 42 to thank for that.
A trip to Rio Julia and her family traveled from New York to visit Rio de Janeiro in Brazil. Julia's cousin Gabriela lived there. They all went to the opening night of the Olympics. The stadium was very crowded. It made Julia nervous. Everyone screamed and cheered. Their seats were far away. Julia could barely see. The music was loud. It made her head hurt. Julia had been happy to visit Rio. Now she just wanted to go home. Gabriela woke Julia up the next morning. "There's another Olympic event today!" she said. Julia did not want to go, but she smiled and got ready. The families walked through shady streets. Gabriela's street ended at a beach. Julia stopped and stared. Tall buildings stood along the beach. Olympic racing boats floated on the water. There was a big mountain behind them. "That's Sugarloaf Mountain," Gabriela said. It was beautiful. The next day, Julia ran to Gabriela's room. "We're going up Sugarloaf Mountain!" she said. They rode a cable car. It hung high above the city. Julia stared out the window. White buildings stood above the green jungle. They went to a big market. Julia tasted a mango. It was not like the mangoes at home. It was juicy and sweet! They went to an Olympic swimming race. Gabriela's brother, Chaz, cheered, "Go Brazil!" "Brazil is not even in this event!" Gabriela said. "Oh." Chaz said. He smiled at Julia. "Go Americа!" It was Julia's last day in Rio. They went to Grandma and Grandpa's. Julia remembered the house. She had visited when she was five. Grandpa had taught her to dance. It felt like home. Grandma made a spicy bean stew. After lunch, they went to an Olympic football game. "The crowd is very noisy," Julia said. "I'm scared." "Don't worry," Grandpa said. "Football fans are one big family." At the stadium, the crowd seemed even louder. Julia held Grandpa's hand. Brazil got the ball. Everyone cheered. Julia got caught up in the game. She cheered, too. Then, Brazil scored a goal. The crowd cheered. Grandpa lifted Julia in the air. They sang a song with the crowd to celebrate.
The big picture of the earliest humans in Britain