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D.1 Séparations, divorces, nouveaux débuts. p.16,2
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CAS 1 Marc veut ouvrir un café. Il se lance comme indépendant et en est le chef d'entreprise. Il n'est donc plus possible de dissocier le café de Marc. Si Marc réalise en janvier un bénéfice de 2.400 euros (c'est-à -dire le chiffre d'affaires de ce mois-là , moins les frais), cette somme ne sera pas considérée comme faisant distinctement partie du patrimoine de l'entreprise. En principe, on peut considérer ces 2.400 euros de bénéfice comme étant le salaire de Marc pour ce mois-là . Il peut décider de la manière dont il utilisera cet argent: épargner pour une maison, réinvestir dans son entreprise en achetant une machine à café supplémentaire… À cet égard, il n'existe pas de règle légale. En cas de dettes, on peut saisir tout ce qui appartient à Marc, et donc l'entièreté de son patrimoine (il n'y a en effet pas de séparation claire), comme la maison qu'il occupe ou la voiture avec laquelle il circule… CAS 2 Elisa et Anne veulent aussi ouvrir un café. Elles sont deux, donc elles créent une société dont elles seront toutes les deux administratrices. Elisa et Anne sont à la fois les fondatrices et les administratrices, mais ce n'est pas toujours le cas. Ainsi, il est tout à fait possible que la mère d'Elisa ou l'ami d'Anne soit administrateur du café. Et si Anne souhaite arrêter, n'importe qui peut en principe reprendre son rôle d’administratrice. La société est en effet une personne morale et non une personne physique, de sorte que le café et Anne ne sont pas une seule et même personne, ce qui, en revanche, était le cas pour Marc et son café. Si Elisa et Anne réalisent en janvier un bénéfice de 2.400 euros (c'est-à -dire le chiffre d'affaires de ce mois-là , moins les frais et moins le salaire des deux administratrices), elles ne peuvent pas utiliser cette somme pour effectuer des dépenses personnelles (c’est leur salaire qui sert à ces dépenses). Elles ne peuvent l'utiliser que pour le café: elles peuvent mettre cet argent de côté, acheter du nouveau matériel…Elisa et Anne ne pourront jamais être déclarées personnellement en faillite, même si la société a d'énormes dettes qu'elle ne peut pas payer. Marc, en revanche, pourrait l'être. Source : www.liantis.be
1. Laquelle des formes juridiques suivantes offre la meilleure protection du patrimoine personnel de l'entrepreneur ? a) Entreprise individuelle b) Société c) Association de fait d) Aucune de ces réponses 2. Quel est l'inconvénient majeur de l'entreprise individuelle ? a) Difficulté de création b) Régime fiscal complexe c) Responsabilité illimitée de l'entrepreneur d) Impossibilité d'avoir des employés 3. Laquelle des affirmations suivantes concernant la société est vraie ? a) Elle n'a pas d'existence juridique propre. b) Le patrimoine des associés est confondu avec celui de la société. c) Elle peut poursuivre ses activités même après le décès d'un associé. d) Elle est toujours soumise à l'impôt sur le revenu des personnes physiques. 4. Dans une association de fait, que se passe-t-il en cas de mauvaise gestion financière de l'un des associés ? a) Seul l'associé responsable est tenu de rembourser les dettes. b) Tous les associés sont solidairement responsables des dettes. c) L'association est automatiquement dissoute. d) Un tribunal désigne un administrateur judiciaire pour gérer les finances. 5. Lequel des éléments suivants est un avantage de la société par rapport à l'entreprise individuelle ? a) Simplicité de création et de gestion b) Responsabilité limitée des associés c) Contrôle total et décisions unilatérales d) Fiscalité moins avantageuse 6. Qu'est-ce qu'une personne morale ? a) Un individu exerçant une activité commerciale en son nom propre b) Une entité juridique distincte de ses membres, dotée de droits et d'obligations c) Un contrat par lequel deux personnes s'engagent à mettre en commun des biens ou leur travail d) Une forme d'entreprise réservée aux professions libérales 7. Laquelle des affirmations suivantes est vraie concernant l'association de fait ? a) Elle nécessite la création d'une personne morale. b) Elle offre une protection du patrimoine personnel des participants. c) Elle peut être considérée comme une entreprise si elle verse des bénéfices à ses membres. d) Elle est soumise aux mêmes obligations comptables que les sociétés. 8. Quel est l'un des avantages de la société en matière de continuité d'activité ? a) Elle est dissoute automatiquement au décès de son fondateur. b) Elle peut être facilement transmise aux héritiers en cas de décès d'un associé. c) Elle cesse son activité si un associé décide de se retirer. d) Elle doit être liquidée en cas d'incapacité de travail d'un associé. 9. Lequel des critères suivants peut influencer le choix entre une entreprise individuelle et une société ? a) Le montant du capital initial b) Le nombre d'employés c) Le secteur d'activité d) La volonté de protéger son patrimoine personnel 10. Quel est l'un des risques majeurs liés à l'absence de séparation des patrimoines dans l'entreprise individuelle ? a) La responsabilité limitée de l'entrepreneur b) La saisie du patrimoine personnel en cas de dettes de l'entreprise c) L'impossibilité de déduire les charges professionnelles de ses revenus d) La difficulté d'obtenir un financement bancaire
1. Battle of Lexington At the Battle of Lexington in 1775, British soldiers and colonial militia faced each other on the village green. A shot was fired, but no one knows for sure who fired first. Even so, this moment is often called the beginning of the American Revolution. Why is the Battle of Lexington still considered the start of the American Revolution, even though it is unclear who fired the first shot? A. It marked the first time colonists and British soldiers fought in open battle B. It proved that colonists planned the war long before the fighting began C. It showed that British soldiers were fully responsible for starting the war D. It confirmed that colonists had already declared independence from Britain 2. Battle of Concord After the fighting at Lexington, British troops marched to Concord to destroy colonial supplies. Instead, colonial militia gathered and fought back, forcing the British to retreat toward Boston. Why is the Battle of Concord considered an important turning point in the early American Revolution? A. It showed that colonial militias could organize and successfully push back British troops B. It proved that the British army had already lost control of all the colonies C. It confirmed that the colonies had officially declared independence from Britain D. It demonstrated that foreign countries were already helping the colonial forces 3. Second Continental Congress Much of the early violent conflict between colonists and British soldiers took place in Boston. After fighting broke out at Lexington and Concord near Boston, colonial leaders met at the Second Continental Congress in 1775 to decide what to do next. Which answer best describes the main actions taken by the Second Continental Congress? A. They created an army, chose a leader, and tried to avoid war with Britain B. They declared independence, wrote the Constitution, and ended the war C. They raised taxes, formed a monarchy, and supported British rule D. They ended slavery, gave women rights, and expanded voting laws 4. Olive Branch Petition In 1775, colonial leaders sent a letter called the Olive Branch Petition to King George III of Britain. Based on this situation, what was the main purpose of the Olive Branch Petition sent to the king? A. To ask the king to restore peace between both sides B. To declare independence from Britain and begin a new nation C. To request help from foreign countries in the war effort D. To organize protests against British taxes across the colonies 5. Battle of Bunker Hill On June 17, 1775, during the Battle of Bunker Hill in Boston, colonial forces fought against the British on a hill overlooking the city. The fighting was intense and led to heavy losses on both sides. Which statement best explains why the Battle of Bunker Hill was an important battle in the war? A. The colonists won the battle, showing they were stronger than British forces B. The colonists won the battle, showing the British that the war would be short and easy for the colonists C. The British won the battle, showing the war would be difficult and costly for both sides D. The British won the battle, showing British forces could defeat the colonists easily 6. Pamphlets During the American Revolution, pamphlets were short printed writings that were inexpensive to produce and often written in everyday language so many colonists could read them. How did these features of pamphlets most affect their role in the American Revolution? A. They helped spread ideas widely, allowing more colonists to form and share opinions B. They limited ideas to educated leaders, keeping most colonists uninvolved C. They replaced newspapers entirely, becoming the only source of information D. They prevented disagreement, causing most colonists to think the same way 7. Thomas Paine’s Common Sense In 1776, Thomas Paine published Common Sense, a widely read piece of writing about the relationship between the colonies and Britain. How did this pamphlet most influence colonial thinking during the American Revolution? A. It encouraged colonists to support independence from Britain B. It convinced colonists to remain loyal to the British government C. It explained how colonial armies should organize attacks D. It described laws colonists were expected to follow 8. Declaration of Independence In July 1776, the Declaration of Independence listed complaints against King George III and explained the colonists’ ideas about government and rights. How do these parts of the Declaration of Independence work together to support the colonists’ decision? A. They connect ideas about rights to real examples, justifying separation from Britain B. They describe past events in detail, showing how the war had already ended C. They list future plans for government, explaining how leaders would be chosen D. They organize military actions, showing how the colonies planned to win 9. Washington Crosses the Delaware and Battle of Trenton In December 1776, George Washington led his army across the Delaware River and launched a surprise attack on Hessian mercenaries in Trenton. Which statement best explains why Washington’s crossing of the Delaware and the attack on Trenton was an important turning point in the war? A. It defeated British forces completely, ending the war in a short time B. It boosted morale, helping discouraged soldiers choose to keep fighting C. It brought foreign allies into the war, adding support for the colonies D. It led to independence, allowing the colonies to form a new nation 10. Battle of Saratoga In 1777, American forces defeated the British at the Battle of Saratoga, a major event during the American Revolution. Which statement best explains why the Battle of Saratoga was an important turning point in the war? A. It brought French support, helping Americans gain a strong advantage in the war B. It ended the war quickly, forcing Britain to surrender all control in the colonies C. It improved army training, helping soldiers become more skilled in future battles D. It changed leadership roles, causing new generals to take control of the army 11. Battle of Yorktown In 1781, American forces surrounded British troops at Yorktown, leading to a major moment in the American Revolution. Which statement best explains why the Battle of Yorktown was an important event in the war? A. French forces helped the Americans win, leading to the end of major fighting in the war B. Italian forces helped the British win, leading to a final victory over the American army C. German forces switched sides and helped the Americans win, leading to a final defeat for British troops D. French forces helped the British win, leading to a complete end of the war in the colonies 12. Treaty of Paris After the Battle of Yorktown, British public opinion turned against the war, and peace negotiations began. In 1783, American leaders signed the Treaty of Paris, which included agreements between the United States and Britain. Which statement best explains how the Treaty of Paris reflected the outcome of the American Revolution? A. Both sides made agreements, recognizing independence B. Americans gained independence, taking land without agreements C. British leaders kept control, ending the war with power D. Both sides refused compromise, continuing the war
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.
6 Kenji SCIENCE Separating Mixtures PART 1
D 1
D 1 Vocab 6A.2 + 3 modal verbs
D.1 Relations parents-ados p.18,1