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900/1-STPM 2023
Quiz by khalijahlatif SK (kEllydjay)
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España 1.900-1.931
V.Zip 2.0 1,801-1,900(Final)
1000語マスター801~900
السؤال 1: باعت شركة 200 وحدة من منتج معين بسعر 30 دولارًا للوحدة. ما هي إيرادات المبيعات؟ A) 6000 دولار B) 5000 دولار C) 7000 دولار D) 8000 دولار ✅ الإجابة: A) 6000 دولار ________________________________________ السؤال 2: إذا باعت شركة 120 وحدة بسعر 25 دولارًا للوحدة، ما هي إيرادات المبيعات؟ A) 3000 دولار B) 3500 دولار C) 4000 دولار D) 5000 دولار ✅ الإجابة: B) 3500 دولار ________________________________________ السؤال 3: باعت شركة 500 وحدة من منتج بسعر 12 دولارًا للوحدة، ما هي إيرادات المبيعات؟ A) 6000 دولار B) 5500 دولار C) 5000 دولار D) 4000 دولار ✅ الإجابة: A) 6000 دولار ________________________________________ السؤال 4: إذا تم بيع 150 وحدة من منتج سعرها 18 دولارًا، ما هي إيرادات المبيعات؟ A) 2700 دولار B) 2500 دولار C) 3000 دولار D) 3500 دولار ✅ الإجابة: A) 2700 دولار ________________________________________ السؤال 5: شركة باعت 400 وحدة بسعر 10 دولارات للوحدة. ما هي إيرادات المبيعات؟ A) 4000 دولار B) 3000 دولار C) 5000 دولار D) 4500 دولار ✅ الإجابة: A) 4000 دولار ________________________________________ السؤال 6: باعت شركة 800 وحدة بسعر 8 دولارات للوحدة. ما هي إيرادات المبيعات؟ A) 6400 دولار B) 7200 دولار C) 8000 دولار D) 8500 دولار ✅ الإجابة: A) 6400 دولار ________________________________________ السؤال 7: إذا كانت إيرادات المبيعات لشركة 1600 دولار، وسعر الوحدة 20 دولارًا، فكم عدد الوحدات المباعة؟ A) 50 B) 80 C) 90 D) 100 ✅ الإجابة: B) 80 ________________________________________ السؤال 8: تم بيع 1000 وحدة من منتج بسعر 5 دولارات للوحدة. ما هي إيرادات المبيعات؟ A) 5000 دولار B) 6000 دولار C) 7000 دولار D) 8000 دولار ✅ الإجابة: A) 5000 دولار ________________________________________ السؤال 9: شركة تبيع 150 وحدة من منتج بسعر 50 دولارًا للوحدة. ما هي إيرادات المبيعات؟ A) 7500 دولار B) 8000 دولار C) 7000 دولار D) 9000 دولار ✅ الإجابة: A) 7500 دولار ________________________________________ السؤال 10: إذا باعت شركة 200 وحدة بسعر 45 دولارًا لكل وحدة، ما هي إيرادات المبيعات؟ A) 8500 دولار B) 9000 دولار C) 9500 دولار D) 10000 دولار ✅ الإجابة: B) 9000 دولار ________________________________________ السؤال 11: شركة تبيع 50 وحدة بسعر 100 دولار للوحدة. ما هي إيرادات المبيعات؟ A) 5000 دولار B) 4000 دولار C) 6000 دولار D) 7000 دولار ✅ الإجابة: A) 5000 دولار ________________________________________ السؤال 12: إذا تم بيع 75 وحدة من منتج بسعر 12 دولارًا للوحدة، ما هي إيرادات المبيعات؟ A) 800 دولار B) 850 دولار C) 900 دولار D) 1000 دولار ✅ الإجابة: C) 900 دولار
Google Form 2 (After the Class) Title: Post-Class Learning Check – Introduction to International Curriculum Description: This quiz is designed to assess your understanding of Session 1. Q1. An international curriculum is designed primarily to: ○ Increase homework ○ Replace national education systems ✅ Prepare learners for global citizenship and international education ○ Reduce classroom teaching Q2. Which of the following is NOT an international curriculum? ○ Cambridge International ○ IB ○ Pearson Edexcel ✅ CBSE Q3. Why did the International Baccalaureate originally develop? ○ To replace national curricula ○ To introduce online learning ✅ To provide a common qualification for internationally mobile students ○ To reduce examinations Q4. Which IB programme was introduced first? ○ PYP ○ MYP ✅ DP ○ CP Q5. The IB currently offers _____ programmes. ○ Two ○ Three ✅ Four ○ Five Q6. What is the major difference between a national curriculum and the IB curriculum? ○ IB teaches fewer subjects. ○ IB is easier. ✅ IB focuses on inquiry, conceptual understanding, and international-mindedness. ○ IB has no assessments. Q7. Why are there more than 8,900 IB programmes but only 6,200 IB World Schools? ○ Every programme is counted twice. ○ Students are counted as programmes. ✅ Many schools offer more than one IB programme. ○ Different countries count programmes differently. Q8. Which statement best explains the purpose of the IB? ○ To prepare students only for university entrance. ○ To prepare students only for employment. ✅ To develop internationally minded, lifelong learners. ○ To prepare students only for examinations. Q9. Match the programme with the age group. Programme Age Group PYP 3–12 MYP 11–16 DP 16–19 CP 16–19 (Use Google Forms "Multiple Choice Grid" or "Matching" add-on if available.) Q10. Which statement best describes your learning after today's session? ○ I still have many doubts. ○ I understood some concepts. ○ I have a good understanding of international curriculum and the IB. ○ I am confident in explaining the differences between national and international curricula. Bonus Reflection Question (Optional – Short Answer) In one or two sentences, describe one feature of the IB curriculum that you would like to adopt in your own teaching practice and explain why.
Crea un quiz basandoti sul seguente testo: La Repubblica Popolare Cinese (中華人民共和國T, 中华人民共和国S, Zhōnghuá Rénmín GònghéguóP ascolta la pronuncia in mandarino standardⓘ), detta anche solo Cina (中國T, 中国S, ZhōngguóP; lett. "Paese di mezzo"),[7] è uno Stato dell'Asia orientale. La Repubblica Popolare Cinese è stata in passato indicata come Cina popolare, al fine di distinguerla dalla Repubblica di Cina, comunemente chiamata Taiwan (o Formosa), indicata invece come Cina nazionalista. Entrambe le entità reclamano il controllo sul territorio complessivo cinese. La Repubblica Popolare Cinese con 1 miliardo e 400 milioni di abitanti nel 2023 è lo stato più popoloso del mondo insieme all'India.[4] La Cina è una repubblica popolare in cui il potere è esercitato dal Partito Comunista Cinese (中国共产党 oppure 中共). Il governo ha sede nella capitale Pechino (北京首都) ed esercita la propria sovranità su ventidue province (省), cinque regioni autonome (自治区), quattro municipalità direttamente controllate (直辖市) (Pechino 北京, Tientsin 天津, Shanghai 上海 e Chongqing 重庆) e due regioni amministrative speciali 特别行政区 (Hong Kong 香港 e Macao 澳门) parzialmente autonome. La Cina rivendica la propria sovranità anche su Taiwan, che a propria volta rivendica la propria sovranità sulla Cina continentale. L'isola è rimasta dal 1949 sotto il controllo del governo della Repubblica di Cina (中華民國 o Taiwan), che precedentemente governava anche la Cina continentale, ed è rivendicata dalla Repubblica Popolare Cinese come provincia di Taiwan. La complessa condizione politica di Taiwan è una delle conseguenze della guerra civile cinese, che ha preceduto la fondazione della Repubblica Popolare Cinese. Con la sua superficie di circa 9 572 900 km², la Cina è il quarto stato più grande del mondo per superficie. Il paesaggio della Cina è vasto e diversificato: va dalle steppe della foresta e i deserti dei Gobi e del Taklamakan nell'arido nord alle foreste subtropicali e umide del sud. L'Himalaya, il Karakorum, il Pamir e il Tian Shan sono le catene montuose che separano la Cina meridionale dall'Asia centrale. Il Fiume Azzurro (长江) e il Fiume Giallo (黄河), rispettivamente il terzo e il sesto più lunghi del mondo, scorrono dall'altopiano del Tibet verso la costa orientale, densamente popolata. La costa della Cina lungo l'oceano Pacifico è lunga circa 14 500 chilometri ed è delimitata dal mare di Bohai, dal mar Giallo, dal mar Cinese Orientale e dal mar Cinese Meridionale. L'antica civiltà cinese, una delle più antiche al mondo, si sviluppò inizialmente nelle pianure comprese tra il Fiume Giallo e il Fiume Azzurro. A partire dall'età del bronzo, verso la fine del II millennio a.C., si ha evidenza di strutture feudali, in cui i nobili si raccoglievano intorno a monarchie ereditarie. Vi sono testimonianze di una casata regnante nella prima metà del I millennio a.C., nota come dinastia Zhou (周朝), il cui declino condusse alla nascita di un discreto numero di regni indipendenti in competizione per il predominio sulla regione (periodo delle Primavere e Autunni, 春秋), con stagioni di conflitto che si fecero particolarmente accese nel periodo che va dall'VIII al III secolo a.C. Nel 221 a.C. lo Stato di Qin sconfisse e conquistò i territori di tutti gli altri Stati combattenti, dando vita al primo impero della storia cinese sotto la guida del primo imperatore cinese Qín Shǐ Huángdì della dinastia Qin (秦朝). Da quel momento il titolo di imperatore della Cina divenne il sinonimo della raggiunta supremazia. La dinastia Qin non durò a lungo, infatti i popoli precedentemente conquistati vennero poco dopo riuniti sotto l'egida della dinastia Han (汉朝, III secolo a.C. - III secolo d.C.). I quattro secoli in cui regnarono i sovrani della dinastia Han sono considerati cruciali per la definizione e l'affermazione della identità culturale cinese, tanto da divenire il termine con cui i cinesi definirono se stessi (con il termine appunto di etnia o popolo han, 汉族). Da allora, la storia cinese ha visto l'alternarsi di periodi di divisione e fasi di unificazione, con conseguenti periodi di frammentazione, contrazione o espansione territoriale, sotto l'egida di diverse dinastie, talora di etnia straniera, come avvenuto nel caso dei mongoli o dei mancesi. L'ultima dinastia fu quella dei Qing, il cui regno si concluse nel 1911 con la fondazione della Repubblica di Cina (中华民国). Dopo la sconfitta dell'Impero giapponese (大日本皇国) durante la seconda guerra mondiale, il Paese fu scosso dalla guerra civile, che vedeva contrapposte le forze nazionaliste del Kuomintang (国民党), il partito che allora deteneva il governo del paese, e le forze facenti capo al Partito Comunista Cinese. Nel 1949 la guerra si concluse con la sconfitta del Kuomintang e la conseguente fuga del governo nazionalista sull'isola di Formosa, nella cui capitale Taipei (台北) ha tuttora sede l'attuale Repubblica di Cina, altresì nota come Taiwan. In seguito alla vittoria conseguita sul continente, il 1º ottobre del 1949 a Pechino le forze comuniste guidate da Mao Zedong proclamarono ufficialmente la nascita della Repubblica Popolare Cinese. Dopo l'introduzione di riforme economiche nel 1978, l'economia cinese è diventata quella dalla crescita più rapida al mondo. A partire dal 2013, è la seconda economia più grande al mondo sia come PIL totale nominale sia per parità di potere d'acquisto; per quanto riguarda solamente il PIL nominale, invece, la Cina ha sorpassato il Giappone, sino ad allora seconda potenza mondiale dal 1987, nel 2010. Nel 2022 il prodotto interno lordo cinese è sui ventimila miliardi di dollari.[8] Essa è anche il più grande esportatore e importatore di merci al mondo. La Cina è ufficialmente uno Stato munito di armi nucleari e ha il più grande esercito permanente del mondo, con il secondo più grande bilancio della difesa. È, inoltre, membro dell'ONU dal 1971, quando ha preso il posto della Repubblica di Cina tra i seggi dei membri permanenti del Consiglio di sicurezza delle Nazioni Unite, e quindi gode del potere di veto. La Cina è anche membro di numerose organizzazioni multilaterali,[9] tra cui l'OMC, l'APEC, il BRICS, l'Organizzazione di Shanghai per la cooperazione, il BCIM[10] e il G20. La Cina, unanimemente riconosciuta come grande potenza dal consesso internazionale, è una potenziale superpotenza secondo un certo numero di accademici e analisti che si occupano di questioni militari, politiche ed economiche. Dissidenti politici e gruppi per i diritti umani hanno denunciato la dittatura del governo cinese per diffuse violazioni dei diritti umani, tra cui repressione politica, repressione delle minoranze religiose ed etniche, censura, sorveglianza di massa e la violenza utilizzata nel reprimere il dissenso, come quella esibita durante le proteste di piazza Tienanmen del 1989.
Lo sai che cos'è il fast fashion con fast fashion? Si intende quel settore dell'abbigliamento che realizza abiti a prezzi super ridotti e che lancia nuove collezioni di continuo pensate proprio per essere usate poco, il tutto sembra figo, no? Vestiti nuovi ogni volta che vuoi. Non è figo, ogni anno nel mondo vengono prodotti oltre 80 miliardi di capi d'abbigliamento e sulla terra siamo 8 miliardi e dopo il brevissimo periodo in cui vengono utilizzati tre capi su quattro finiscono in discarica o vengono inceneriti e soltanto un quarto viene riciclato Si stima che in un anno una persona compri circa 15 kg di vestiti. Eppure la sensazione è quella di non aver mai in mente da mettere e questo è esattamente quello che l'industria del fast fashion vuole. Lo sai che per produrre una t-shirt ci vogliono circa 3.900 litri di acqua l'equivalente di quello che beve una persona in cinque anni, un capo d'abbigliamento ha un impatto sull'ambiente in ogni fase della sua produzione, dalla produzione delle fibre, all'assemblaggio dal trasporto allo smaltimento, ma come è possibile che un top un prodotto che è stato coltivato come fibra trasformato in filato cucito dall'altra parte del mondo tinto confezionare e spedito costi meno di un panino? La risposta è semplice materie prime scadenti a basso costo abbinate quasi sempre allo sfruttamento dei lavoratori, la maggior parte dei capi di abbigliamento fast fashion viene prodotta in paesi come bangladesh, vietnam cambogia, spesso le condizioni di lavoro in questi paesi sono terribili e i diritti umani vengono calpestati, se paghiamo un top 7,95 significa che qualcun altro ha pagato il prezzo al posto nostro. Riassumendo il fast fashion nè non è sostenibile e probabilmente non ci rende più felici, ma l'abbigliamento è anche un modo per esprimere la propria personalità e creatività quindi ecco alcuni consigli. 1 comprare meno e curare meglio certo spendere tanto per una maglietta ci sembra assurdo ma ricordiamoci che una t-shirt a 4 euro probabilmente durerà poco o niente e soprattutto siamo sicuri di averne bisogno di 10? 2 scegliere bene i marchi alcuni siti possono aiutarti a scoprire nuovi brand etici e sostenibili e magari anche nuovi stili 3 riutilizzare e riparare, scambiamo i vestiti che non usiamo più quelle amiche quegli amici oppure vendiamoli sui mercatini online, ma soprattutto non buttiamoli appena salta un bottone, proviamo piuttosto a ripararli Come ultimo consiglio resiste l'impulso da shopping compulsivo, la maggior parte degli acquisti, soprattutto online in periodo di saldi vengono fatti d'impulso, fermati un momento e pensa a tutto ciò che quel vestito rappresenta, come è stato fatto, chi lo ha creato quanto ha inquinato e chiediti se ti serve davvero e ora che sai riconoscerlo cosa ne pensi del fast fashion?
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.