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Equivalenze intervallo di tempo
Quiz by Laura Iannotti
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Crea un quiz con le seguenti domande. Inserisci anche la spiegazione. Domande Vero/Falso: 1. Vero o Falso: Se moltiplichiamo entrambi i membri di una disequazione per un numero negativo, il segno dell'ineguaglianza cambia. o Risposta: Vero o Spiegazione: Quando moltiplichiamo o dividiamo entrambi i membri di una disequazione per un numero negativo, il segno dell'ineguaglianza si inverte. 2. Vero o Falso: Una disequazione può avere solo una soluzione. o Risposta: Falso o Spiegazione: Una disequazione può avere zero, una o infinite soluzioni, a seconda dei valori coinvolti. 3. Vero o Falso: Se sommiamo o sottraiamo la stessa quantità da entrambi i membri di una disequazione, la soluzione rimane invariata. o Risposta: Vero o Spiegazione: Aggiungere o sottrarre la stessa quantità da entrambi i membri di una disequazione non cambia la relazione tra le soluzioni. 4. Vero o Falso: Se abbiamo una disequazione del tipo 2x>102x>10, la soluzione è x<5x<5. o Risposta: Vero o Spiegazione: Dividendo entrambi i membri per 22, otteniamo x>5/2x>5/2, che può essere semplificato a x>2.5x>2.5 o x>5/2x>5/2. 5. Vero o Falso: Una disequazione può avere solo numeri interi come soluzioni. o Risposta: Falso o Spiegazione: Le soluzioni di una disequazione possono essere numeri razionali o reali, non solo numeri interi. 6. Vero o Falso: Una disequazione del tipo 3x−2<53x−2<5 ha x>7/3x>7/3 come soluzione. o Risposta: Falso o Spiegazione: La soluzione corretta è x<7/3x<7/3 poiché 3x−23x−2 deve essere minore di 55, non maggiore. 7. Vero o Falso: Una disequazione del tipo 4x+7≥3x+54x+7≥3x+5 ha una soluzione unica. o Risposta: Vero o Spiegazione: Sottraendo 3x3x da entrambi i lati otteniamo x+7≥5x+7≥5, che semplificato diventa x≥−2x≥−2, quindi ha una soluzione unica. 8. Vero o Falso: Una disequazione quadratica è un tipo di disequazione di primo grado. o Risposta: Falso o Spiegazione: Una disequazione quadratica coinvolge il quadrato di una variabile e può essere di secondo grado o superiore, mentre una disequazione di primo grado coinvolge solo variabili elevate alla prima potenza. 9. Vero o Falso: Una disequazione del tipo 2(x−3)<82(x−3)<8 può essere risolta dividendo entrambi i membri per 22. o Risposta: Vero o Spiegazione: Dividendo entrambi i membri otteniamo x−3<4x−3<4, che può essere semplificato a x<7x<7 dopo l'aggiunta di 33 ad entrambi i membri. 10. Vero o Falso: Se abbiamo una disequazione del tipo x≤4x≤4 e x≥3x≥3, allora la soluzione è x=4x=4. o Risposta: Falso o Spiegazione: La soluzione è 3≤x≤43≤x≤4, il che significa che xx può essere qualsiasi numero tra 33 e 44, inclusi tutti i valori decimali in questo intervallo. Domande a Risposta Multipla: 11. Qual è la soluzione della disequazione 2x+5>112x+5>11? a) x<3x<3 b) x>3x>3 c) x<8x<8 d) x>8x>8 o Risposta: b) x>3x>3 o Spiegazione: Sottraendo 55 da entrambi i lati otteniamo 2x>62x>6, quindi x>3x>3. 12. Quale delle seguenti è una soluzione della disequazione 3x−1≤83x−1≤8? a) x=3x=3 b) x=1x=1 c) x=0x=0 d) x=4x=4 o Risposta: d) x=4x=4 o Spiegazione: Aggiungendo 11 ad entrambi i lati otteniamo 3x≤93x≤9, quindi x≤3x≤3. 13. Quale delle seguenti disequazioni è equivalente a 2(x+1)>62(x+1)>6? a) 2x>62x>6 b) 2x+2>62x+2>6 c) x+1>3x+1>3 d) x>2x>2 o Risposta: c) x+1>3x+1>3 o Spiegazione: Distribuendo 22 otteniamo 2x+2>62x+2>6, quindi x+1>3x+1>3. 14. Qual è la soluzione della disequazione 5x−4<3x+75x−4<3x+7? a) x<11x<11 b) x>11x>11 c) x<−11x<−11 d) x>−11x>−11 o Risposta: d) x>−11x>−11 o Spiegazione: Sottraendo 3x3x da entrambi i lati otteniamo 2x−4<72x−4<7, quindi 2x<112x<11 e infine x>−11x>−11. ……. 15 Qual è la soluzione della disequazione 2x+3≥5x−12x+3≥5x−1? a) x≤−1x≤−1 b) x≥−1x≥−1 c) x<2x<2 d) x>2x>2 o Risposta: c) x<2x<2 o Spiegazione: Sottraendo 5x5x da entrambi i lati otteniamo −3x+3≥−1−3x+3≥−1, quindi −3x≥−4−3x≥−4. Dividendo entrambi i lati per −3−3, ricordando di invertire il segno, otteniamo x<2x<2. 16 Quale delle seguenti è una soluzione della disequazione 4x−2≤2x+64x−2≤2x+6? a) x≤−2x≤−2 b) x≥−2x≥−2 c) x<2x<2 d) x>2x>2 o Risposta: b) x≥−2x≥−2 o Spiegazione: Sottraendo 2x2x da entrambi i lati otteniamo 2x−2≤62x−2≤6, quindi 2x≤82x≤8 e infine x≥−2x≥−2. 17 Quale delle seguenti è la soluzione della disequazione 3(x−2)>93(x−2)>9? a) x>3x>3 b) x>5x>5 c) x<3x<3 d) x<5x<5 o Risposta: b) x>5x>5 o Spiegazione: Dividendo entrambi i lati per 33, otteniamo x−2>3x−2>3, quindi x>5x>5. 18 Qual è la soluzione della disequazione 2x+4≤102x+4≤10? a) x≤2x≤2 b) x≥2x≥2 c) x<2x<2 d) x>2x>2 o Risposta: a) x≤2x≤2 o Spiegazione: Sottraendo 44 da entrambi i lati otteniamo 2x≤62x≤6, quindi x≤3x≤3. Tuttavia, dovremmo tenere conto che 22 è positivo, quindi la soluzione è x≤2x≤2. 19 Quale delle seguenti disequazioni è equivalente a 2x≤82x≤8? a) x≥4x≥4 b) x≤4x≤4 c) x>4x>4 d) x<4x<4 a. Risposta: b) x≤4x≤4 b. Spiegazione: Dividendo entrambi i lati per 22, otteniamo x≤4x≤4. 20 Quale delle seguenti è una soluzione della disequazione 5(x−3)>105(x−3)>10? a) x<−1x<−1 b) x>−1x>−1 c) x>5x>5 d) x<5x<5 a. Risposta: c) x>5x>5 b. Spiegazione: Dividendo entrambi i lati per 55, otteniamo x−3>2x−3>2, quindi x>5x>5.
Place value, number notation, fractions, decimals, percentages, average, simpl interest, ratio, proportion, factors, multiples, properties of 2D and 3D shapes, area of shapes, addition and subtraction of fractions, angles, perfect squares, cost and selling price, equivalent fractuions, circle, addition of decimals, perimeter of shapes, mean, median, mode, algebra, increment, discount, multiplication of decimals, positive and negative integers, order of fractions, area of circle, time interval and difference, odd and even numbers, pythagoras theorem, average speed, distance and time, probability, roman numerals
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.
What is a rubric? A tool comprising a set of criteria (with possible levels of performance quality on the criteria) developed to assess learners’ work, from written to oral to visual. It is used tomeasureperformance,suchastheprocess of doing something (e.g.,playing a musical instrument, making a speech) or products of the learners’ work (e.g., concept map, laboratory report, bookshelf) (Brookhart, 2013). BENEFITS OF USING RUBRICS Support authentic assessment Reflects how well learners are able to apply knowledge inthe real-world context. Communicate expectations Gives learners an idea of what is expected of them. It is especially useful when the rubrics are communicated to the learners before they are assessed. Improve performance Explicit criteria and performance level descriptions allow learners to understand the desired performance. Learners are able to assess themselves by referring to the specific criteria and performance-level descriptions. Provide informative feedback Instructors are able to provide constructive feedback to learners on their weaknesses and strengths. Promote thinking andlearning 4 Provide informative feedback Instructors are able to provide constructive feedback to learners on their weaknesses and strengths. Learners are able to review and revise their work,thus reflecting on their learning experiences. Ensure fairness Learner performance assessed fairly given its objectivity. It helps avoid disputes between learners and instructors about the scores/grades achieved. TYPES OF RUBRIC ANALYTIC It consists of individual criterion with corresponding descriptor of performance. HOLISTIC It consists of performance descriptors that are placed together to refeclet to overalll performance. ANATOMY OF ANANALYTIC RUBRIC Rating scales with corresponding scores or weights The row represents the criteria for the desired performance, while the column represents the evaluation score. Under the rating scale (corresponding weights orscorescanbeassigned),theperformance descriptors are explicitly stated ANATOMYOF AHOLISTICRUBRIC Descriptions: It comprises the rating scale (corresponding weights or scores can be assigned) in the row while the combined desired performance descriptors are placed in the column. Description of the task The purpose of the assignment is to assess learner’s cognitive and analytic skills in applying knowledge gained and constructed throughout the course Diffusion of Innovation,bywatching the Surrogates movieand writing ananalytical review of the movie in the context of innovation diffusion.Iwant to provide learners with informative feedback on their cognitive and analytic skills such as the following: applying the concepts of innovation diffusion,making judgmentson the scenes related to innovation diffusion identified from the movie,selecting and critiquing theories of innovation diffusion and making connections between the theories,aswell asarguingand proposing necessary solutions to the problemss hown in the movie. ESTABLISHING ALTERNATIVEASSESSMENTINHIGHEREDUCATION VALIDITYAND RELIABILITYOF RUBRICS. Validity Measuring what is supossedto be measured. Reability Yielding consists results. Instruments that are used in the alternative assessment must be aligned to the learning outcomes and measure well what it intends to measure (valid) and produce consistent scores (reliable). The valid instrument will manifest the true ability (latent trait) of learners and permit appropriate inferences to be made about a specific group of people for specific purposes. TYPES OF VALIDITY FACE VALIDITY Simple form of validity thatapplies a superficial and subjective assessment whether the instrument measures what it is supposed to measure. CONTENT VALIDITY Refers to the extent to which the items on a measure assess the same content or how wellthe content material was sampled inthe measure. CONSTRUCT VALIDITY Refers to the extent to which the test may be said to measure a theoretical construct or trait. CONCURRENT VALIDITY Refers to the extent to which scores onanewmeasure are related to scores from a criterion measure administered at the same time. PREDICTIVE VALIDITY Refers to the uses of the scores from the new measure to predict performance on a criterion measure administered ata later time. STEPS TO CONSIDER WHEN ESTABLISHING CONTENT VALIDITY Calculate the level of expert agreeement for the content validity, get expert to verfy. Interview the expert ,make meta contentdata análisis from literatura. STEPS TO CONSIDER WHEN ESTABLISHING CONSTRUCCT VALIDITY Administer the instrument for alll learners, revise any item necccesay, run an apropriates statistical analiysis, administerthe instrument to learners as a pilot test . CONSTRUCTMAP Morepreciseconceptthan construct. Ranges from one extreme to another(fromhightolow,small tolarge,positivetonegative,or strongtoweak). Identifiesthepositionofthe respondentsinthisrange. Representativenessofsampling (questions and ability of respondents). EXAMPLEO FACONSTRUCTMAP:AFFECTIVE LEVELOF AFFECTIVE VARIABLES EXAMPLESOFITEMSIN MEASURINGTEAM WORKING SKILLS 5. Characterisation Learnersvolunteerstodothe groupworks. 4. Organisation Learners are willing to help others,althoughitisnottheir scopeoftask. 3. Valuing Learners respect other team members’opinionwhendoing thediscussion. 2. Responding Learnergivescooperationwhen neededingroupworks. 1. Receiving Learneracceptsthediversityof races and nationalities among groupmembers. EXAMPLEOFACONSTRUCTMAP:PSYCHOMOTOR LEVELOF PSYCHOMOTOR VARIABLES EXAMPLESOFITEMSIN MEASURING DIGITAL SKILLS 7.Origination Learnerscanmodifytheirowndevicesto performbetter. 6.Adaptation Learnerscansolveandtroubleshootthe problemwhileusingthecomputer. 5.ComplexOvertResponse Learnerscanusethecomputercompetently. 4.Mechanism Learners can use the computer independently,butstillmakeminorerrors. 3.GuidedResponses Learnerscanusethecomputer,butstill needguidance. 2.Set Learnersarereadytousethecomputer. 1.Perception Learnerscanobservehowtousecomputer. EXAMPLEOFACONSTRUCTMAP:COGNITIVE LEVELOF COGNITIV E VARIABLES EXAMPLESOFITEMS IN MEASURING THINKINGSKILLS 6. Creating Learners are able to suggest anewmodelorframeworkof learningdigitalcommunity. 5. Evaluating Learners are able to judge the impactofthescenariotowards educationperspective. 4. Analysing Learnerscandifferentiate the factsusingafew theories. 3. Applying Learnerscansolveproblems usingthefactsgiven. 2. Understanding Learnersareabletoexplainthe factsusingtheirownwords. 1. Remembering Learnersonlymemorisethe. Direction of Increasing “X” Learners Learners with high “X” Learners with mid range “X” Learners with low “X” Responses to Item Item response indicate highest level of X Item response indicate higher level of X Item response indicate lower level of X The construct map shows the lower ability students are in line with the lower level of items. This shows that when educators plan to develop an instrument, it Item response indicate lowest level of X Direction of Decreasing “X” is crucial to create an item difficulty thatrepresents learners’ ability. Learners’ ability Learners who engage in level characterisation Learners who engage in level organisation Learners who engage in level valuing Learners who engage in level responding Learners who engage in level receiving Direction of Decreasing“X” MEASURINGCONSTRUCTVALIDITY Unlike content validity, this construct validity can be analysed using statistical analysis. Use Exploratory FactorAnalysis [EFA], Confirmatory FactorAnalysis [CFA] or Unidimensionality to confirm all items are measuring the right construct and the raw variance explained for the latent variables is sufficient. Gap initem map also can show accuracy in construct validity. RELIABILITY The degree to which test scores are consistent over repeated administrations of the same/ equivalent test and therefore considered dependable and repeatable for an individual learner.A test thatproduces highly consistent and stable results (i.e. relative free from random error) is said to be highly reliable. TYPESOFRELIABILITY Test-retest demonstrates the stability of a measure over time 01 Internal consistency most of the items within a rating scale of a concept show consistency of scoring. Inter-rater the extent to which two or more independent raters are consistent in observing, recording and scoring data (should be 70% or higher agreement) 04 Intra-rater relies on one rater to rate an object or event twice (70% or higher of agreement) FACTORSAFFECTING VALIDITYANDHOWTO INCREASEVALIDITY? FACTORS AFFECTING VALIDITY HOWTO INCREASE VALIDITY? 1. Inaccuracy of items in measuringtheoutcomes 1. Vetting session to get reviewsfromtheexpert. 2. Pooritemsdevelopment 2. Followtheformatandtips indevelopinggooditems. 3. Unclearinstructions 3. Do pilot testing to measuretheusabilityof thetest. 4. Interveningevents 4. Controltheinternalthreats validityfactors. 5. Itemsdifficultyisnot suitableforthelearners 5. Create a construct map toensurethereisanitem thatrepresentslearners ability. FACTORS AFFECTING RELIABILIT Y HOWTOINCREASERELIABILITY? 1. TestLength 1. Thetestlengthshouldbeappropriate withtestdifficulty. 2. Test retest interval 2. Suggesteddurationisbetween3 weeksto2months. 3. Variability of scores 3. Doconstructmaptoensuretheitems aresuitablewithlearners’ability. 4. Guessing 4. Penalisetheguessinganswers.You alsocandetecteitherthelearnersare guessing or not using the statistical analysis named guessing analysis andpersonfitanalysis. 5. Inconsistency score from different raters 5. Appointtheratertomarkcertain questionsforalllearners(Thisalways happen when you have more than onesectionandhavemorethanone lecturer). CONCLUSION Coming back to the issue of validity and reliability in assessment, there is a need for educators to put an effort to ensurethattheitemsintheformofquestionsorinstructions arenotonlyclearbutalsoabletomeasurewhatitisintended tomeasurebasedontherelatedlearningoutcomes. Establishingvalidityandreliabilityofinstrumentscan provide educators with some indications of the quality of the measuring tools being used. Valid and reliable instruments enabletheeducatorstocontinuouslyusethemeasuringtools withoutreservation. Reliablenot valid Precisenot Accurate Reliableand valid Preciseand Accurate NotReliable butvalid NotPrecisebut Accurate NotReliable butNotvalid NotPrecisebut NotAccurate 94
We found you were making a quiz on the subject of "What is a rubric? A tool comprising a set of criteria (with possible levels of performance quality on the criteria) developed to assess learners’ work, from written to oral to visual. It is used tomeasureperformance,suchastheprocess of doing something (e.g.,playing a musical instrument, making a speech) or products of the learners’ work (e.g., concept map, laboratory report, bookshelf) (Brookhart, 2013). BENEFITS OF USING RUBRICS Support authentic assessment Reflects how well learners are able to apply knowledge inthe real-world context. Communicate expectations Gives learners an idea of what is expected of them. It is especially useful when the rubrics are communicated to the learners before they are assessed. Improve performance Explicit criteria and performance level descriptions allow learners to understand the desired performance. Learners are able to assess themselves by referring to the specific criteria and performance-level descriptions. Provide informative feedback Instructors are able to provide constructive feedback to learners on their weaknesses and strengths. Promote thinking andlearning 4 Provide informative feedback Instructors are able to provide constructive feedback to learners on their weaknesses and strengths. Learners are able to review and revise their work,thus reflecting on their learning experiences. Ensure fairness Learner performance assessed fairly given its objectivity. It helps avoid disputes between learners and instructors about the scores/grades achieved. TYPES OF RUBRIC ANALYTIC It consists of individual criterion with corresponding descriptor of performance. HOLISTIC It consists of performance descriptors that are placed together to refeclet to overalll performance. ANATOMY OF ANANALYTIC RUBRIC Rating scales with corresponding scores or weights The row represents the criteria for the desired performance, while the column represents the evaluation score. Under the rating scale (corresponding weights orscorescanbeassigned),theperformance descriptors are explicitly stated ANATOMYOF AHOLISTICRUBRIC Descriptions: It comprises the rating scale (corresponding weights or scores can be assigned) in the row while the combined desired performance descriptors are placed in the column. Description of the task The purpose of the assignment is to assess learner’s cognitive and analytic skills in applying knowledge gained and constructed throughout the course Diffusion of Innovation,bywatching the Surrogates movieand writing ananalytical review of the movie in the context of innovation diffusion.Iwant to provide learners with informative feedback on their cognitive and analytic skills such as the following: applying the concepts of innovation diffusion,making judgmentson the scenes related to innovation diffusion identified from the movie,selecting and critiquing theories of innovation diffusion and making connections between the theories,aswell asarguingand proposing necessary solutions to the problemss hown in the movie. ESTABLISHING ALTERNATIVEASSESSMENTINHIGHEREDUCATION VALIDITYAND RELIABILITYOF RUBRICS. Validity Measuring what is supossedto be measured. Reability Yielding consists results. Instruments that are used in the alternative assessment must be aligned to the learning outcomes and measure well what it intends to measure (valid) and produce consistent scores (reliable). The valid instrument will manifest the true ability (latent trait) of learners and permit appropriate inferences to be made about a specific group of people for specific purposes. TYPES OF VALIDITY FACE VALIDITY Simple form of validity thatapplies a superficial and subjective assessment whether the instrument measures what it is supposed to measure. CONTENT VALIDITY Refers to the extent to which the items on a measure assess the same content or how wellthe content material was sampled inthe measure. CONSTRUCT VALIDITY Refers to the extent to which the test may be said to measure a theoretical construct or trait. CONCURRENT VALIDITY Refers to the extent to which scores onanewmeasure are related to scores from a criterion measure administered at the same time. PREDICTIVE VALIDITY Refers to the uses of the scores from the new measure to predict performance on a criterion measure administered ata later time. STEPS TO CONSIDER WHEN ESTABLISHING CONTENT VALIDITY Calculate the level of expert agreeement for the content validity, get expert to verfy. Interview the expert ,make meta contentdata análisis from literatura. STEPS TO CONSIDER WHEN ESTABLISHING CONSTRUCCT VALIDITY Administer the instrument for alll learners, revise any item necccesay, run an apropriates statistical analiysis, administerthe instrument to learners as a pilot test . CONSTRUCTMAP Morepreciseconceptthan construct. Ranges from one extreme to another(fromhightolow,small tolarge,positivetonegative,or strongtoweak). Identifiesthepositionofthe respondentsinthisrange. Representativenessofsampling (questions and ability of respondents). EXAMPLEO FACONSTRUCTMAP:AFFECTIVE LEVELOF AFFECTIVE VARIABLES EXAMPLESOFITEMSIN MEASURINGTEAM WORKING SKILLS 5. Characterisation Learnersvolunteerstodothe groupworks. 4. Organisation Learners are willing to help others,althoughitisnottheir scopeoftask. 3. Valuing Learners respect other team members’opinionwhendoing thediscussion. 2. Responding Learnergivescooperationwhen neededingroupworks. 1. Receiving Learneracceptsthediversityof races and nationalities among groupmembers. EXAMPLEOFACONSTRUCTMAP:PSYCHOMOTOR LEVELOF PSYCHOMOTOR VARIABLES EXAMPLESOFITEMSIN MEASURING DIGITAL SKILLS 7.Origination Learnerscanmodifytheirowndevicesto performbetter. 6.Adaptation Learnerscansolveandtroubleshootthe problemwhileusingthecomputer. 5.ComplexOvertResponse Learnerscanusethecomputercompetently. 4.Mechanism Learners can use the computer independently,butstillmakeminorerrors. 3.GuidedResponses Learnerscanusethecomputer,butstill needguidance. 2.Set Learnersarereadytousethecomputer. 1.Perception Learnerscanobservehowtousecomputer. EXAMPLEOFACONSTRUCTMAP:COGNITIVE LEVELOF COGNITIV E VARIABLES EXAMPLESOFITEMS IN MEASURING THINKINGSKILLS 6. Creating Learners are able to suggest anewmodelorframeworkof learningdigitalcommunity. 5. Evaluating Learners are able to judge the impactofthescenariotowards educationperspective. 4. Analysing Learnerscandifferentiate the factsusingafew theories. 3. Applying Learnerscansolveproblems usingthefactsgiven. 2. Understanding Learnersareabletoexplainthe factsusingtheirownwords. 1. Remembering Learnersonlymemorisethe. Direction of Increasing “X” Learners Learners with high “X” Learners with mid range “X” Learners with low “X” Responses to Item Item response indicate highest level of X Item response indicate higher level of X Item response indicate lower level of X The construct map shows the lower ability students are in line with the lower level of items. This shows that when educators plan to develop an instrument, it Item response indicate lowest level of X Direction of Decreasing “X” is crucial to create an item difficulty thatrepresents learners’ ability. Learners’ ability Learners who engage in level characterisation Learners who engage in level organisation Learners who engage in level valuing Learners who engage in level responding Learners who engage in level receiving Direction of Decreasing“X” MEASURINGCONSTRUCTVALIDITY Unlike content validity, this construct validity can be analysed using statistical analysis. Use Exploratory FactorAnalysis [EFA], Confirmatory FactorAnalysis [CFA] or Unidimensionality to confirm all items are measuring the right construct and the raw variance explained for the latent variables is sufficient. Gap initem map also can show accuracy in construct validity. RELIABILITY The degree to which test scores are consistent over repeated administrations of the same/ equivalent test and therefore considered dependable and repeatable for an individual learner.A test thatproduces highly consistent and stable results (i.e. relative free from random error) is said to be highly reliable. TYPESOFRELIABILITY Test-retest demonstrates the stability of a measure over time 01 Internal consistency most of the items within a rating scale of a concept show consistency of scoring. Inter-rater the extent to which two or more independent raters are consistent in observing, recording and scoring data (should be 70% or higher agreement) 04 Intra-rater relies on one rater to rate an object or event twice (70% or higher of agreement) FACTORSAFFECTING VALIDITYANDHOWTO INCREASEVALIDITY? FACTORS AFFECTING VALIDITY HOWTO INCREASE VALIDITY? 1. Inaccuracy of items in measuringtheoutcomes 1. Vetting session to get reviewsfromtheexpert. 2. Pooritemsdevelopment 2. Followtheformatandtips indevelopinggooditems. 3. Unclearinstructions 3. Do pilot testing to measuretheusabilityof thetest. 4. Interveningevents 4. Controltheinternalthreats validityfactors. 5. Itemsdifficultyisnot suitableforthelearners 5. Create a construct map toensurethereisanitem thatrepresentslearners ability. FACTORS AFFECTING RELIABILIT Y HOWTOINCREASERELIABILITY? 1. TestLength 1. Thetestlengthshouldbeappropriate withtestdifficulty. 2. Test retest interval 2. Suggesteddurationisbetween3 weeksto2months. 3. Variability of scores 3. Doconstructmaptoensuretheitems aresuitablewithlearners’ability. 4. Guessing 4. Penalisetheguessinganswers.You alsocandetecteitherthelearnersare guessing or not using the statistical analysis named guessing analysis andpersonfitanalysis. 5. Inconsistency score from different raters 5. Appointtheratertomarkcertain questionsforalllearners(Thisalways happen when you have more than onesectionandhavemorethanone lecturer). CONCLUSION Coming back to the issue of validity and reliability in assessment, there is a need for educators to put an effort to ensurethattheitemsintheformofquestionsorinstructions arenotonlyclearbutalsoabletomeasurewhatitisintended tomeasurebasedontherelatedlearningoutcomes. Establishingvalidityandreliabilityofinstrumentscan provide educators with some indications of the quality of the measuring tools being used. Valid and reliable instruments enabletheeducatorstocontinuouslyusethemeasuringtools withoutreservation. Reliablenot valid Precisenot Accurate Reliableand valid Preciseand Accurate NotReliable butvalid NotPrecisebut Accurate NotReliable butNotvalid NotPrecisebut NotAccurate 94 ". Would you like to continue making it or start afresh?
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