course Phy: 201 This document consists of a series of questions in QA (question-answer) format. •The idea of this format is to give you instant feedback on each question, as soon as you have completed the answer.
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Given Solution: Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rate is the amount that something changes in a certain period of time. This changed 3 meters every second. confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred. More specifically • The rate of change of A with respect to B is defined to be the quantity (change in A) / (change in B). An object which moves 12 meters in 3 seconds changes its position by 12 meter during a change in clock time of 3 seconds. So the question implies • Change in position = 12 meters • Change in clock time = 3 seconds When we divide the 12 meters by the 3 seconds we are therefore dividing (change in position) by (change in clock time). In terms of the definition of rate of change: • the change in position is the change in A, so position is the A quantity. • the change in clock time is the change in B, so clock time is the B quantity. So (12 meters) / (3 seconds) is (change in position) / (change in clock time) which is the same as average rate of change of position with respect to clock time. Thus • average velocity is average rate of change of position with respect to clock time. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Need to be a little more specific. Got the right idea in mind, just need to specify everything and get more detailed. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q003. Is object position dependent on time or is time dependent on object position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: object position is dependent on time because the longer the time interval the further the object is going to move. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Missed the average rate of change definition, didn’t mention anything about velocity or acceleration. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If there’s anything you don’t understand, be sure to address it in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Speed cannot be a negative number. So therefore, the objects speed is 2 meter/second. The objects velocity is -2 meters/second in the negative direction. I thought of this as a concordance graph. The problem said -6 meters so I thought of it in the negative direction on the graph. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative. Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity. In general distance has no direction, while velocity does have direction. Putting it loosely, position is just how fast something is moving; velocity is how fast and in what direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got the right idea with the right number answers. However, need to address definitions and what you are explaining. ------------------------------------------------ Self-critique rating: 4 ********************************************* Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: velocity is the change in position with respect to the change in clock time. vAve = ‘ds/ ‘dt. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Average velocity is rate of change of position with respect to clock time. Change in position is `ds and change in clock time is `dt, so average velocity is expressed in symbols as • vAve = `ds / `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: when writing ‘ds and ‘dt you use a triangle in front of the symbol. The triangle stands for the Greek symbol delta which means change in. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1. `d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d. You may use either `d or Delta when submitting work and answering questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 50 meters, the rate of change in change in A with respect to change in B. So the average rate of change was given, you just have to find out the distance travelled. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. The definition of rate of change states that the rate of change of A with respect to B is (change in A) / (change in B), which we abbreviate as `dA / `dB. `dA stands for the change in the A quantity and `dB for the change in the B quantity. For the present problem we are given the rate at which position changes with respect to clock time. The definition of rate of change is stated in terms of the rate of change of A with respect to B. • So we identify the position as the A quantity, clock time as the B quantity. The basic relationship • ave rate = `dA / `dB can be algebraically rearranged in either of two ways: • `dA = ave rate * `dB or • `dB = `dA / (ave rate) Using position for A and clock time for B the above relationships are • ave rate of change of position with respect to clock time = change in position / change in clock time • change in position = ave rate * change in clock time • change in clock time = change in position / ave rate. In the present situation we are given the average rate of change of position with respect to clock time, which is 5 meters / second, and the change in clock time, which is 10 seconds. Thus we find • change in position = ave rate * change in clock time = 5 cm/sec * 10 sec = 50 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘ds = vAve * ‘dt; the original equation is vAve = ‘ds/ ‘dt. So you would cross multiply and divide to solve for ‘ds. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the change in a quantity we multiply the rate by the time interval during which the change occurs. The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval: • `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: • We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour). • When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: average velocity is the change in velocity with respect to clock time. Time interval is the change in clock time that something took place, ‘dt displacement is the change in time on the side of the scale (displacement), ‘ds is the change in position. Rate is the change in A with respect to the change in B. Any of these quanitites could be used in the rate equation. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we would multiply both sides by ‘dt to get vAve* ‘dt = ‘ds. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: average velocity is the change in position with respect to change in clock time. aAve is average velocity, ‘ds is displacement, ‘dt is change in clock time. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For most of us our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): wasn’t quite on it. Understood the formula, just didn’t understand what the question was asking. ------------------------------------------------ Self-critique rating:2 ********************************************* Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: you multiply by ‘dt to get vAve* ‘dt = ‘ds. Then you will divide both sides by vAve to get ‘dt = ‘ds/vAve. confidence rating: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Didn’t quite get this one the first time. Needed some help to figure out the algebra. Keep practicing algebra like this. ------------------------------------------------ Self-critique rating: 1 ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: average velocity is vAve, displacement is ‘ds, and clock time is ‘dt. If you know at least two of the variables you can always solve for that one and then plug in the numbers to get the number to that variable. confidence rating: 4 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. • If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. This is equivalent to the calculation `dt = `ds / vAve. • We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval. When dealing with displacement, velocity and time interval, we can always check our thinking by making the analogy with a simple example involving miles, hours and miles/hour. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: You should submit the above questions, along with your answers and self-critiques. You may then begin work on the Questions, Problem and Exercises, as instructed below. The Process of Responding and Self-Critiquing Answering and self-critiquing problems and questions On every item in a qa document, you are to provide your best attempt at a solution, and if a solution to the question or problem is given you are to self-critique as appropriate (see below for more about self-critique). If you believe you are unable to solve a problem, you should indicate as specifically as possible what you do and do not understand about each phrase in the problem, and about the problem as a whole. Every solution should include documentation, meaning that you should indicate your reasoning and an outline of the steps you used in obtaining the solution. Your instructor's main function is not to point out errors in your solutions or evaluate your solutions, but to address your needs as they are indicated in your self-critiques. Self-critique, in those cases where your solution differs from the given solution, is the most important part of the Query/qa process. Self-critique focuses your attention on essential details, and it provides the instructor the information required to help you. The purpose of the self-critique is to demonstrate to your instructor what you do and do not understand, so your instructor can help you address any specific idea or procedure that might be giving you trouble. Your instructor can and will help if you provide the necessary information. How to self-critique The first thing required is that you recognize when your solution does not match the given solution. If your solution differs from the given solution you should then include statements that indicate the differences, and that document what you do and do not understand about that solution. Your self-critique should document understanding, and not simply assert understanding (i.e., you should show that you understand, not just say that you do). A good place to start with a self-critique is to think in terms of the following questions: • What specifically do you understand, what do you think you might understand, and what do you not understand about question and the given solution? • If you don't understand the entire question and the given solution, break the solution into phrases and ask the same questions about each phrase. • What questions do you have about the problem, or about parts of the problem? Try to be as specific as possible. The nature of a good self critique In a good self-critique you need identify the specific things you do and do not understand in the given solution, and either demonstrate your understanding or ask specific questions about what you don't understand. You need to detail the points on which your solution differs from the given solution. Deconstruct the given solution and explain in detail what you do and do not understand about every part. Once you have defined and detailed what you do and do not understand, your instructor can help you in the appropriate manner. If you fail to do this, your instructor might not have enough information to help. A self-critique should include specific questions about the problem or the given solution. If you didn't self-critique in the prescribed manner You are welcome to insert additional self-critiques, including specific questions, into a copy of this document for my further response. Mark your insertions with &&&& so your instructor can quickly identify them. "