#$&* course Phy201+L 002. Velocity*********************************************
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Given Solution: Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should be more thorough in my explanation of my answer. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Rate is defined as the change in quantity of the same measurement. So for distance there is a change in 12 meters, and for time there is a change in 3 seconds for the above question. Then using this information we can determine velocity, or the rate of change in distance, in respect to time. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I think again my answers should be more thorough when compared to the given solution. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q003. We are still referring to the situation of the preceding questions: Is object position dependent on time or is time dependent on object position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The objects position is dependent on time, as it is at a constant velocity, so the longer the time the farther away from the beginning point the object will be. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. We are still referring to the situation of the preceding questions: So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I feel that I covered all of the explanations. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If theres anything you dont 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: Since the distance is a negative 6 meters that might mean that there is no velocity. However, velocity does not matter in terms of direction. So to determine velocity, -6 meters is divided by 3 seconds, to give you -2 meters/second. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative. Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity. In general distance has no direction, while velocity does have direction. Putting it loosely, speed is just how fast something is moving; velocity is how fast and in what direction. In this case, the average velocity is vAve = `ds / `dt = -6 m / (3 s) = -2 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should put more ideas into my explanations. ------------------------------------------------ Self-critique rating:3 ********************************************* Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which its position changes, then what expression stands for the average velocity vAve of the object during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds/dt. confidence rating #$&*:32; 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 these on paper you use the Greek symbol delta, which means change in. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1. `d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d. You may use either `d or Delta when submitting work and answering questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If it travels 5 meters/second for 10 seconds it will have traveled 50 meters. Using the velocity relationship, we can determine that if a object travels for one second, it will have traveled 5 meters. So multiplying 5 by 10 will give us 50 meters. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: You would re-write the equation so that you multiply vAve by dt which then equals ds, giving you the change in position. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the change in a quantity we multiply the rate by the time interval during which the change occurs. The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour). When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Velocity is the rate, or the change in A/ the change in B. Time interval and displacement are the two quantities that contribute to us finding the rate for the problem. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To solve for ds we simply bring the dt across to the other side of the equals sign by multiplying it by both sides, cancelling it on the side where it already exists, and keeping it on the side where it does not exist. This gives us the equation vAve * dt = ds. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This shows us that when we multiply dt by vAve, it cancels out the change in time by cross multiplication, leaving us with the units of change in distance. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique rating: ********************************************* 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: To solve for dt, we go about the same process of multiplying it by both sides. Once dt is on the other side of the equation, we can divide both sides by vAve leaving us with the equation dt = ds/vAve. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This result shows us that once we have ds/ vAve, the units will cancel out, leaving us with the units for change in time. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Problems related to qa Guidelines for solving problems and answering questions: Include all steps in your solution. Every quantity which has units must be given in terms of those units. If a question involves an average rate of change of one quantity with respect to another, explain specifically, as best you can, how your solution is related to the definition of an average rate of change. This definition is in terms of quantites A and B; be sure you identify quantity A and quantity B. Jot down notes like 'quantity A stands for ... with unit of ...', and 'quantity B stands for ... with unit of ... '. This will help you avoid a great deal of confusion. Your explanations may be concise and may use reasonable abbreviations, but must clearly show your thinking. You need to actually work out your algebra and be prepared to explain it, including the details of the algebra of your units. 1. If the position of an object changes from 34 cm at clock time 4.6 seconds to 87 cm at clock time 5.3 seconds, then during this interval what is the average rate of change of its position with respect to clock time? 87-34 = 53 cm (change in distance) 5.3-4.6= 0.7 seconds (change in time) 53 cm/ 0.7 seconds= 75.7 cm/sec (velocity) 2. If the velocity of an object changes from 12 cm/second at clock time 6.9 seconds to 20 cm/s at clock time 15.3 seconds, then what is the rate of change of its velocity with respect to clock time? 20 - 12= 8 cm/sec ( change in velocity) 15.3 - 6.9= 8.4 seconds ( change in time ) 8 cm/ sec /8.4 seconds = 0.95 cm/s/s (change in velocity with respect to time) 3. What is your best estimate of the average velocity of the object in #2, for the given time interval? 16 cm/second. 4. If the average rate of change of position with respect to clock time during a certain interval is 24 meters / second, and if the interval lasts for 5 seconds, then what quantity can you determine by applying the definition of an average rate of change? Find this quantity and explain in detail how you found it. 24 meters/ second * 5 seconds = 120 meters. 5. What is wrong with saying the average velocity = position / clock time? Most times velocity is not constant, it is a constantly shifting number so saying just that velocity is equal to position/clock time may not be sufficient in most cases. Text-related questions: 1. What is the percent uncertainty in a measured time interval of 3.4 seconds, given that the timing mechanism has an uncertainty of +- .1 second? What is the percent uncertainty in a time interval of .87 seconds, measured using the same mechanism? When using this mechanism, how does the percent uncertainty in measuring a time interval depend on the duration of that interval? 2. What is the uncertainty in the following reported measurements, and what is the percent uncertainty in each? 5.8 centimeters 2350 kilometers 350. seconds 3.14 3.1416 3. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm? 4. (Principles of Physics students are invited to solve this problem, but are not required to do so): What is the approximate uncertainty in the area of a circle, based on a reported radius of 2.8 * 10^4 cm? 5. What is your height in meters, and your ideal mass in kilograms? How much uncertainty do you think there is in each, and why? " 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Problems related to qa Guidelines for solving problems and answering questions: Include all steps in your solution. Every quantity which has units must be given in terms of those units. If a question involves an average rate of change of one quantity with respect to another, explain specifically, as best you can, how your solution is related to the definition of an average rate of change. This definition is in terms of quantites A and B; be sure you identify quantity A and quantity B. Jot down notes like 'quantity A stands for ... with unit of ...', and 'quantity B stands for ... with unit of ... '. This will help you avoid a great deal of confusion. Your explanations may be concise and may use reasonable abbreviations, but must clearly show your thinking. You need to actually work out your algebra and be prepared to explain it, including the details of the algebra of your units. 1. If the position of an object changes from 34 cm at clock time 4.6 seconds to 87 cm at clock time 5.3 seconds, then during this interval what is the average rate of change of its position with respect to clock time? 87-34 = 53 cm (change in distance) 5.3-4.6= 0.7 seconds (change in time) 53 cm/ 0.7 seconds= 75.7 cm/sec (velocity) 2. If the velocity of an object changes from 12 cm/second at clock time 6.9 seconds to 20 cm/s at clock time 15.3 seconds, then what is the rate of change of its velocity with respect to clock time? 20 - 12= 8 cm/sec ( change in velocity) 15.3 - 6.9= 8.4 seconds ( change in time ) 8 cm/ sec /8.4 seconds = 0.95 cm/s/s (change in velocity with respect to time) 3. What is your best estimate of the average velocity of the object in #2, for the given time interval? 16 cm/second. 4. If the average rate of change of position with respect to clock time during a certain interval is 24 meters / second, and if the interval lasts for 5 seconds, then what quantity can you determine by applying the definition of an average rate of change? Find this quantity and explain in detail how you found it. 24 meters/ second * 5 seconds = 120 meters. 5. What is wrong with saying the average velocity = position / clock time? Most times velocity is not constant, it is a constantly shifting number so saying just that velocity is equal to position/clock time may not be sufficient in most cases. Text-related questions: 1. What is the percent uncertainty in a measured time interval of 3.4 seconds, given that the timing mechanism has an uncertainty of +- .1 second? What is the percent uncertainty in a time interval of .87 seconds, measured using the same mechanism? When using this mechanism, how does the percent uncertainty in measuring a time interval depend on the duration of that interval? 2. What is the uncertainty in the following reported measurements, and what is the percent uncertainty in each? 5.8 centimeters 2350 kilometers 350. seconds 3.14 3.1416 3. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm? 4. (Principles of Physics students are invited to solve this problem, but are not required to do so): What is the approximate uncertainty in the area of a circle, based on a reported radius of 2.8 * 10^4 cm? 5. What is your height in meters, and your ideal mass in kilograms? How much uncertainty do you think there is in each, and why? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy201+L 002. `ph1 query 2
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Given Solution: Average velocity is defined as the average rate of change of position with respect to clock time. The average rate of change of A with respect to B is (change in A) / (change in B). Thus the average rate of change of position with respect to clock time is ave rate = (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Why can it not be said that average velocity = position / clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The velocity is very subject to change over a certain period, so saying just that velocity is equal to period over time would not be correct, as the velocity will likely increase and decrease over a given period. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The definition of average rate involves the change in one quantity, and the change in another. Both position and clock time are measured with respect to some reference value. For example, position might be measured relative to the starting line for a race, or it might be measured relative to the entrance to the stadium. Clock time might be measure relative to the sound of the starting gun, or it might be measured relative to noon. So position / clock time might, at some point of a short race, be 500 meters / 4 hours (e.g., 500 meters from the entrance to the stadium and 4 hours past noon). The quantity (position / clock time) tells you nothing about the race. There is a big difference between (position) / (clock time) and (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: Explain in your own words the process of fitting a straight line to a graph of y vs. x data, and briefly discuss the nature of the uncertainties encountered in the process. For example, you might address the question of how two different people, given the same graph, might obtain different results for the slope and the vertical intercept. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: You can have a different equation but still come out with the same line, given that you can use different vertical intercepts to convey your line. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: (Principles of Physics students are invited but not required to submit a solution) Give your solution to the following, which should be in your notes: Find the approximate uncertainty in the area of a circle given that its radius is 2.8 * 10^4 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the uncertainty in the measurement. The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises).. With this uncertainty estimate, we find that the area is between a lower area estimate of pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. The area we would get from the given radius is about halfway between these estimates, so the uncertainty in the area is about half of the difference. We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2. Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 * 10^8 cm uncertainty in area is about 4% of the area. The area of a circle is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. ** STUDENT COMMENT: I don't recall seeing any problems like this in any of our readings or assignments to this point INSTRUCTOR RESPONSE: The idea of percent uncertainty is presented in Chapter 1 of your text. The formula for the area of a circle should be familiar. Of course it isn't a trivial matter to put these ideas together. STUDENT COMMENT: I don't understand the solution. How does .176 * 10^9 become 1.76 * 10^8? I understand that there is a margin of error because of the significant figure difference, but don't see how this was calculated. INSTRUCTOR RESPONSE: .176 = 1.76 * .1, or 1.76 * 10^-1. So .176 * 10^9 = 1.76 * 10^-1 * 10^9. Since 10^-1 * 10^9 = 10^(9 - 1) =10^8, we have .176 * 10^9 = 1.76 * 10^8. The key thing to understand is the first statement of the given solution: Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This is because any number between 2.75 and 2.85 rounds to 2.8. A number which rounds to 2.8 can therefore lie anywhere between 2.75 and 2.85. The rest of the solution simply calculates the areas corresponding to these lower and upper bounds on the number 2.8, then calculates the percent difference of the results. STUDENT COMMENT: I understand how squaring the problem increases uncertainty and I understand the concept of a range of uncertainty but I am having trouble figuring out how the range of 2.75 * 10^4 and 2.85*10^4 were established for the initial uncertainties in radius. INSTRUCTOR RESPONSE: The key is the first sentence of the given solution: 'Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.' You know this because you know that any number which is at least 2.75, and less than 2.85, rounds to 2.8. Ignoring the 10^4 for the moment, and concentrating only on the 2.8: Since the given number is 2.8, with only two significant figures, all you know is that when rounded to two significant figures the quantity is 2.8. So all you know is that it's between 2.75 and 2.85. STUDENT QUESTION I honestly didn't consider the fact of uncertainty at all. I misread the problem and thought I was simply solving for area. I'm still not really sure how to determine the degree of uncertainty. INSTRUCTOR RESPONSE Response to Physics 121 student: This topic isn't something critical to your success in the course, but the topic will come up. You're doing excellent work so far, and it might be worth a little time for you to try to reconcile this idea. Consider the given solution, the first part of which is repeated below, with some questions (actually the same question repeated too many times). I'm sure you have limited time so don't try to answer the question for every statement in the given solution, but try to answer at least a few. Then submit a copy of this part of the document. Note that a Physics 201 or 231 student should understand this solution very well, and should seriously consider submitting the following if unsure. This is an example of how to take a solution phrase by phrase and self-critique in the prescribed manner. ##&* ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* We regard 2.75 *10^4 cm as the lower bound and 2.85 *10^4 cm as the upper bound on the radius. 2.75 is .05 less than 2.8, and 2.85 is .05 greater than 2.8, so we say that the actual number is 2.8 +- .05. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* Thus we express the actual radius as (2.8 +- .05) * 10^4 cm, and we call .05 * 10^4 cm the uncertainty in the measurement. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area of a circle is pi r^2, with which you should be familiar (if for no reason other than that you used it and wrote it down in the orientation exercises). With this uncertainty estimate, we find that the area is between a lower area estimate of pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and and upper area estimate of pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The difference between the lower and upper estimate is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area we would get from the given radius is about halfway between these estimates, so the uncertainty in the area is about half of the difference. We therefore say that the uncertainty in area is about 1/2 * 1.76 * 10^8 cm^2, or about .88 * 10^8 cm^2.Note that the .05 * 10^4 cm uncertainty in radius is about 2% of the radius, while the .88 * 10^8 cm uncertainty in area is about 4% of the area. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* The area of a circle is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. Do you understand what this is saying, and why it is so? If not, tell me what you think you understand, what you are pretty sure you don't understand, and what you think you might understand but aren't sure. ##&* STUDENT QUESTION I said the uncertainty was .1, which gives me .1 / 2.8 = .4. INSTRUCTOR RESPONSE A measurement of 2.8 can be taken to imply a number between 2.75 and 2.85, which means that the number is 2.8 +- .05 and the uncertainty is .05. This is the convention used in the given solution. (The alternative convention is that 2.8 means a number between 2.7 and 2.9; when in doubt the alternative convention is usually the better choice. This is the convention used in the text. It should be easy to adapt the solution given here to the alternative convention, which yields an uncertainty in area of about 8% as opposed to the 4% obtained here). Using the latter convention, where the uncertainty is estimated to be .1: The uncertainty you calculated would indeed be .04 (.1 / 2.8 is .04, not .4), or 4%. However this would be the percent uncertainty in the radius. The question asked for the uncertainty in the area. Since the calculation of the area involves squaring the radius, the percent uncertainty in area is double the percent uncertainty in radius. This gives us a result of .08 or 8%. The reasons are explained in the given solution. NOTE FOR UNIVERSITY PHYSICS STUDENTS (calculus-based answer): Note the following: A = pi r^2, so the derivative of area with respect to radius isdA/dr = 2 pi r. The differential is thereforedA = 2 pi r dr. Thus an uncertainty `dr in r implies uncertainty `dA = 2 pi r `dr, so that `dA / `dr = 2 pi r `dr / (pi r^2) = 2 `dr / r. `dr / r is the proportional uncertainty in r. We conclude that the uncertainty in A is 2 `dr / r, i.e., double the uncertainty in r. STUDENT QUESTION I looked at this, and not sure if I calculated the uncertainty correctly, as the radius squared yields double the uncertainty. I know where this is in the textbook, and do ok with uncertainty, but this one had me confused a bit. INSTRUCTOR RESPONSE: In terms of calculus, since you are also enrolled in a second-semester calculus class: A = pi r^2 The derivative r^2 with respect to r is 2 r, so the derivative of the area with respect to r is dA / dr = pi * (2 r). If you change r by a small amount `dr, the change in the area is dA / dr * `dr, i.e., rate of change of area with respect to r multiplied by the change in r, which is a good commonsense notion. Thus the change in the area is pi * (2 r) `dr. As a proportion of the original area this is pi ( 2 r) `dr / (pi r^2) = 2 `dr / r. The change in the radius itself was just `dr. As a proportion of the initial radius this is `dr / r. The proportional change in area is 2 `dr / r, compared to the proportional change in radius `dr / r. That is the proportional change in area is double the proportional change in radius. STUDENT COMMENT I used +-.1 instead of using +-.05. I understand why your solution used .05 and will use this method in the future. INSTRUCTOR RESPONSE Either way is OK, depending on your assumptions. When it's possible to assume accurate rounding, then the given solution works. If you aren't sure the rounding is accurate, the method you used is appropriate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: What is your own height in meters and what is your own mass in kg (if you feel this question is too personal then estimate these quantities for someone you know)? Explain how you determined these. What are your uncertainty estimates for these quantities, and on what did you base these estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I determined my height in meters based on my height in inches (75). Using the formula 1 inch= 2.54 cm, giving me 190.5 cm. Then using this number I multiplied it by 0.01 to give me my height in meters (1.905 m). I already know that my weight in kg is 100 kg because I have a scale that can give me metric units. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Presumably you know your height in feet and inches, and have an idea of your ideal weight in pounds. Presumably also, you can convert your height in feet and inches to inches. To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm. Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value. Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm. For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in. in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm. STUDENT SOLUTION 5 feet times 12 inches in a feet plus six inches = 66 inches. 66inches * 2.54 cm/inch = 168.64 cm. 168.64 cm * .01m/cm = 1.6764 meters. INSTRUCTOR COMMENT: Good, but note that 66 inches indicates any height between 65.5 and 66.5 inches, with a resulting uncertainty of about .7%. 168.64 implies an uncertainty of about .007%. It's not possible to increase precision by converting units. STUDENT SOLUTION AND QUESTIONS My height in meters is - 55 = 65inches* 2.54cm/1in = 165cm*1m/100cm = 1.7m. My weight is 140lbs* 1kg/2.2lbs = 63.6kg. Since 55 could be anything between 54.5 and 55.5, the uncertainty in height is ???? The uncertainty in weight, since 140 can be between 139.5 and 140.5, is ?????? INSTRUCTOR RESPONSE Your height would be 5' 5"" +- .5""; this is the same as 65"" +- .5"". .5"" / 65"" = .008, approximately, or .8%. So the uncertainty in your height is +-0.5"", which is +-0.8%. Similarly you report a weight of 140 lb +- .5 lb. .5 lb is .5 lb / (140 lb) = .004, or 0.4%. So the uncertainty is +-0.5 lb, or +- 0.4%. STUDENT QUESTION I am a little confused. In the example from another student her height was 66 inches and you said that her height could be between 65.5 and 66.5 inches. but if you take the difference of those two number you get 1, so why do you divide by .5 when the difference is 1 INSTRUCTOR RESPONSE If you regard 66 inches as being a correct roundoff of the height, then the height is between 65.5 inches and 66.5 inches. This makes the height 66 inches, plus or minus .5 inches. This is written as 66 in +- .5 in and the percent uncertainty would be .5 / 66 = .007, about .7%. If you regard 66 inches having been measured only accurately enough to ensure that the height is between 65 inches and 67 inches, then your result would be 66 in +- 1 in and the percent uncertainty would be 1 / 66 = .015 or about 1.5%. STUDENT QUESTION If a doctor were to say his inch marker measured to the nearest 1/4 inch, would that be the uncertainty? Meaning, would I only have to multiply that by .0254 to find the uncertainty in meters, dividing that by my height to find the percent uncertainty? INSTRUCTOR RESPONSE That's pretty much the case, though you do have to be a little bit careful about how the rounding and the uncertainty articulate. For example I'm 72 inches tall. That comes out to 182.88 cm. It wouldn't make a lot of sense to say that I'm 182.88 cm tall, +- .64 cm. A number like 182.88 has a ridiculously high number of significant figures. It wouldn't quite be correct to just round up and say that I'm 183 cm tall +- .64 cm. We might be able to say that I'm 183 cm tall, +- .76 cm, but that .76 cm again implies more precision than is present. We would probably end up saying that I""m 183 cm tall, +- 1 cm. Better to overestimate the uncertainty than to underestimate it. As far as the percent uncertainty goes, we wouldn't need to convert the units at all. In my case we would just divide 1/4 in. by 72 in., getting about .034 or 3.4%. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: How far the ball rolled along each book. The time interval the ball requires to roll from one end of each book to the other. How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average velocity on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine the average velocity on each book you can use the velocity at the beginning and the end of each book to obtain the average velocity. To determine how quickly the speed is changing you can use the time intervals and velocity at each time interval to determine how quickly the velocity is changing. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: How far the ball rolled along each book. The time interval the ball requires to roll from one end of each book to the other. How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average velocity on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine the average velocity on each book you can use the velocity at the beginning and the end of each book to obtain the average velocity. To determine how quickly the speed is changing you can use the time intervals and velocity at each time interval to determine how quickly the velocity is changing. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy201+L 003. Velocity Relationships*********************************************
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Given Solution: vAve = `ds / `dt. The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec. Thus vAve is in m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If vAve is measured in cm/sec and dt is measured in seconds, then ds is measured in cm. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm. STUDENT QUESTION I dont get how sec and sec would cancel each other out INSTRUCTOR RESPONSE cm / s * s means(cm/s) * s, which is the same as(cm / s) * (s / 1). Multiplying numerators and denominators we have(cm * s) / (s * 1) or just(cm * s) / s, which is the same as cm * (s / s) = cm * 1 = cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By multiplying the unit cm/sec by the unit sec, you eliminate the unit of seconds by cross multiplication, giving you only the unit of centimeters. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm. STUDENT RESPONSE: For some reason this question just isn't making sense to me. INSTRUCTOR RESPONSE: In a self-critique you need to address the given solution in detail. A general statement such as yours gives me no information on what you understand. I need this information as a basis for helping you with what you don't understand. In order to give me the information I need you should be addressing each statement, and each phrase, to show me what you do and do not understand. The given solution can be broken into individual statements: 1. When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. 2. When we multiply fractions we will multiply numerators and denominators. 3. We obtain cm * sec / ( sec * 1). 4. This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm. Do you understand Statement 1? If not, have you written out the expressions cm/sec and sec/1 in standard form? (You might want to review the link given at the end of the Typewriter Notation exercise from Orientation, which should be posted at your access page). As best you can communicate it, what do you and do you not understand about this statement? Do you understand Statement 2? If not, what do you and do you not understand about this statement? Do you understand Statement 3? If not, have you written out the multiplication of cm/sec and sec/1 on paper? The multiplication is (cm / sec) * (sec / 1). Again, if you aren't sure how to write this out, refer to the link at the end of the Typewriter Notation exercise. Do you understand Statemet 4? If not do you understand that (sec / sec) * (cm / 1) is equal to sec * cm / (sec * 1), which is in turn equal to (cm * sec) / (sec * 1)? If not, specifically what do you and do you not understand? If you don't understand anything, then you should start with a review of basic fractions, a topic which is very much neglected in the typical curriculum in U.S. schools. Then you should return to these questions and give your best answers. A good link, current as of Sept. 2010: http://www.themathpage.com/arith/multiply-fractions-divide-fractions.htm You should submit a copy of question `q003, your solution, the given solution and this note. Insert your answers and/or additional specific questions and mark with &&&& before and after each insertion, then submit using the Submit Work Form. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If vAve is measured in km/sec and ds is measured in km, then dt is measured in seconds. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds. STUDENT SOLUTION LACKING DOCUMENTATION seconds INSTRUCTOR RESPONSE You should show the reasoning; we know in advance that `dt will be in seconds, but be sure you understand how to get there from the given units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By dividing the unit km/sec into the unit km, you cancel out the unit of km by cross multiplication, leaving you only the unit of seconds. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: S= 10 - 4 = 6 meters. T= 5-2= 3 seconds. 6 meters/3 seconds= 2 meters/sec. The change in ds position is 6 meters. The change in dt clock time is 3 seconds. We combine these quantities by dividing seconds by meters to give us meters per second, our average velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time? What expression therefore symbolizes the average velocity between the two clock times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds= s2- s1 dt= t2- t1 The expression that therefore represents the average velocity is ds/dt, or vAve. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. The symbolic expression for the average velocity is therefore vAve = `ds / `dt = (s2 - s1) / (t2 - t1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rise of the triangle represents the change in position coordinate of the first point too the second, which is 6 m (10-4). The run of the triangle represents the change in clock time, 3 sec (5-2). confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the triangle is 6 meters/3 seconds= 2 meters/second, or velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sense the graph of position vs. clock time would be rise over run, that would give us the equation of ds/dt, or velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change `ds in position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position / change in clock time, which is `ds / `dt. This is equal to the average rate of change of position with respect to clock time, which is the definition of average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph would be increasing at an increasing rate, as the car is picking up speed as it is going down hill. The slope of my graph is therefore increasing. The slope of my graph shows that, the longer the time period the greater my slope will be. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate (an alternative description would be that the graph is increasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get? Cm/hr. 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? Seconds. 3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive. How is it that the change in position can be negative? Something can be moving in opposite direction of the reference point. 4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity? vAve. 5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph? As clock time increases the slope increases. 6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? Meters, Kilometers, feet, yards. Seconds, minutes, hours. Meters/second, kilometers/minute. Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.54 km. 2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter. 1.80 + 1.425 + 0.534 = 3.759 m. 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 - 3.842= -0.002 kg. Questions related to key systems 1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? 2. Questions/problems for University Physics Students 1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are measured in the counterclockwise direction from the positive x axis. 2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and describe how it appears to support your result. 3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y direction? What do you think is the uncertainty in its direction? " 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get? Cm/hr. 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? Seconds. 3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive. How is it that the change in position can be negative? Something can be moving in opposite direction of the reference point. 4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity? vAve. 5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph? As clock time increases the slope increases. 6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? Meters, Kilometers, feet, yards. Seconds, minutes, hours. Meters/second, kilometers/minute. Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.54 km. 2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter. 1.80 + 1.425 + 0.534 = 3.759 m. 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 - 3.842= -0.002 kg. Questions related to key systems 1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? 2. Questions/problems for University Physics Students 1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are measured in the counterclockwise direction from the positive x axis. 2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and describe how it appears to support your result. 3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y direction? What do you think is the uncertainty in its direction? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!
#$&* course Phy201+L 003. Velocity Relationships*********************************************
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Given Solution: vAve = `ds / `dt. The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec. Thus vAve is in m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If vAve is measured in cm/sec and dt is measured in seconds, then ds is measured in cm. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm. STUDENT QUESTION I dont get how sec and sec would cancel each other out INSTRUCTOR RESPONSE cm / s * s means(cm/s) * s, which is the same as(cm / s) * (s / 1). Multiplying numerators and denominators we have(cm * s) / (s * 1) or just(cm * s) / s, which is the same as cm * (s / s) = cm * 1 = cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By multiplying the unit cm/sec by the unit sec, you eliminate the unit of seconds by cross multiplication, giving you only the unit of centimeters. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm. STUDENT RESPONSE: For some reason this question just isn't making sense to me. INSTRUCTOR RESPONSE: In a self-critique you need to address the given solution in detail. A general statement such as yours gives me no information on what you understand. I need this information as a basis for helping you with what you don't understand. In order to give me the information I need you should be addressing each statement, and each phrase, to show me what you do and do not understand. The given solution can be broken into individual statements: 1. When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. 2. When we multiply fractions we will multiply numerators and denominators. 3. We obtain cm * sec / ( sec * 1). 4. This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm. Do you understand Statement 1? If not, have you written out the expressions cm/sec and sec/1 in standard form? (You might want to review the link given at the end of the Typewriter Notation exercise from Orientation, which should be posted at your access page). As best you can communicate it, what do you and do you not understand about this statement? Do you understand Statement 2? If not, what do you and do you not understand about this statement? Do you understand Statement 3? If not, have you written out the multiplication of cm/sec and sec/1 on paper? The multiplication is (cm / sec) * (sec / 1). Again, if you aren't sure how to write this out, refer to the link at the end of the Typewriter Notation exercise. Do you understand Statemet 4? If not do you understand that (sec / sec) * (cm / 1) is equal to sec * cm / (sec * 1), which is in turn equal to (cm * sec) / (sec * 1)? If not, specifically what do you and do you not understand? If you don't understand anything, then you should start with a review of basic fractions, a topic which is very much neglected in the typical curriculum in U.S. schools. Then you should return to these questions and give your best answers. A good link, current as of Sept. 2010: http://www.themathpage.com/arith/multiply-fractions-divide-fractions.htm You should submit a copy of question `q003, your solution, the given solution and this note. Insert your answers and/or additional specific questions and mark with &&&& before and after each insertion, then submit using the Submit Work Form. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If vAve is measured in km/sec and ds is measured in km, then dt is measured in seconds. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds. STUDENT SOLUTION LACKING DOCUMENTATION seconds INSTRUCTOR RESPONSE You should show the reasoning; we know in advance that `dt will be in seconds, but be sure you understand how to get there from the given units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: By dividing the unit km/sec into the unit km, you cancel out the unit of km by cross multiplication, leaving you only the unit of seconds. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: S= 10 - 4 = 6 meters. T= 5-2= 3 seconds. 6 meters/3 seconds= 2 meters/sec. The change in ds position is 6 meters. The change in dt clock time is 3 seconds. We combine these quantities by dividing seconds by meters to give us meters per second, our average velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time? What expression therefore symbolizes the average velocity between the two clock times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds= s2- s1 dt= t2- t1 The expression that therefore represents the average velocity is ds/dt, or vAve. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. The symbolic expression for the average velocity is therefore vAve = `ds / `dt = (s2 - s1) / (t2 - t1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rise of the triangle represents the change in position coordinate of the first point too the second, which is 6 m (10-4). The run of the triangle represents the change in clock time, 3 sec (5-2). confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the triangle is 6 meters/3 seconds= 2 meters/second, or velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Sense the graph of position vs. clock time would be rise over run, that would give us the equation of ds/dt, or velocity. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change `ds in position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position / change in clock time, which is `ds / `dt. This is equal to the average rate of change of position with respect to clock time, which is the definition of average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph would be increasing at an increasing rate, as the car is picking up speed as it is going down hill. The slope of my graph is therefore increasing. The slope of my graph shows that, the longer the time period the greater my slope will be. confidence rating #$&*:32; 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate (an alternative description would be that the graph is increasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get? Cm/hr. 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? Seconds. 3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive. How is it that the change in position can be negative? Something can be moving in opposite direction of the reference point. 4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity? vAve. 5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph? As clock time increases the slope increases. 6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? Meters, Kilometers, feet, yards. Seconds, minutes, hours. Meters/second, kilometers/minute. Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.54 km. 2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter. 1.80 + 1.425 + 0.534 = 3.759 m. 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 - 3.842= -0.002 kg. Questions related to key systems 1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? 2. Questions/problems for University Physics Students 1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are measured in the counterclockwise direction from the positive x axis. 2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and describe how it appears to support your result. 3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y direction? What do you think is the uncertainty in its direction? " 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get? Cm/hr. 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? Seconds. 3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive. How is it that the change in position can be negative? Something can be moving in opposite direction of the reference point. 4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity? vAve. 5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph? As clock time increases the slope increases. 6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? Meters, Kilometers, feet, yards. Seconds, minutes, hours. Meters/second, kilometers/minute. Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.54 km. 2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter. 1.80 + 1.425 + 0.534 = 3.759 m. 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 - 3.842= -0.002 kg. Questions related to key systems 1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? 2. Questions/problems for University Physics Students 1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are measured in the counterclockwise direction from the positive x axis. 2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and describe how it appears to support your result. 3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y direction? What do you think is the uncertainty in its direction? " 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If we were to multiply the unit cm / sec by the unit sec / hr what unit would we get? Cm/hr. 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? Seconds. 3. In this qa we began with the definition of average velocity. We have seen that this definition is consistent with our idea of speed as distance traveled divided by time required. However our definition of velocity can produce a negative result, which was not the case for speed. The negative result will, for example, occur when the change in position is negative while the change in clock time is positive. How is it that the change in position can be negative? Something can be moving in opposite direction of the reference point. 4. Having defined average velocity, we used symbols to stand for the initial and final positions and clock times. What expression did this give us for the average velocity? vAve. 5. Having defined average velocity, we then represented a pair of initial and final positions and clock times on a graph of position vs. clock time. What did we learn about the graph? As clock time increases the slope increases. 6. What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? Meters, Kilometers, feet, yards. Seconds, minutes, hours. Meters/second, kilometers/minute. Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.54 km. 2. Find 1.80 m + 142.5 cm + 5.34 * 10^5 micro m, to the correct number of significant figures. You should know that 'micro' indicates multiplication by 10^-6, so a micro m is 10^-6 meter. 1.80 + 1.425 + 0.534 = 3.759 m. 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 - 3.842= -0.002 kg. Questions related to key systems 1. Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? 2. Questions/problems for University Physics Students 1. Find the x and y components of two displacements, one of 4.0 meters at 60 degrees and the other of 3.0 meters at 135 degrees. Both angles are measured in the counterclockwise direction from the positive x axis. 2. If you drive 2.6 km north, then 4 km east before turning 45 deg to your left and traveling an addition 3.1 km, then what is the magnitude and direction of your displacement from the original position to the final position? Sketch a figure depicting this motion, describe the figure and describe how it appears to support your result. 3. What do you think is the uncertainty in the magnitude of a vector whose components are given as 0.7 cm in the x direction and 5.932 cm in the y direction? What do you think is the uncertainty in its direction? " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!