course Phy 121 1/31 9 Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.
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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 explained how to determine the correct answer, but failed to provide an example that would allow someone to visualize the distance traveled per second. ------------------------------------------------ Self-critique rating #$&*:3 ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The preceding problem is related to the concept of rate because it shows the average rate of change for a distance traveled per unit of time. We figure out rate by taking the change in one unit divided by the change in another. confidence rating #$&*: 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 explained how to find rate as well as determined the correct rate, but did not provide as in depth of a description to solve the answer with step by step explanations. ------------------------------------------------ Self-critique rating #$&*: 2 ********************************************* Question: `q003. Is object position dependent on time or is time dependent on object position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I think that object position is dependent on time as it will continue to change even if and object isn’t. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average rate of change is determined by figuring out the average rate of change for one quantity and dividing it by the average rate of change for a second quantity. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If there’s anything you don’t understand, be sure to address it in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If an object is displaced -6 meters in 3 seconds then the average speed of the object would be 2 meters per second. I determined this by dividing the distance traveled by the time it took to travel that far. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative. Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity. In general distance has no direction, while velocity does have direction. Putting it loosely, position is just how fast something is moving; velocity is how fast and in what direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Velocity is how fast and object is moving and cannot be negative, therefore we can’t really calculate the velocity. Distance does not necessarily have a direction, but velocity has a direction. Could you still determine the velocity of an object moving backwards??? Ex. A care in reverse??? ------------------------------------------------ Self-critique rating #$&*:3 ********************************************* Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve= ‘ds/ ‘dt confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): The average velocity is the rate of change of position with respect to the time on the clock. This information is used to put together the equation vAve= ‘ds/’dt ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When writing `ds and `dt on your paper you do so by drawing a symbol that looks like a triangle. confidence rating #$&*: 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): The triangle symbol is referred to as the Delta symbol. ------------------------------------------------ Self-critique rating #$&*: ok ********************************************* Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To determine the average rate, we must determine the average of quantity a and the average of quantity b. Once we have determined the two quantities, we divide the change in a/change in b. However, since we already know the average rate of the object and for how long the object is traveling at that rate, we will not divide for this problem. Instead, we will multiply. So instead of the change in a/change in b, we will be following the formula change in a * change in b. This is written as 5 * 10 = 50. The object will move 50 meters. confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To write the expression for ‘ds, it is ‘ds= vAve * ‘dt. In other words, to find the change in position, you must multiply the average velocity by the time interval. confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique rating #$&*: ********************************************* Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve is equal to the velocity , ‘dt is equal to the time interval, and ‘ds is the change in position. So to figure out the velocity we use the following formula: vAve= ‘ds/’dt. confidence rating #$&*: 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): OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The algebraic steps used to solve for ‘ds is as follows: We must first get ‘ds by itself. To do this, we need to multiply each side by ‘dt. This will allow us to get ‘ds by itself on one side. On the other side of our equation it would give us vAve * ‘dt. The complete equation would be vAve*’dt= ‘ds confidence rating #$&*: 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): ok ------------------------------------------------ Self-critique rating #$&*: ok ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The previous problem tells us the following about rate: The change in position is determined by multiplying the speed by time it took to travel. This would give us the total miles traveled for a trip. confidence rating #$&*:: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique rating #$&*: OK ********************************************* Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve for ‘dt, we must multiply each side ‘dt. This will give us vAve * ‘dt= ‘ds. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): We must multiply each side by ‘dt so that we can get ‘ds by itself. ------------------------------------------------ Self-critique rating #$&*: 3 ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Each is related and the equation can be reworked with parts of the data to solve for what information is asked. Velocity, displacement and clock time are all dependent on one another. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): You can always check your work by thinking about mules, hours and miles per hour. We generally know that to figure out hot long it will take to make a trip at a certain speed you divide the total distance by the speed. If we divide the miles we need to travel total and the number of miles we can travel in an hour it will give us the number of hours needed to make the trip. ------------------------------------------------ Self-critique rating #$&*: 3 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? The rate of change is equal to the change in a/change in b. The change in a is found be 87-34= 53 The change in b is found b 5.3-4.6= 0.7 So 53/0.7= 75.71 cm per second 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? A values are 12 and 20. The change is a values = 8. B values are 6.9 and 15.3 . The change in b values = 8.4 To determine the rate of change 8/8.4= 0.95 cm per second 3. What is your best estimate of the average velocity of the object in #2, for the given time interval? 1.2 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. To determine the average rate of change we will call the interval of 24 m/s our change in position (a) and 5 seconds our change in time (b). When we divide 24/5 it gives us 4.8. 5. What is wrong with saying the average velocity = position / clock time? This statement should be sure to include the average velocity= the change in position/ the change in clock time. Text-related questions (to be submitted after completing text assignment) 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? The percent uncertainty of a time interval with 3.4 seconds and an uncertainty of +-.1 second is as follows: .1/3.4=0.029 0.029 * 100 = 2.9 % .1/.87=.11 .11* 100 = 11% I think the percent uncertainty depends on the duration of the interval because it all depends on how many significant numbers are in the time and that determines what your uncertainty is. 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 1. uncertainty =+- .1 percent uncertainty = 1.7% 2. uncertainty = +-10 percent uncertainty = .43 % 3. uncertainty = +-5 percent uncertainty = 1.4% 4. uncertainty = +-.1 percent uncertainty = 3.18% 5. uncertainty = +-.1 percent uncertainty = 3.1831% 3. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm? The uncertainty for the length is + .1 and the uncertainty for the width is +.01 4. What is the approximate uncertainty in the area of a circle, based on a reported radius of 2.8 * 10^4 cm? The approximate uncertainty of the area of a circle would be +.1 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? My height in meters is 1.65m and my weight is 81.081 kg. I think that there is + .1 uncertainty. Depending on shoes or clothing worn when measure or weighed it could change results from what they actually are.