#$&* course If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
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Given Solution: Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: In terms of speed rate is defined as the speed at which one measured quantity (meters) happens in relation to another measured amount such as time (seconds). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK I left out some details you included in the solution but my grasp of the concept here is fairly firm, becoming more so by the day. ------------------------------------------------ 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: In this case, time is dependent on the objects position. We are not measuring here the number of seconds elapsed within a displacement of position, but rather we are trying to find the magnitude (???? Am I using this word right here????) of displacement in relation to the amount of time it takes. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Okay so because time itself acts alone and continues indefinitely it is the independent variable, and because meters is measured in relation to time, it must be the dependent variable. Therefore, object position is dependent on time ------------------------------------------------ Self-critique rating: OK
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Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If theres anything you dont understand, be sure to address it in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object? What is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If my reasoning is correct, velocity is defined as being the speed of something, the average velocity and average speed will be the same. vAve = `ds (6 m) / `dt (3 sec) = vAve (2 m / sec) Average rate of change of A with respect to change of B = Change in A / Change in B confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Okay so I got the average velocity but the average speed is actually something different. Speed relates simply to distance travelled and how quickly this is done. Velocity refers a change in POSITION which could be forward or back, positive or negative. Speed however cannot be negative, do the two are in fact different. Ohhh and there IS NO SPEED!! I missed that the 6 is negative, so my answer was technically wrong but I used the proper formula. The -6 means there can definitely not be a measurement of speed since the distance changes in a negative direction, something only velocity can account for. ------------------------------------------------ Self-critique rating: OK
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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): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would use the triangle symbol which I recall from my problem sets means change in, this must also be the greek symbol for delta. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1. `d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d. You may use either `d or Delta when submitting work and answering questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The rate here is given, which means the formula for calculating must be changed slightly. Here, `ds = vAve (5 m / sec) * `dt (10 sec) = `ds (50 m) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: OK ********************************************* Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds = vAve * `dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the change in a quantity we multiply the rate by the time interval during which the change occurs. The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour). When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The vAve is a measurement of the average velocity, which relates to the change in quantity A divided by the change in quantity B. If we need to find the change in quantity B or A, the formula is altered slightly to multiply change in quantity A by vAve. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): It might help for me to write this out in my own words so, the vAve relates to position change, change in position is change in A (or `ds) and change in time is change in B (or `dt) ------------------------------------------------ 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: Starting with our basic formula of vAve = `ds / `dt we can also find `ds or `dt. In this case, to find `ds we multiply both sides by `dt which eliminates it from one side leaving us with `dt * vAve = `ds Using this formula we can calculate the change in position in relation to the average rate of velocity and change in time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The three work together and each value combined with another can be used to find the remaining 3rd value. As vAve increases, so does the amount of distance covered, but the time will count in the same way, there will be more distance covered however in a shorter amount of time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For most of us our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Multiplying the average velocity by the duration of time results in the change in position during that given time interval. ------------------------------------------------ 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: Multiply both sides by `dt, `dt on the right cancels out leaving us with `dt * vAve = `ds. Divide both sides by vAve to get `dt = `ds / vAve confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we take the change in position and divide it by the average velocity, we will obtain the change in time confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Everyone drives cars so it is an easy relation to make between the three concepts presented. To find the time it will take you for a trip (`dt) we must divide the total distance or displacement (`ds) by the average rate of velocity (vAve) in this case, miles per hour. ------------------------------------------------ Self-critique rating: ok You should submit the above questions, along with your answers and self-critiques. ********************************************* Question: `q015. A ball falls 20 meters from rest in 2 seconds. What is the average velocity of its fall? Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average velocity is the amount of distance covered with respect to a given amount of time. In this case, the vAve is found by dividing the magnitude of the displacement by the amount of time elapsed, therefore vAve = `ds (20 m) / `dt (2 sec) = vAve (10 m / sec) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q016. A car moves at an average speed of 20 m/s for 6 seconds. How far does it move? Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average velocity is a relative amount of distance covered with respect to an amount of time. In this case, the vAve and the time (`dt) is already known, we must find the magnitude of displacement (`ds). `ds = vAve ( 20 m / sec) * `dt (6 sec) = `ds (120 m) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q017. An object's position changes by amount `ds during a time interval `dt. What is the expression for its average velocity during this interval? Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The average velocity is found by dividing the change in position by the change in time. For example, 3 meters (`ds) in 3 seconds (`dt) means an average velocity (vAve) of 1 m / sec vAve = `ds / `dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique rating: ok 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? 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? 3. What is your best estimate of the average velocity of the object in #2, for the given time interval? 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. 5. What is wrong with saying the average velocity = position / clock time? 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 rating: ????I did not see an answer prompt for this section so I am assuming this is for out notebook only, I have worked the problems out there but if you would like to see them I can send them in a separate submission. Please let me know how to proceed. Some questions follow???? ????In number 1 (QA) the question should be asking about average velocity instead of the average change in position. The change in position is already given `ds (87cm - 34 cm = 53cm) when divided by the change in time `dt ( 5.3 - 4.6 = 0.7 sec) reveals an average rate of velocity vAve (53 cm/ 0.7 sec = 75.71 cm / sec)?????