#$&* course Phy 121 6/14 at 4:45 PM 002. Velocity*********************************************
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Given Solution: Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will span 3 meters, corresponding to the distance moved in 1 second, on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The concept of a rate is when a change in a certain quantity is divided by the change in a different quantity. Here, we took the change in meters and divided it by the change in seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 ------------------------------------------------ Self-critique rating: OK ********************************************* 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: I think time is dependent on object position. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer was wrong. I thought that the change in time would depend on where the object was positioned. However, now I realize that time is independent of position because it runs no matter if the object is moving or not. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q004. We are still referring to the situation of the preceding questions, which concern average velocity: So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I don’t think that I missed any concepts in my explanations of average rate. The average rate at which position is changing with respect to clock time simply means you must find the displacement of position and divide it by the clock time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. Your solution: The average velocity here is -6 m/ 3 s = -2 m/s. Average velocity can be positive or negative. However, average speed cannot be negative because it does not make since for there to be < 0 meters traveled in > 0 seconds. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative. Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity. In general distance has no direction, while velocity does have direction. Putting it loosely, speed is just how fast something is moving; velocity is how fast and in what direction. In this case, the average velocity is • vAve = `ds / `dt = -6 m / (3 s) = -2 m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which its position changes, then what expression stands for the average velocity vAve of the object during this time interval? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve = ‘ds / ‘dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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? Your solution: To write ‘ds and ‘dt on your paper you use the Greek capital letter Delta. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1. `d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d. You may use either `d or Delta when submitting work and answering questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The object moves 50 meters. We know this by setting up the equation vAve * ‘dt = ‘ds, so 5 m/s * 10 s = 50 m. A rate is obtained by taking the change of one quantity and dividing it by the change of another. This is related to the concept of a rate because the average velocity of 5 m/s is found by taking the change of position with respect to clock time, and we have to understand this fully in order to use the correct equation to find the displacement. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: To find ‘ds we have to multiply the rate by the ‘dt that the change occurs during. For example, vAve * ‘dt = ‘ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the change in a quantity we multiply the rate by the time interval during which the change occurs. The velocity is the rate, so we obtain the change in position by multiplying the velocity by the time interval: • `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: • We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour). • When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance. Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem. Your solution: vAve is the average rate at which a position changes with respect to clock time. Therefore, we know that vAve = ‘ds / ‘dt to solve the problem. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): 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: To find ‘ds by using the equation vAve = `ds / `dt: First, vAve * `dt = `ds / `dt * `dt Because ‘dt / ‘dt = 1, we have now solved for ‘ds vAve * ‘dt = ‘ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not show the final equation to find ‘ds from both sides. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? Your solution: The preceding result relates to our intuition about those meanings because when we multiply the average velocity by the time interval, we obtain the displacement during that interval. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I feel as if I could have given a specific example such as car travel like you did in order for my answer to make more sense. ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result? Your solution: vAve = `ds / `dt. vAve * `dt = `ds / `dt * `dt ‘dt / ‘dt = 1 so vAve * `dt = `ds (vAve * ‘dt) / (vAve) = (‘ds / vAve) ‘dt = ‘ds / vAve confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? Your solution: This result is related to our intuition about the meanings of the terms average velocity, displacement and clock time when we compare it to figuring out how long a trip will take by car at a certain speed. In this case we divide the distance by the speed in order to get trip time. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ 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 = change in position / change in time vAve = 20 m / 2 s = 10 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `q016. A car moves at an average speed of 20 m/s for 6 seconds. How far does it move? Your answer should begin with the definition of average velocity, in terms of the definition of an average rate of change. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Average velocity is the change in position with respect to clock time. Therefore, vAve = ‘ds / ‘dt 20 m/s = ‘ds/ 6s Multiply both sides by 6s so, 120 m = ‘ds confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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: Average velocity is the change in position with respect to clock time. Therefore the expression for this is 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? The average rate of change of its position with respect to clock time is the same as finding average velocity. The general definition to the average rate of change is change of A with respect to B = change in A / change in B. In this situation quantity A is the change in position and quantity B is the change in clock time. Furthermore we know that vAve = (87cm - 34cm) / (5.3s - 4.6s) = (53cm) / (.7s) = 75.71 cm/s. 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? The general definition to the average rate of change is change of A with respect to B = change in A / change in B. In this situation quantity A is the change in velocity and quantity B is the change in clock time. Furthermore we know that the average rate of change = (20 cm/s - 12 cm/s) / (15.3s - 6.9s) = (8 cm/s) * (1 / 8.4s) = (8 cm / 8.4 s^2) = .95 cm/s^2. 3. What is your best estimate of the average velocity of the object in #2, for the given time interval? To find the average velocity of the object in #2 we must find the positions for both velocities given. Velocity = distance traveled / time. Therefore to find the 2 positions we perform the following. For initial position: 12 cm/s = s / 6.9sec so 12 cm/s * 6.9s = 82.8 cm. For final position: 20 cm/s = s / 15.3 sec so 20 cm/s * 15.3 s = 306 cm. Now we have the correct things to find the average velocity, also known as the average rate of change of position with respect to clock time. The general definition to the average rate of change is change of A with respect to B = change in A / change in B. In this situation quantity A is the change in position and quantity B is the change in clock time. Furthermore we know that vAve = (306cm - 82.8cm) / (15.3s - 6.9s) = ( 223.2 cm) / (8.4 s) = 26.57 cm/s. 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. The general definition to the average rate of change is change of A with respect to B = change in A / change in B. In this situation quantity A is the change in position and quantity B is the change in clock time. The quantity that we can determine is the change in position or ‘ds. Furthermore we know that 24 m/s = (‘ds) / (5s). We multiply both sides by 5s to get ‘ds = 120 m. 5. What is wrong with saying the average velocity = position / clock time? Average velocity is considered an example of average rate of change because it is defined in terms of displacement. The general definition to the average rate of change is change of A with respect to B = change in A / change in B. In this situation quantity A is the change in position and quantity B is the change in clock time. Furthermore, vAve = change in position / change in 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? a) (.1 / 3.4) * 100 = 2.9% b) (.1 / .87) * 100 = 11.5% c) When using this mechanism there is much higher percent uncertainty with a shorter time interval. Furthermore, there is much lower percent uncertainty with a longer time interval. 2. What is the uncertainty in the following reported measurements, and what is the percent uncertainty in each? • 5.8 centimeters o 5.8 cm can be taken to imply a number between 5.75 cm and 5.85 cm, which means that the number is 5.8 cm +- .05 and the uncertainty is .05. Therefore (.05 / 5.8 cm) * 100 = .86% uncertainty • 2350 kilometers o (.05 / 2350 km) * 100 = .002 % uncertainty • 350. Seconds o (.05 / 350 s) * 100 = .014% uncertainty • 3.14 o (.05 / 3.14) * 100 = 1.59% uncertainty • 3.1416 o (.05 / 3.1416) * 100 = 1.59% uncertainty 3. What is the uncertainty in the area of a rectangle, based on reported length 23.7 cm and width 18.34 cm? Area of a rectangle = length * width. 23.7 cm +- .05 cm = length and 18.34 cm +- .05 = width Area for lower values= 23.65 cm * 18.29 cm = 432.56 cm^2 Area for upper values= 23.75 cm * 18.39 cm = 436.76 cm^2 Difference between lower and upper areas = 4.204 cm^2 Uncertainty= (1/2) * 4.204 cm^2 = 2.102 cm^2 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? I am 5’6” tall. So, 5 ft * 12 inches = 60 in + 6 in = 66in. 66 in * 2.54 cm/inch = 168.64 cm 168.64 cm * .01m/cm = 1.6764 meters. I weigh 150 lbs. So, 150 lbs * (1 kg / 2.21 lbs) = 67.9 kg. 66” is halfway between 65.5” and 66.5” so uncertainty is 66” +- .5” .5”/66” = .0076 or .76%.. Uncertainty is +- .5” and it is +-.76%. 150 lbs is halfway between 149.5 lbs and 150.5 lbs so uncertainty is 150 lbs +- .5 lbs. .5 lbs / 150 lbs = .003 or .3%. Uncertainty is +- .5 lbs and it is +- .3% ------------------------------------------------ Self-critique rating: I really had a hard time on the text-related questions that involved uncertainty. Especially on number 2 because I understand how to find percent uncertainty by taking the ratio of the uncertainty to the measured value and multiplying it by 100; however, I do not understand how to find the uncertainty when just given a measured value. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: I really had a hard time on the text-related questions that involved uncertainty. Especially on number 2 because I understand how to find percent uncertainty by taking the ratio of the uncertainty to the measured value and multiplying it by 100; however, I do not understand how to find the uncertainty when just given a measured value. " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!