#$&* course Phy 121 this is a resubmission. my bad, wrong form.Tue 09 Nov 2010 11:31:47 PM EST 003. Velocity Relationships
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Given Solution: vAve = `ds / `dt. The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec. Thus vAve is in m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: ds = 1 cm/sec * 1 sec = 1 cm / 1 sec * 1 sec = 1 cm = cm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm. STUDENT QUESTION I don’t get how sec and sec would cancel each other out INSTRUCTOR RESPONSE cm / s * s means (cm/s) * s, which is the same as (cm / s) * (s / 1). Multiplying numerators and denominators we have (cm * s) / (s * 1) or just (cm * s) / s, which is the same as cm * (s / s) = cm * 1 = cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ds = 1 cm/sec * 1 sec = 1 cm / 1 sec * 1 sec = 1 cm = cm to explain sec/sec = 1/1 = 1 -> identity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ 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: dt = 1 km / 1 km/sec = 1 km / (1 km / 1 sec) = 1 km * 1 sec / 1 km = 1 sec = sec confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds. STUDENT SOLUTION LACKING DOCUMENTATION seconds INSTRUCTOR RESPONSE You should show the reasoning; we know in advance that `dt will be in seconds, but be sure you understand how to get there from the given units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 km / (1 km / 1 sec) = 1 km * 1 sec / 1 km = 1 sec = sec You always flip the dividend fraction to form an inverse product (mathematically equivalent). Anyways, 1/(x/y) = 1 * y/x = (x/y)^-1 (I think). Eighth grade stuff here I think. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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_0 = 4 m s_f = 10 m Δs = s_f - s_0 = 10 m - 4 m = 6 m t_0 = 2 s t_f = 5 s Δt = t_f - t_0 = 5 s - 2 s = 3 s v_ave = Δs / Δt = 6 m / 3 s = 2 m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: s_0 = s1 s_f = s2 Δs = s_f - s_0 = s2 - s1 t_0 = t1 t_f = t2 Δt = t_f - t_0 = t2 - t1 v_ave = Δs / Δt = (s2 - s1) / (t2 - t1) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. The symbolic expression for the average velocity is therefore vAve = `ds / `dt = (s2 - s1) / (t2 - t1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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 is equivalent to the change in position (displacement) and the run is equivalent to the change in time. As the slope they represent the average velocity. s_0 = 4 m s_f = 10 m Δs = s_f - s_0 = 10 m - 4 m = 6 m Ξ rise t_0 = 2 s t_f = 5 s Δt = t_f - t_0 = 5 s - 2 s = 3 s Ξ run v_ave = Δs / Δt = 6 m / 3 s = 2 m/s Ξ rise/run confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: v_ave = Δs / Δt = 6 m / 3 s = 2 m/s Ξ slope = rise/run = 6 m / 3 s = 2 m/s v_ave = slope confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: A steeper slope implies greater velocity because there is more displacement in a given time, i.e. the object is moving fast. A shorter slope implies lesser velocity because there is less displacement in a given time, i.e. the object is moving slow. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Since the rise between two points on a graph of velocity vs. clock time represents the change `ds in position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position / change in clock time, which is `ds / `dt. This is equal to the average rate of change of position with respect to clock time, which is the definition of average velocity. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* 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: Before the car rolls, it's velocity is 0. As the car rolls down the hill, it accelerates in a linear motion so long as the angle of the hill doesn't change. Gravity accelerates it, friction slows it, air could slow it but probably too little at these velocities to profit any from concerning it. If the hill is a PERFECTLY straight line (unlikely), then a graph of velocity will be linear; increasing at a constant rate. If the hill is curved so that the top is steeper, then the graph will start out steep and work its way out to a time asymptote. If the hill is curved so that the bottom is steeper, then the graph's slope will start short and work its way steeper The graph actually might look similar to the hill turned upside down. -- That is speculative though and requires checking before I'll vote on it. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate (an alternative description would be that the graph is increasing and concave up). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: 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 / sec) / (sec / hr) = cm / sec * hr / sec = cm hr / (sec^2) 2. If we were to divide the unit cm by the unit cm / sec what unit would we get? = cm / (cm / sec) = cm * sec / cm = sec 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? If velocity is the change in position over the change in time, then either a negative position or a negative time (but not both; discrete mathematics teaches that it is called an exclusive or, commonly abbreviated as xor and bearing the symbol ⊕) must be the cause. If a negative position is given it must mean the object is moving backward with time; reverse direction. If a negative time is given, it must mean that the object is moving forward and the time is moving backwards ---- what a strange incident. Guess logic serves to tell it probably won't be consumed very frequently in the Δd/-Δt format. The sign indicates direction along a single axis, or as my imagination can only wonder, some product of axes Δ(v_x * v_y * v_z) / Δt. 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? 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? 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? Questions related to text 1. What is 3.2 km + 340 meters, to the correct number of significant figures? 3.2 km + 340 meters = 3.2 km + .34 km = 3.54 km using significant figures (round, floor): 3.5 using significant figures (ceil): 3.6 * I think this is right. I really think using significant figures limits accuracy quite a bit sometimes. Why would this be important??? 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 m + 142.5 cm + 5.34 * 10^5 μm conversion is needed, looks like a base meter is best for this 1.80 m = 1.80 m 142.5 cm = 142.5 cm * 10^-2 m/cm = 142.5 cm / 100 m/cm = 142.5 cm * 1/100 m/cm =142.5 / 100 * cm*m/cm = 1.425 m 5.34 * 10^5 μm = (5.34 * 10^5 μm) * 10^-6 m/μm = 5.34 * 10^5 * 10^-6 μm * m/μm = 5.34 * 10^-1 m = 5.34 m / 10 = 0.534 m now better 1.80 m + 1.425 m + 0.534 m = 3.759 m (already in significant figures, if you use the max significant figures (1.425); if you use the least, it would only be 3 digits) 3. Find 3.84 kilograms - 3842 grams, to the appropriate number of significant figures. 3.84 kilograms - 3842 grams = 3.84 * 1000 grams - 3842 grams = 3840 grams - 3842 grams = -2 grams = -2.000 grams 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, but you only need 1. the difference is 8. you could say 8.0 if you needed the digit to count for accuracy (but we all know it's the same number, zero or not). 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?"