#$&* course Phy 201 2/17/11 around 2:46pm 003. Velocity Relationships
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Given Solution: vAve = `ds / `dt. The units of `ds are m and the units of `dt are sec, so the units of `ds / `dt must be m / sec. Thus vAve is in m/s. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 'ds will be measured in cm because vAve=`ds/`dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:ok ********************************************* Question: `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (cm/sec)*(sec/1) (cm*sec)/(sec*1)= The seconds cancel out and we are left with cm 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:OK ********************************************* 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 will be measured in seconds. to get the km/(km/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:OK ********************************************* Question: `q005. Explain the algebra of dividing the unit km / sec into the unit km. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: km/(km*sec)= (km/1) * (sec/km) = (km*sec)/(1*km) the kilometers will cancel out till just seconds are left confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds = 10-4= 6 meters `dt= 5-2 = 3 seconds change in position/ change in clock time equal the average velocity. The average velocity is (6/3)= 2m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:ok ********************************************* Question: `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time? What expression therefore symbolizes the average velocity between the two clock times. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: `ds= (s2-s1) `dt=(t2-t1) vAve= (s2-s1)/(t2-t1) or vAve= `ds/`dt confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. The symbolic expression for the average velocity is therefore * vAve = `ds / `dt = (s2 - s1) / (t2 - t1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating:ok ********************************************* Question: `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: rise is the change in the y values (position). 10-4= 6 meters the run is the change in the x values(time) 5-2= 3 seconds the slope = rise/run or 6/3= 2 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:OK ********************************************* Question: `q009. What is the slope of this triangle and what does it represent? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The slope of the triangle is the rise/run or 6/3= 2 the slope represents the average velocity of 2m/s confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The slope of this graph is 6 meters / 3 seconds = 2 meters / second. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: OK ********************************************* Question: `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the slope in the graph is found by dividing the run from the rise. The run is the change in clocktime and the rise is the change in position. vAve= `ds/`dt the greater the slope the greater the velocity confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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:ok ********************************************* Question: `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph will gradually increase from left to right. As the clock time moves on the cars position will keep increasing. The slope is increasing. a positive slope indicates that the velocity is increasing. 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 I failed to mention that the slope of the graph is increasing at an increasing rate. ------------------------------------------------ Self-critique rating:ok" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!