course PHY 121 Ӄڪ~Wȉmfzassignment #003
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22:35:11 `q001. Note that there are 11 questions in this assignment. vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?
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RESPONSE --> vAve will be in meters per second confidence assessment: 3
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22:35:50 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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RESPONSE --> Right, must have misread the units of measurement self critique assessment: 3
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22:35:57 12-11-2007 22:35:57 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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NOTES ------->
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22:38:07 `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?
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RESPONSE --> It would be measured in cm, because seconds cancel out. confidence assessment: 3
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22:38:25 12-11-2007 22:38:25 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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NOTES ------->
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22:39:29 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> We are trying to solve for cetimeters, so to get that we must multiply each side by seconds to obtain our units or distance. confidence assessment: 3
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22:39:32 12-11-2007 22:39:32 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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NOTES ------->
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22:40:06 12-11-2007 22:40:06 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.
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NOTES -------> ok same as my thought
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22:41:35 `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?
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RESPONSE --> 'dt would be measured in seconds, the only other choice after completing the math in which km cancel one another out confidence assessment: 3
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22:41:47 12-11-2007 22:41:47 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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NOTES -------> ok
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22:44:03 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> since the rate would be in 'ds/'dt, it would make vave km/sec, so if we want to solve for an unknown ('dt) then to solve for time, we must divide both sides by the distance confidence assessment: 3
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22:44:34 12-11-2007 22:44:34 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.
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NOTES -------> our main objective being to solve for time
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22:47:53 `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?
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RESPONSE --> avg vel = final pos - inititial pos/time elapsed, so 6/3=2m/sec. change in 'ds is 6meters and dt' is 3sec. take difference of initial from final and divide by difference in clock time. confidence assessment: 3
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22:48:08 12-11-2007 22:48:08 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.
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NOTES -------> same thought
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22:49:49 `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?
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RESPONSE --> change 'ds = s2 - s1 and change 'dt = t2 - t1 confidence assessment: 3
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22:51:03 We see that 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. What expression therefore symbolizes the average velocity between the two clock times.
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RESPONSE --> avg vel = final pos - initial pos divided by time elapsed self critique assessment: 3
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22:56:05 `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?
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RESPONSE --> rise = 5/2 while the run = 10/4. The rise represents the seconds while the run represents the meters. confidence assessment: 2
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22:57:58 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.
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RESPONSE --> ok, this makes more sense with the rise being equal to double the run with 10m - 4m while the run is represented by change in clock time. rise/run = 6m/3s self critique assessment: 3
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22:59:23 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> 6m/3s = 2m/s and it represents the displacement of position with respect to clock time confidence assessment: 3
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22:59:34 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> ok self critique assessment: 3
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23:01:00 `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?
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RESPONSE --> Because a steeper slope implies faster change in position with respect to clock time which implies greater velocity. confidence assessment: 3
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23:01:14 12-11-2007 23:01:14 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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NOTES -------> ok
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23:03:45 `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?
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RESPONSE --> As we move out the x axis with displacement, we also are moving up the y axis in clock time making the slope increase at a constant rate. slope would be increasing. confidence assessment: 3
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23:05:05 12-11-2007 23:05:05 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.
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NOTES -------> I thought it would be increasing at an increasing rate, but figured since we didn't know by how much the velocity was increasing that I would just say it was constant. However, if it is constantly increasing velocity, then it must be increasing at an increasing rate.
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