course H³ö‘öûÄzøñª† —°‰µëµÈú¯Ñ‡Ëòassignment #003
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15:38:43 `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 would be written as meters/seconds confidence assessment: 3
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15:39:13 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 --> true self critique assessment: 3
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15:40:00 `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 --> the displacement would just be in cm confidence assessment: 3
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15:40:13 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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RESPONSE --> correct self critique assessment: 3
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15:41:15 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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RESPONSE --> when you multiply the units cm/sec be sec then the seconds cancel each other just leaving cm self critique assessment: 3
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15:43:03 `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 confidence assessment: 3
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15:43:16 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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RESPONSE --> correct self critique assessment: 3
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15:43:57 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> by dividing the unit km/sec be the unit km you are left with the unit sec confidence assessment: 3
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15:44:19 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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RESPONSE --> correct self critique assessment: 3
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15:46:01 `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 --> 3meters/sec, the 'ds is 6 meters, and 'dt is 3 seconds. we take 6meters/3sec to give you 2m/s confidence assessment: 3
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15:46:18 `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 --> correct confidence assessment: 3
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15:47:06 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 --> s1/t1 and s2/t2 self critique assessment: 2
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15:49:09 `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 --> the rise is 6 meters, the run is 3 seconds, the rise indicates the distance traveld and the run represents the time interval confidence assessment: 3
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15:49:22 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 --> correct self critique assessment: 2
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15:49:41 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> the slope is 2 confidence assessment: 3
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15:50:02 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> 2 meters/second not just 2 self critique assessment: 3
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15:52:53 `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 --> it is the same as just meters divide by the time if the distance is greater than the time it takes to travel then that means the velocity is greater confidence assessment: 3
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15:53:07 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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RESPONSE --> correct self critique assessment: 3
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15:55:02 `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 --> if the car rolls down the hill the velocity increases the futher it travels down the hill the more it increase the less time required to travel. the slope of the graph would be increasing at an increasing rate. The slope of the graph is increasing confidence assessment: 3
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15:55:20 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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RESPONSE --> correct response self critique assessment: 3
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