course phy 201 qӤᝳ}ᩏQۺ assignment #003 003. Velocity Relationships Physics I 02-07-2008
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13:50:01 `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 --> meters per sec m/s confidence assessment: 3
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13:50:50 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 --> i thought it sad 'ds was meters not centimeters i just mis read the text i guess self critique assessment: 2
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14:16:53 `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 --> 'ds is measured in cm otherwise where did we get the cm in cm/sec if we only started off with sec. confidence assessment: 2
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14:17:12 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 --> exactly self critique assessment: 2
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14:23:56 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> cm/sec * sec means you have to multiply fractions cm/sec*sec=cmsec/sec=cm because the secs cancelled each other out. confidence assessment: 3
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14:25:35 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 --> i think thats what i said its def what i ment self critique assessment: 2
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14:31:34 `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 --> sec because km/sec=km/? so km/sec /km = km/? / km km/sec can multiplied by 1/km leaving sec=? because the km's cancell each other out confidence assessment: 3
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14:32:04 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 --> i didn't do the ( ) self critique assessment: 2
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14:35:00 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> km/sec will be muliplied by sec/1 to isolate the km confidence assessment: 2
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14:36:24 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 --> I just said i forgot the( ) and i did it again. i know the fraction has to be recriprocated but i dont think i conveyed that self critique assessment: 1
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14:42:48 `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 --> 'ds=10-4=6meters beacuse 'ds means the change in position 'dt=5-2=3sec because 'dt means the change in elapsed time so vAve='ds/'dt=6m/3s=2m/s confidence assessment: 3
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14:43:33 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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RESPONSE --> I STILL left out the ( ) but i got the rest right self critique assessment: 2
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14:46:50 `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 --> so its s2-s1 / t2-t1 = 'ds/'dt = vAve is this what your asking for confidence assessment: 0
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14:49:07 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 --> vAve = 'ds/'dt self critique assessment: 1
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15:03:56 `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 slope of the rise and run of this triangle are the rise/run or rise equals time while run equals position these are the y and x axises accordingly the slope in this case is 2 meters / sec found by 10-4 = 6meters 5-2=3sec 6m/3s=2m/sec confidence assessment: 1
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15:05:46 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> ok so i was on the right track i just went too far so the slope is the rise / run which is 6m/3s = 2m/s confidence assessment: 3
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15:06:04 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> got it self critique assessment: 2
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15:13:47 `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 represents the velocity which shows the rate at which the object accelerates. the greater the slope (rise/run) the greater the velocity because the greater the change in position related to the change in elapsed time confidence assessment: 1
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15:14:46 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 --> i think i was headed in that direction but im not sure that it was conveyed in my answer i think i got side tracked. self critique assessment: 1
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15:25:24 `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 --> the graph of a velocity increasing is a gently ""sloping"" line rising or falling from left to right or right to left depending on the + or - of the slope. we always assume an object rolling downwards is an increasing velocity when it starts at rest. so this is an increasing at an increasing rate. my slope is + so it is increasing. there are too many factors attributing to this like friction, terrain, tires, gas pedal, foot pressure, ect. so there is no way unless cruise control is used and since it was not stated so we don't asume. confidence assessment: 0
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15:25:59 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 --> right self critique assessment: 0
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course phy 201 Ȥx꼦zzLassignment #003 003. `Query 3 Physics I 02-08-2008
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01:57:52 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> The answer is 3.76m we get this by 1.80m + 1.425m + .534m=3.76m confidence assessment: 2
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