course phy 201 ßҌņwŭassignment #003
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11:01:30 Given the initial velocity, final velocity and time duration of a uniformly accelerating object, how do we reason out the corresponding acceleration and change in the position of an object?
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RESPONSE --> (final velocity minus initial velocity)/time duration gives you the average velocity then average velocity * time duration gives the distance the object changed
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11:02:08 ** COMMON ERRONEOUS STUDENT RESPONSE: To find the average acceleration, we divide the change in veolocity by the time interval. To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. You are not given the average velocity or the change in velocity. You have to first determine the average velocity; then your strategy will work. Since acceleration is constant you can say that average velocity is the average of initial and final velocities: vAve = (v0 + vf) / 2. Change in velocity is `dv = vf - v0. Now we can do as you say: To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. **
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11:06:39 In terms of the meanings of altitudes, area, slope and width, how does a velocity vs. clock time trapezoid represent change in position and acceleration?
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RESPONSE --> it allows you to visualize the results by giving it shape you can get the averages from the lengths of the sides
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11:06:45 ** If you multiply the average altitude by the width (finding the area) of the trapezoid you are multiplying the average velocity by the time interval. This gives you the displacement during the time interval. The rise of the triangle represents the change in velocity and the run represents the time interval, so slope = rise / run represents change in velocity / time interval = acceleration. **
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Jқ~^婭 assignment #001 }bVͫZw{喽 Physics I Vid Clips 06-27-2007
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11:15:40 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> velocity is not constant, it would be increasing
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11:15:46 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases. A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **
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11:16:21 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> the time it takes for the ball to travel the first half and the second half of the ramp, and compare the times
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11:16:31 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity. INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **
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11:17:04 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> it would be a positive curved line with an increasing slope
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11:17:10 ** If the ramp has an increasing slope, the velocity would increase at an increasing rate and the graph would curve upward, increasing at an increasing rate. If the ramp has a decreasing slope, like a hill that gradually levels off, the graph would be increasing but at a decreasing rate. On a straight incline it turns out that the graph would be linear, increasing at a constant rate, though you aren't expected to know this at this point. All of these answers assume an absence of significant frictional forces such as air resistance. **
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11:17:49 A ball rolls down ramp which curves upward at the starting end and otherwise rests on a level table. What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> the velocity would increase greatly at the beginning and then it would accelerate at a lower rate
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11:17:58 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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11:18:11 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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11:18:17 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right. Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**
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ժ{Qت assignment #004 }bVͫZw{喽 Physics I Class Notes 06-27-2007
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19:48:21 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> position and clock time
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19:49:38 ** If we first calculate velocities from the position vs. clock time data we get decreasing velocities. If we graph these velocities vs. midpoint clock times we get a graph which appears to be well-fit by a straight line. This is evidence that the acceleration of the water surface is uniform. If we calculate average accelerations based on average velocities and midpoint clock times we get a lot of variation in our results. However since acceleration results depend on velocities and changes in clock times, and since the velocities themselves were calculated based on changes in clock times, our results are doubly dependent on the accuracy of our clock times. So these fluctuating results don't contradict the linearity of the v vs. t graph. We also find that the position vs. clock time data are very well-fit by a quadratic function of clock time. If the position vs. clock time graph is quadratic then the velocity is a linear function of clock time (University Physics students note that the derivative of a quadratic function is a linear function) and acceleration is constant (University Physics students note that the second derivative of a quadratic function is constant). **
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RESPONSE --> ok
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xM[T}zʫWи assignment #004 }bVͫZw{喽 Physics I Class Notes 06-27-2007
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19:55:07 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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RESPONSE --> by finding the average velocity
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19:55:25 How do we reason out the process of determining acceleration from rest given displacement and time duration?
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|txYD{w assignment #003 003. Velocity Relationships Physics I 06-27-2007
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19:14:11 `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 m/s confidence assessment: 3
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19:15:18 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 the question said that ds was measured in meters not centimeters self critique assessment: 0
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19:17:36 `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 must be measured in cm b/c vAve is measured in cm/sec and 'dt is measured in sec, you cross out the sec and are left with cm confidence assessment: 3
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19:17:47 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 --> self critique assessment: 3
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19:19:15 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> cm/sec* sec, you cross out the secs b/c they are the same cm/sec*sec/1=cm, cross out top to bottom confidence assessment: 3
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19:19:23 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 --> self critique assessment: 3
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19:21:45 `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 sec, bc you invert vAve were its being divided 'dt=km/km/sec 'dt=km*sec/km confidence assessment: 3
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19:22:10 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 --> self critique assessment: 3
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19:23:24 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> km/km/sec is the same thing as taking km*sec/km, when dividing two you can invert the fraction and multiply the two together confidence assessment: 3
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19:23:34 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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RESPONSE --> self critique assessment: 3
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19:25:55 `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 --> vAve= s2-s1/t2-t1 vAve= 10-4/5-2 vAve=6/3 vAve=2 confidence assessment: 3
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19:26:04 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 --> self critique assessment:
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19:26:37 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 --> forgot to hit response, i forgot to put on meters per second self critique assessment: 0
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19:28:10 `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 --> delta s/ delta t or change in s over change in t s2-s1/t2-t1 confidence assessment: 3
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19:28:25 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 --> self critique assessment: 3
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19:37:01 `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 --> 2, slope=change in x/change in t confidence assessment: 2
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19:37:30 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 --> self critique assessment: 3
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19:39:13 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> sorry answered that one on the last one the slope would be 2. confidence assessment: 2
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19:39:26 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> self critique assessment: 3
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19:41:13 `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 --> slope of the x vs. t is equal to the velocity. the greater the slope the greater the velocity. confidence assessment: 2
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19:41:22 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 --> self critique assessment: 3
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19:44:28 `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 slope of the graph would be increasing bc as the car rolls from rest it would be increasing speed and time so it would be increasing at an increasing rate confidence assessment: 2
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19:44:43 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 --> self critique assessment: 3
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v\zlx·} assignment #003 003. `Query 3 Physics I 06-28-2007
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16:12:39 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 --> these added together to be 3.8m, there is only 1 significant figure due to the fact of the 142.5 cm having only 1 significant figure. confidence assessment: 3
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16:12:51 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> self critique assessment: 1
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16:13:32 06-28-2007 16:13:32 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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NOTES -------> only doing college physics
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16:13:34 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> confidence assessment: 3
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16:13:42 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE --> self critique assessment: 3
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