course phy 201
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12:22:17 How does the shape of the corresponding position vs. clock time graph, with its upward curvature, show us that the time required to travel the first half of the incline is greater than that required to travel the second half?
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RESPONSE --> ok
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12:22:23 ** STUDENT RESPONSE: The shape of the graph, with its upward curve, shows us that it is increasing at an increasing rate over the duration of time. The average velocity between initial time and t1 is `ds/`dt = 1/1 or 1 m/s. From t1 to t2, `ds/`dt = (4-1)/1 = 3 m/s. The next 1 second increment produces a velocity of 5 m/s, and finally from t3 to t4, velocity is shown to be 7 m/s. It is simple to visually see that from the first half of the time duration, the slope is less, than for the second half of the time duration. Also, finding the average s, (16-0)/2 = 8, we find that the average or halfway distance is reached at 3 seconds, or three-quarters of the way through the total time duration, and in the next second, the distance doubles, therefore it does take longer to travel the first half of the incline, and much less time to travel the second half. It is easier to see than to explain. INSTRUCTOR COMMENT: Everything you say is correct. The halfway position is indicated by the position on the vertical or y axis halfway between initial and final position. The instant at which the halfway position is attained is the corresponding coordinate on the horizontal or t axis. The time required for the change in position from the initial to the halfway position is represented by the interval between the two corresponding clock times. On a graph which is concave down, this position will occur more than halfway between initial and final clock times. ** ANOTHER GOOD STUDENT RESPONSE: We could mark on the graph the initial and final points, then move over to the y axis and mark the position at the intial point and at the final point. We could then mark the y coordinate halfway between these and move over to the graph to obtain the graph point which corresponds to the halfway point. We can then construct a line segment from the graph point corresponding to the initial point, to the graph point corresponding to the halfway point, and finally to the graph point corresponding to the final point. Since the graph increases at an increasing rate the slope of the second segment is greater than that of the first. Since the change in the y coordinate is the same for both, it follows that the run of the second segment is shorter than the first. Since the run corresponds to the time interval, we see that the time interval corresponding to the second half is shorter than that corresponding to the first half. **
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`£½†êæç÷囧ǎבÙwĸÚÓrô assignment #002 ÕÚ˜¢x÷WÚíÀ•È’ÏŽÜJ}¤ªeÏڱ̦sÒÔ| Physics I Class Notes 06-02-2006
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12:27:34
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12:28:32 How does the shape of the corresponding position vs. clock time graph, with its upward curvature, show us that the time required to travel the first half of the incline is greater than that required to travel the second half?
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RESPONSE --> the shape of an upward curve graph shows an increase at increasing rate/duration of time.
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12:28:42 ** STUDENT RESPONSE: The shape of the graph, with its upward curve, shows us that it is increasing at an increasing rate over the duration of time. The average velocity between initial time and t1 is `ds/`dt = 1/1 or 1 m/s. From t1 to t2, `ds/`dt = (4-1)/1 = 3 m/s. The next 1 second increment produces a velocity of 5 m/s, and finally from t3 to t4, velocity is shown to be 7 m/s. It is simple to visually see that from the first half of the time duration, the slope is less, than for the second half of the time duration. Also, finding the average s, (16-0)/2 = 8, we find that the average or halfway distance is reached at 3 seconds, or three-quarters of the way through the total time duration, and in the next second, the distance doubles, therefore it does take longer to travel the first half of the incline, and much less time to travel the second half. It is easier to see than to explain. INSTRUCTOR COMMENT: Everything you say is correct. The halfway position is indicated by the position on the vertical or y axis halfway between initial and final position. The instant at which the halfway position is attained is the corresponding coordinate on the horizontal or t axis. The time required for the change in position from the initial to the halfway position is represented by the interval between the two corresponding clock times. On a graph which is concave down, this position will occur more than halfway between initial and final clock times. ** ANOTHER GOOD STUDENT RESPONSE: We could mark on the graph the initial and final points, then move over to the y axis and mark the position at the intial point and at the final point. We could then mark the y coordinate halfway between these and move over to the graph to obtain the graph point which corresponds to the halfway point. We can then construct a line segment from the graph point corresponding to the initial point, to the graph point corresponding to the halfway point, and finally to the graph point corresponding to the final point. Since the graph increases at an increasing rate the slope of the second segment is greater than that of the first. Since the change in the y coordinate is the same for both, it follows that the run of the second segment is shorter than the first. Since the run corresponds to the time interval, we see that the time interval corresponding to the second half is shorter than that corresponding to the first half. **
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12:30:02 object?
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RESPONSE --> we can take acceleration multiplied by time to get velocity. We could then take the average of the velocity.
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12:31:06 ** The reasoning process is as follows: To get the change in velocity you multiply the average rate at which velocity changes (i.e., the acceleration) by the time interval. This change in velocity will be added to the initial velocity to get the final velocity. Since the rate of change of velocity is constant, we can average initial and final velocities to get average velocity. Since velocity is rate of change of position, meaning that ave velocity = change in position / change in clock time, then position change (i.e., displacement) is the product of average velocity and the time interval (i.e., the period of time `dt). In symbols using v0, vf and `dt: vAve = (v0 + vf) / `dt because accel is uniform. vAve = `ds / `dt (this is the definition of vAve and applies whether accel is uniform or not) `ds = vAve * `dt = (vf + v0) / 2 * `dt. FORMULA VS. EXPLANATION: Distance is x. x=1/2 A t ^2 + the initial velocity times time INSTRUCTOR COMMENT: That's a (correct) formula, not a reasoning process. Be sure you know the difference, which will be important on some of the tests. The formula is the end result of a reasoning process, but it is not the process. I emphasize the process at the begiinning, as you can tell from the fact that we devote over a week establishing and reasoning through these concepts. The formulas can be taught in a day; really understanding motion takes longer. **
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RESPONSE --> ok
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12:32:08 position vs. clock time?
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RESPONSE --> we could construct a graph by finding the position points instead of using velocity.
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12:32:38 ** STUDENT RESPONSE: We determine position changes over specified time intervals by subtracting the initial time from the final time for that specified time period. Then take the square of the time that we just determined and multiple it by half the acceleration. You could construct a graph by simply finding all the position points and plotting those on the graph instead of the velocity. INSTRUCTOR COMMENTS: We can of course do this, except that the graph doesn't directly tell us the acceleration. To find that we would first have to find the rate of velocity change, represented by the slope of the graph. The geometrical picture is very important. What we do is find the area of the trapezoid formed by the graph between the two clock times. The altitudes represent velocities, so the average altitude of a trapezoid represents the average velocity during the corresponding time interval. Since the width of the trapezoid represents the time interval, multiplying the altitude by the width to get the area represents the product of average velocity and time interval, which is the displacement corresponding to the time interval. Typically we will do this for several consecutive trapezoids, in order to get the change in position from the start of the first trapezoid to the end of the last. If this is done over a series of adjacent trapezoids, we get a series of areas, which are position changes. We then construct a graph of position vs. clock time for the given clock times by adding each successive position change to the previous position. **
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RESPONSE --> ooohhh
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12:33:22 In terms of the meanings of altitudes, area and width, how does a velocity vs. clock time trapezoid represent change in position?
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12:35:17 ** The trapezoid as constructed has a single width, which represents the time interval. The average height represents the average velocity. So ave ht multiplied by width represents the product of ave vel and time interval, which gives us displacement. Good Student Response: Change in position can easily be determined from a graph of velocity versus clock time. Using the idea of a trapezoid or triangle you take the average height, which is the average velocity and multiply it by the average width, which is the time interval.. This will give us the area, which is the position change.
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RESPONSE --> ok, so the average height is the average velocity, and the average width is the time interval. all of this can give us the position change.
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12:35:49 How can a series of velocity vs. clock time trapezoids help us to calculate and visualize position vs. clock time information?
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RESPONSE --> the greater the area of the trapezoid, then the greater the position change.
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12:35:58 ** The position change for a given time interval corresponds to the area of the v vs. t trapezoid which covers that time interval. The greater the area of the v vs. t trapezoid, then, the greater the change in position. Starting with the first position vs. clock time point, we increase the clock time and change the position coordinate according to the dimensions of each new trapezoid, adding each new trapezoid area to the previous position. Assuming the trapezoids are constructed on equal time intervals, then, the greater the average altitude of the trapezoid the greater the position change, so that the average height of the triangle dictates the slope of the position vs. clock time graph. ANOTHER INSIGHTFUL STUDENT RESPONSE: Well the series would be trapezoids for each time interval. Using each interval and plugging it in to the equation mentioned in the previous question you would come up with different positions. The total position change up to a given clock time is easily found by adding the position changes during all the time intervals up to that clock and calculating the areas of successive trapezoids gives us a series of successive displacements, each added to the previous position, so that trapezoid by trapezoid we accumulate our change in position **
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RESPONSE --> ok
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¿xÄø®˜þÇ¿ñ—öCŦ‹öàåú„© assignment #003 ÕÚ˜¢x÷WÚíÀ•È’ÏŽÜJ}¤ªeÏڱ̦sÒÔ| Physics I Class Notes 06-02-2006
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12:37:08 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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12:39:06 ** 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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12:39:33 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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12:40:33 ** 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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CwFåŸéË¥¢Ó¦»õ™ÒËà¬Ôî…ŒJæÛ† assignment #003 ÕÚ˜¢x÷WÚíÀ•È’ÏŽÜJ}¤ªeÏڱ̦sÒÔ| Physics I Class Notes 06-02-2006
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12:42:49 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 --> we divide by the change in velocity by the time interval to get an average acceleration. and to find the displacement we muliply the average velocity by the duration of time?
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12:44:28 ** 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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RESPONSE --> ok
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12:45:37 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 --> the average height is equal to the average velocity and the average width is equal to the time interval. if we multiply the average height and the average width we find the displacement.
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12:46:53 ** 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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RESPONSE --> oooh ok
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