course phy121
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17:32:47 How do flow diagrams help us see the structure of our reasoning processes?
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RESPONSE --> By using the flow diagrams, we can map out the relationships among the variables. If we are given 1dt, a and v0, we can begin reasoning their relationships in order to solve for the remaining variables. from `dt*a we get `dv, from v0+`v we get vf, and so on.
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17:32:54 ** They help us to visualize how all the variables are related. Flow diagrams can also help us to obtain formulas relating the basic kinematic quantities in terms of which we have been analyzing uniformly accelerated motion. **
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RESPONSE --> ok.
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17:35:30 How do the two most fundamental equations of uniformly accelerated motion embody the definitions of average velocity and of acceleration?
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RESPONSE --> The two most fundamental equations provide us with the information needed to define average velocity and accerlation. Uniformly accelerated motion can be analyzed and understood by its five quantities of v0, vf, a, `ds and `dt. From these quanitities we can define vAve and a.
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17:37:14 ** Velocity tells us the rate at which the position changes whereas the acceleration tells us the rate at which the velocity is changing. If acceleration is uniform ave velocity is the average of initial and final velocities. The change in position is found by taking the average velocity vAve = (vf+ v0) / 2 and multiplying by the'dt to get the first fundamental equation `ds = (v0 + vf)/2 * `dt. The acceleration is accel = rate of change of velocity = change in velocity / `dt = (vf - v0) / `dt. In symbols this equation is a = (vf + v0) / `dt. Algebraic rearrangement gives us this equation in the form vf = v0 + a `dt. This form also has an obvious interpretation: a `dt is the change in velocity, which when added to the initial velocity gives us the final velocity. **
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RESPONSE --> Ok, I am very familiar with these fundamental equations: `ds=(vo+vf)/2*`d vf = v0+a*`dt I provided a much briefer explanation, but I do understand the relationship here.
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17:44:10 How can we interpret the third fundamental equation of uniformly accelerated motion?
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RESPONSE --> For trhe third equation, we use the 2nd equation for vf, vf=v0+a*`dt, and then place this equation for vf, in the equation for `ds. `ds = (v0+vf)/2*`dt, replacing with 2nd equation for vf `ds = (v0+v0+a`dt)/2*`dt we then solve for this equation `ds = 2v0+a`dt)/2*`dt `ds = v0`dt+.5a`dt^2
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17:44:32 ** The third equation says that `ds = v0 `dt + .5 a `dt^2. This means that the displacement `ds arises independently from initial velocity v0 and acceleration a: v0 `dt is the displacement of an object with uniform velocity moving at velocity v0, and 1/2 a `dt^2 the distance moved from rest by a uniformly accelerating object. The two contributions are added to get the total `ds. **
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RESPONSE --> ok.
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17:45:47 Why can we not directly reason out the basic 'impossible situation'?
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RESPONSE --> We can not reason out the basic impossible situation because given `ds, a and v0, there is no direction relationship among these variables. By substituting the fundamental equations, and using the relationships among those, we can solve for this situation.
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17:45:56 ** In this situation we know v0, a and `ds. From v0 and a we cannot draw any conclusions, and the same is true for v0 and `ds and also for a and `ds. No pair of variables allows us to draw any additional information. **
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RESPONSE --> ok.
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17:47:25 What strategy will we use to reconcile the basic 'impossible situation'? WE cann write down the 2 most fundamental equations and see what we do know.
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RESPONSE --> 'ds = (v0+vf)/2*`dt vf=v0 +a*`dt From these two equations we can eliminate one of the variables, either `dt or vf and solve for the missing variable. This will allow us to use the relationships we know to determine the missing variables.
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17:48:00 ** We can use the fourth equation vf^2 = v0^2 + 2 a `ds to obtain vf, then knowing the values of v0, vf, a and `ds we easily find `dt either by direct reasoning or by using one of the fundamental equations. **
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RESPONSE --> Ok, I answered with the step prior to using the 4th equation.
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17:50:56 What is the difference between understanding uniformly accelerated motion and analyzing it with the use of equations?
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RESPONSE --> Understanding the concept is simply enough, an ojbect accelerates at a uniform motion. By analyzing it with the equations, we can visualize how the relationship takes place among velocity, change in time, change in position, acceleration, and gain a fuller understanding of uniform motion. WIth uniform motion, vAve is the average between the final and initial velocities, or, the final will be twice the velocity of the initial. There are just principles and concepts that make more sense once we see the relationships among the equations.
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17:52:21 ** You can use equations without understanding much of anything. To use the equations you don't even need to understand things like average velocity or change in velocity. You just have to be able to identify the right numbers and plug them in, which is an important task in itself but which doesn't involve understanding of the physical concepts behind the equations. **
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RESPONSE --> This is interesting, b/c seeing the relationships in the equations, definitely helped me understand the concept and vice versa. They support each other. If we had just been shown the equations and told to memorize, that would have been a different case, and I don't think they would hold much meaning. Anyone can memorize something, but actually applying it, using it in problems, switching around the variables, helps to see the actual relationship of the variables.
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17:54:04 How do we extrapolate our acceleration vs. ramp slope data to obtain an estimate of the acceleration of gravity?
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RESPONSE --> we plot our actual points of acceleration from our data, and then graph a line that represents the average of the acceleration points. We do not have this ""average"" go through any point directly, just as close to the points as possible. From this slope of the ""average"" of the points, we can extrapolate the acceleration of gravity.
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17:54:11 ** We can sketch a straight line as close as possible to our data points. Then we use the average slope of that graph; this average slope is the acceleration of gravity. **
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RESPONSE --> ok.
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17:55:54 How do the unavoidable timing errors due to the uncertainty in the computer timer affect our estimate of the acceleration of gravity?
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RESPONSE --> The timing errors greatly affect our estimate of the acceleration of gravity since it includes both an error in reaction time and also in anticipation. These systematic errors have an increasing effect on acceleration, as opposed to the linear effect we should be getting from the data.
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17:56:12 ** STUDENT ANSWER: This error causes the slope to increase at an increasing rate rather than form a linear line. INSTRUCTOR COMMENT: Good answer. A systematic error would do that. Even random, non-systematic errors affect the placement of points on the graph, and this tends to affect the slope of the straight line approximating the graph, and also to reduce the accuracy of this slope. **
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RESPONSE --> ok.
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17:57:27 How could the slight slope of the table on which the ramp rests, if not accounted for, affect our graph of acceleration vs. ramp slope but not our estimate of the acceleration of gravity?
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RESPONSE --> The slope of the table would cause our slope measuremement to be inaccurate, but it would be throughout all our data points. It may shift the slope either to the right or the left of the data points, but would do so consistently and be the same throughout, so all points would be affected by the same shift.
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17:57:45 ** GOOD STUDENT ANSWER: The only effect this systematic error has on the graph is to change the m coordinate of each point by an amount equal to the slope of the table, which is always the same.Since it is the graph slope that comprises our final result, a small table slope would have no effect on our conclusions. **
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RESPONSE --> Ok.
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18:00:26 How could anticipation of the instant at which a cart reaches the end of the ramp, but not of the instant at which it is released, affect our graph as well as our estimate of the acceleration of gravity?
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RESPONSE --> When the cart is released, since the timer has no control over this, the timer typically is reacting to the release. There is a delay in reaction time from when the timer witnesses the release, and then clicks on the computer timer. While the cart is rolling down the ramp, however, the timer can see it is approaching the end, and will therefore slightly anticipate when the cart will make contact with the end. In the first instant, we have a delay in time, from the actual, so we're starting ""later"". In the second case, we are hitting the timer too ""early"", so the total change in time is less than it should be, but since we know the distance covered, and the slope, those variables do not change, so we are getting a false acceleration.
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18:00:44 ** GOOD STUDENT ANSWER: The timer is started with a slight delay due to the reaction time of the person doing the timing. This would be OK if the individual's reaction time caused the individual to stop the timer with the same delay. However, the person doing the timing often anticipates the instant when the cart reaches the end of the ramp, so that the delay is not added onto the end time as it was to the starting time. The anticipating individual often triggers the timer slightly before the cart reaches the end, compounding the error even further and also causing the graph to curve rather than remain linear. **
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RESPONSE --> ok.
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Ħr{ assignment #007 y Ս Physics I Class Notes 02-11-2006
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18:49:40 Why do we say that the first equation of uniformly accelerated motion expresses the definition of average velocity, while the second expresses the definition of acceleration?
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RESPONSE --> The first equation, `ds=(v0+vf)/2*`dt expresses the definition of average velocity, since for unifom acceleration vAve = (v0+vf)/2. We can find the average velocity from the change in position over the change in time. The 2nd equation, vf=v0+a*`dt expresses acceleration, since it shows that acceleration is the change in velocity over the change in time (vf-v0)/`dt.
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18:50:18 ** When acceleration is uniform average velocity is the average of initial and final velocities, (vf + v0) / 2. Average velocity is `ds / `dt (whether accel is uniform or not). Setting `ds / `dt = (vf + v0) / 2 we obtain `ds = (vf + v0) / 2 * `dt, which is the first equation of uniformly accelerated motion. So the definition of average velocity is equivalent to the definition of average velocity. Average acceleration is aAve = `dv / `dt = (vf - v0) / `dt. Since for uniform acceleration the acceleration is constant, we can just say that in this case a = (vf - v0) / `dt. This equation is easily rearranged to give the second equation of motion, vf = v0 + a `dt. **
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RESPONSE --> OK.
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18:54:55 Why, for uniform acceleration, is vAve = (vf + v0) / 2, while this is not usually true for nonuniform acceleration?
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RESPONSE --> For uniform acceleration, velocity is accelerating at a uniform pace, the average acceleration will be the midpoint between its initial and final velocities since for every change in time, the object is accelerating at the same velocity.
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18:55:09 ** If acceleration is uniform then the v vs. t graph is linear, so that the average velocity over any time interval must be equal to the velocity at the midpoint of that interval. It follows that the average velocity must be midway between the initial and final velocities. (vf + v0) / 2 is the average of the initial and final velocities, and is therefore halfway between the v0 and vf. **
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RESPONSE --> ok.
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18:56:23 In commonsense terms, why does change of velocity over a given distance, with a given uniform acceleration, differ with initial velocity?
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RESPONSE --> If an object is accelerating, even if it is accelerating uniformly, it is still increasing its velocity as it travels over time. It is not standing still and so its velocity will increase as time passes, the change in the velocity, however, will be constant over the time intervals.
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18:57:03 ** If the uniform acceleration is the same in both cases, then assuming that both initial velocity and acceleration are positive, a greater initial velocity will result in a shorter time interval to cover the given distance. A shorter time interval leaves less time for velocity to change, resulting in a smaller change in velocity. **
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RESPONSE --> Ok. I did not go into that it would take less time to cover the distance as velocity increases.
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՜xa^{~ assignment #003 y Ս Physics I Vid Clips 02-11-2006
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19:12:23 Video Clip 7: Slope triangle, instantaneous acceleration, tangent line How do we represent the calculation of the rate which velocity changes on graph of velocity vs. time?
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RESPONSE --> We represent the calculation of the rate which velocity changes on a graph of velocity vs. time by using the slope triangles at the point we are interested in, and another point along the graph.
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19:12:31 ** Correct response from student: We can represent the rate of velocity change by calculating the slope. Slope is rise/run which is change in velocity/change in time **
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RESPONSE --> ok.
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19:16:22 If we know the velocity at every clock time, that how we find the precise rate at which velocity changes adding given clock time?
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RESPONSE --> At a given clock time, we can use the slope triangles to find the slope (rise/run) between the point we are interested in and a point further along the graph. From this slope triangle, we can calculate the tangent line to this triangle which will be equivalent to the precise rate at which velocity changes at the given clock time.
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19:17:14 ** STUDENT RESPONSE: I am not sure I understand, however I think you would make a triangle between the two times and velocities--i.e., between the two points on the v vs. t graph. Then find slope. INSTRUCTOR CRIQITUE: You would do this, but you would calculate the rate or slope over smaller and smaller intervals containing the clock time at the specified instant. You would try to determine the limiting value as the interval approaches 0. This limiting value is the instantaneous rate as well as the slope of the tangent line. **
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RESPONSE --> Ok, I feel I understand this in theory, but would have problems understanding this in actuality. An example of how we would determine this would be good.
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19:20:52 How do we depict the instantaneous rate of velocity change on a graph of velocity vs. time?
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RESPONSE --> I feel this was answered in the previous question. Once again, I am trying to visualize determining the slope of the triangle as we approach the given time, but I am unable to put my hands around it in terms of the tangent line of the slope. Please provide an example with an actual problem. That might help me. I feel there is just something I'm not understanding, but it's hard to know what exactly.
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19:20:57 ** See the preceding answer. **
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RESPONSE --> ok.
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19:26:18 Physics video clips 08: Finding displacement: define displacement, then calculate by common sense, by formula, by area What is the difference between displacement and distance?
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RESPONSE --> Displacement is the change in position of an object over time. Displacement, unlike distance, can be negative or positive since it is showing a change in position, and does not involve distance, which is only positive. If we know how long it took an object to make a change in position, and how fast the object was moving, we can then calculate how much change in position the object experienced. The formula for `ds = vAve/`dt. This formula also applies to the area on a graph between in the initial velocity, and final velocity, over the time interval. The area will resemble a quadrilateral, if velocity is constant, it will resemble a square or rectangle, if velocity changes it will resemble a trapezoid. The velocity would represent the heigt of the quadrilaterial, and the change in time represents its width, so the change in position would equal the area of this change, or height x width. As noted above, displacement is the change in position of an object, it can be positive or negative, where as distance is only a positive value.
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19:26:27 ** Displacement tells the change in position, which is the same as distance but displacement can have a negative or positive value whereas distance is always positive. **
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RESPONSE --> ok.
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19:29:33 How do we calculate displacement from velocity and time interval? What is the common sense of this calculation?
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RESPONSE --> We calculate displacement from the average velocity*change in time. SInce velocity changes over time, we multiply the change in time by the average velocity to get the change in position.
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19:29:44 ** Correct response from student: The vAve multiplied by the time interval. If the vAve tells us how far it travels one second, then if we multiply it by the time interval it will tell us how far it will travel in that amt. of time. **
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RESPONSE --> ok.
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19:30:47 How do we use a graph to help visualize the calculation of displacement when velocity is constant?
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RESPONSE --> When velocity is constant, we can use the graph to calculate the area over a time interval. The height is equal to the velocity, while the width is equal to the change in time. We find the height * width to find the area, which is the change in displacement.
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19:31:06 ** Since the vel. is constant, a graph of velocity vs. clock time over a given time interval will form a rectangle whose 'height' represents the velocity and whose width represents the time interval. Multiplying velocity * time interval gives displacement. A graph of position vs. clock time will be a straight-line graph, and the displacement corresponding to two clock times will be the change in the y coordinate between those clock times. **
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RESPONSE --> ok
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19:32:01 What aspect of a v vs. t graph tells us the displacement?
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RESPONSE --> In a v vs t graph, the area under the velocity points during the time interval tells us the displacement.
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19:32:09 ** The area of the rectangle formed between the two given clock times will indicate the displacement (see preceding comment). **
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RESPONSE --> ok."