course phy 201 ???wh??|??c??Z?assignment #003003. Velocity Relationships
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15:16:25 `questionNumber 30000 `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 seconds confidence assessment: 2
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15:17:12 `questionNumber 30000 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 --> ok self critique assessment: 2
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15:18:01 `questionNumber 30000 `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 --> cm confidence assessment: 2
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15:20:08 `questionNumber 30000 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> sec|cm ------------- |sec = cm confidence assessment: 2
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15:20:25 `questionNumber 30000 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 --> ok self critique assessment: 3
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15:22:35 `questionNumber 30000 `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 confidence assessment: 3
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15:22:57 `questionNumber 30000 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 --> ok self critique assessment: 3
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15:24:27 `questionNumber 30000 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> |km ---------- km|sec = sec^-1 confidence assessment: 3
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15:24:51 `questionNumber 30000 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 --> ok self critique assessment: 1
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15:26:43 `questionNumber 30000 `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 --> 6 m='ds 3 sec='dt = 2 m/sec =Vavg confidence assessment: 3
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15:27:03 `questionNumber 30000 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 --> none self critique assessment: 3
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15:28:08 `questionNumber 30000 `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 --> 'ds=(s2-s1) ------------------- 'dt=(t2-t1) confidence assessment: 3
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15:28:55 `questionNumber 30000 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 --> 'ds ------- 'dt self critique assessment: 3
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15:37:16 `questionNumber 30000 `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 --> rise= the change in position run= the change in time confidence assessment: 3
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15:37:33 `questionNumber 30000 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 --> ok self critique assessment: 3
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15:39:11 `questionNumber 30000 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> slope=45^(1/2)vand represents the change in position with respect to time confidence assessment: 3
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15:39:33 `questionNumber 30000 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> ok self critique assessment: 0
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15:41:51 `questionNumber 30000 `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 --> in order to have a steeper slope, you need more distance covered per second, and second don't change therefore the only thing that can chage is the amount of distance covered in a sec. confidence assessment: 3
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15:42:04 `questionNumber 30000 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 --> ok self critique assessment: 3
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15:47:03 `questionNumber 30000 `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 --> it's a graph from lower left to upper right, and it incresing at a decreasing rate the slope is increasing the slope is steeper at first, then get less steep as time passes confidence assessment: 2
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15:47:43 `questionNumber 30000 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 --> ok, but is the a real or idea hill self critique assessment: 2
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?k???I?????????g?~? assignment #002 002. Velocity Physics I 01-28-2007
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16:51:53 `questionNumber 20000 `q001. Note that there are 14 questions in this assignment. If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.
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RESPONSE --> 3 meters per sec change in distance over change in time confidence assessment: 3
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16:52:11 `questionNumber 20000 Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will have 3 meters, corresponding to the distance moved in 1 second, on the average.
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RESPONSE --> ok self critique assessment: 3
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16:53:10 `questionNumber 20000 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> a rate is a comparasion of the changing quality confidence assessment: 1
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16:53:23 `questionNumber 20000 A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
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RESPONSE --> ok self critique assessment: 1
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16:54:19 `questionNumber 20000 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> position is dependent on time, even if the the object didn't move time will change confidence assessment: 3
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16:54:33 `questionNumber 20000 Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else.
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RESPONSE --> ok self critique assessment: 3
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16:54:55 `questionNumber 20000 `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
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RESPONSE --> none confidence assessment: 3
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16:55:05 `questionNumber 20000 You should always self-critique your work in this manner. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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RESPONSE --> ok self critique assessment: 3
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16:57:54 `questionNumber 20000 `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
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RESPONSE --> 6 meters in three second = 2 meters per sec the negative describes position from a starting place you can't go negative meters per sec confidence assessment: 3
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16:58:10 `questionNumber 20000 Speed is the average rate at which distance changes, and distance cannot be negative. Therefore speed cannot be negative. Velocity is the average rate at which position changes, and position changes can be positive or negative.
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RESPONSE --> ok self critique assessment: 3
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16:59:11 `questionNumber 20000 `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
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RESPONSE --> vAve= 'ds/'dt confidence assessment: 3
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16:59:21 `questionNumber 20000 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok self critique assessment: 3
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16:59:49 `questionNumber 20000 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> 'ds ------ 'dt confidence assessment: 3
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17:00:05 `questionNumber 20000 You use the Greek capital Delta symbol Delta. `d is often used here because the symbol for Delta is not interpreted correctly by some Internet browsers. You should get in the habit of thinking and writing Delta when you see `d. You may use either `d or Delta when submitting work and answering questions.
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RESPONSE --> ok self critique assessment: 2
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17:02:20 `questionNumber 20000 `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate?
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RESPONSE --> 5 m|10 sec ------------------- = 50 m sec by having the rate and the time you can figure out the distance confidence assessment: 3
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17:03:10 `questionNumber 20000 In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which one quantity changes with respect to another, and the change in the second quantity, how do we obtain the resulting change in the first?
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RESPONSE --> multiply the rate by the time elapsed self critique assessment: 3
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17:17:48 `questionNumber 20000 `q009. If vAve stands for the rate at which the position of the object changes (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?
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RESPONSE --> vAve * delta t = delta s confidence assessment: 3
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17:18:00 `questionNumber 20000 To find the change in a quantity we multiply the rate by the time interval during which the change occurs. We therefore obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: Just as when we multiply pay rate by time (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour), when we multiply vAve, in cm / sec or meters / sec or whatever, by `dt in seconds, we get displacement in cm or meters, or whatever, depending on the units of distance used.
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RESPONSE --> ok self critique assessment: 3
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17:18:49 `questionNumber 20000 `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem problem.
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RESPONSE --> vAve= (delta s/delta t) confidence assessment: 3
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17:19:19 `questionNumber 20000 vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok self critique assessment: 3
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17:22:01 `questionNumber 20000 To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt.
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RESPONSE --> vAve=(delta s/delta t) mutliply both sides by delta t self critique assessment: 3
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17:23:50 `questionNumber 20000 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> if you know how much time past and distance travelled you know the rate, if you know the rate and one other, distance or time, you can figure out the missing one confidence assessment: 2
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17:24:03 `questionNumber 20000 Our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
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RESPONSE --> ok self critique assessment: 1
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17:27:36 `questionNumber 20000 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
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RESPONSE --> divide both side by delta s invert both side and you get (delta s/vAve) = delta t confidence assessment: 3
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17:28:19 `questionNumber 20000 To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.
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RESPONSE --> ok self critique assessment: 2
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17:29:01 `questionNumber 20000 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> there are related, if you have two of the three you can figure out the other one confidence assessment: 3
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17:29:15 `questionNumber 20000 If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
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RESPONSE --> ok self critique assessment: 3
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