course Phy 201 Íñ•¥ˆÝ—˜d‹Ê×¾Ô¤èšÓã§Ëassignment #002
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15:14:14 `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 --> d=rt 12=r4 r=3 m/sec If an car moves from one stop light to another, which is 12m away, and does this in 4 seconds, then the rate that the car moves is 3 m/sec confidence assessment: 3
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15:14:35 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 --> This is more commonsense image, but i understand the concept self critique assessment: 3
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15:17:06 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> rate is similar to the average change in speed. It is the change in distance over the change in time. Which is just how we solved this problem confidence assessment: 3
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15:17:13 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 --> self critique assessment:
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15:17:54 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> Object position is dependent on time confidence assessment: 3
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15:18:03 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 --> i understand self critique assessment: 3
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15:18:39 `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 --> I don't think i missed any confidence assessment: 3
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15:18:54 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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15:20:48 `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 --> I'm not sure if this question is worded wrong. But the math is still the same. d=rt 6=r3 r=2m/sec Divide that 6 meters into three sections for each second. For ever two meters it takes one sec to travel confidence assessment: 3
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15:21:35 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 --> I was not thinking in positive or negative. I thought displacement was also an absolute number. Speed cannot be negative. Velocity can be. self critique assessment: 2
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15:21:59 `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 --> 'dr confidence assessment: 3
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15:22:21 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 --> 'dr='ds/'dt self critique assessment: 3
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15:23:15 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> These are derivatives. I am not sure how you would write these differently, or what it is asking confidence assessment: 0
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15:24:35 `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 --> 'd=50 meters Rate is the change in one value over the change in another. In this case we are given a rate and multiply it by time confidence assessment: 3
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15:26:20 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 --> You should add or subtract based on the change in the second to get the first change self critique assessment: 2
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15:27:11 `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 --> 'ds=('ds/'dt)('dt) confidence assessment: 3
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15:27:27 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 --> I understand this method. self critique assessment: 3
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15:28:43 `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 --> All these values are changes from an intial and a final value. Thus using these changes, we can divide or multiply to find the change in velocity, time or displacement confidence assessment: 3
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15:28:56 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 --> Yes self critique assessment: 3
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15:29:50 `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
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RESPONSE --> vAve='ds/'dt Multiply both sides by 'dt. 'ds=vAve x 'dt confidence assessment: 3
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15:30:08 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 --> this is a bit more detailed. But exactly the same method self critique assessment: 3
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15:32:45 `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 --> the result shows that by dividing change by change one can derive velocity. Also give two changes, one can solve for any value confidence assessment: 3
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15:33:03 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: 3
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15:34:33 `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 --> vAve = `ds / `dt Multiply by 'dt on both sides you now have 'dt x vAve='ds Now divide both sides both sides by vAve you now have 'dt='ds/vAve confidence assessment: 3
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15:34:40 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 --> yes self critique assessment: 3
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15:35:18 `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 --> Again, give any two intervals or changes one can solve for anything. Whether it be displacement, velocity, and or time confidence assessment: 3
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15:35:23 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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