course Phy 231 ~xZassignment #002
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00:03:16 `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 --> 12m / 4sec = 3 m/sec. For each second that time elapses, on average the object moves 3 meters. confidence assessment: 3
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00:04:08 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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00:04:44 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> We found the objects velocity, and velocity is the rate of change of the object's position. confidence assessment: 3
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00:04:54 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: 3
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00:05:27 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> The objects position is dependent on time. In many equations we have y(t) and in this form, we see that when t changes, the function y changes. Not the other way around. confidence assessment: 3
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00:05:34 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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00:06:09 `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 --> rate = change in position/ change in time. velocity = rate of change in objects position confidence assessment: 3
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00:06:18 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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00:13:10 `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 --> Average velocity = displacement/time = 6m/3sec = 2 m/sec. For each second, on average the object displaces 2 meters. I can't calculate average speed in this problem because the displacement was 6 but that might not have been the distance traveled. Maybe the object went 9 meters and then back 3 meters. In that case the distance traveled is 12 meters with a displacement of 6 meters. There is not enogh info. confidence assessment: 3
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00:13:28 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:
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00:15:06 `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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00:15:26 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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00:16:48 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> Well you could just write them ds and dt but usually I write them delta s and delta t. The symbol for delta is a little triangle. confidence assessment: 3
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00:17:31 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: 3
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12:18:16 `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 --> I don't know how far it moves. It could have gone forward and then back and in that case the object would move 0 meters. There's not enough info. confidence assessment: 2
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12:20:23 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 --> This problem is related to rate in that we are given the rate of change with respect to time and the time interval and we are expected to find the position of change. self critique assessment: 3
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12:22:01 `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 = vAve * dt confidence assessment: 3
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12:22:50 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 --> We can make sure this equation is correct by matching up the units and see if we obtain the wanted units. self critique assessment: 3
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13:07:36 `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 --> A rate is a certain quantity with respect to another quantity. vAve is the rate of change in position ds with respect to change in time dt. The formula is: vAve = ds/dt, vAve = 6m/3sec = 2m/sec confidence assessment: 3
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13:08:20 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 --> vAve is the average rate at which position changes. self critique assessment: 3
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13:10:45 `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 --> When we have vAve = ds/dt, we multiply both sides by dt to obtain ds = vAve * dt. confidence assessment: 3
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13:11:02 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 --> OK self critique assessment: 3
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13:31:50 `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 --> displacement(change in objects position) = vAve(rate of change of objects position) * clock time(change in clock time). I am sorry I'm not sure what this question is asking for. confidence assessment: 1
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13:34:53 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 --> If we are talking about m/sec of a runner who is in a race. We can multiply the average velocity of the runner in m/s by the number of seconds it took the runner to run the race, we can find the distance of the race in meters. self critique assessment: 2
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13:36:42 `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, we can multiply both sides by dt to obtain vAve * dt = ds, and then divide both sides by vAve to obtain dt = ds/vAve. confidence assessment: 3
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13:36:52 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: 3
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13:38:25 `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 --> Using the runner example again, if we know the length of the race and the average velocity in which the runner ran the race, we can divide length/velocity to obtain how long the runner took to run the race. confidence assessment: 3
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13:38:39 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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