course Phy 201 مeեHassignment #002
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20:26:39 `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 --> 12/4 = 3 meters/second If a street is, hypothetically, 12 meters long and it takes 4 seconds to travel down it, one can commonsensi""cly"" figure out that for every second one will travel 3 meters. confidence assessment: 3
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20:26:49 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 --> self critique assessment: 3
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20:27:35 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> Rate is the change in one quantity/change in another quantity. Velocity is merely a more specific ""rate"" formula, since it incorporates the idea of distance changes divided by time changes. confidence assessment: 3
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20:27:39 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: 3
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20:30:54 `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. My rationale: x is normally the independent variable, and in this case x is time; so, y is dependent on x, meaning distance (or position) is dependent on time. confidence assessment: 3
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20:31:03 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 --> self critique assessment: 3
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20:31:28 `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 really didn't miss any concepts. confidence assessment: 3
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20:31:39 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 --> I understand. self critique assessment: 3
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20:34:07 `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 --> avg. velocity=change in distance/change in time =-6/3 =--2 meters/second If the idea here is a position change in relation to time, it is obvious that division is needed; hence, one can furthermore conclude that -6 units change in 3 (anything time measurement) will be -2; then, fill in the units (meters/second) and that is the answer. confidence assessment: 3
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20:34:57 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 understand the difference between speed and velocity; velocity is sensitive to direction, whereas speed is not (hence the reason velocity can be negative--a direction--whereas general speed cannot). self critique assessment: 2
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20:37:10 `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 --> 'ds/'dt=vAve confidence assessment: 3
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20:37:15 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 --> self critique assessment: 3
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20:38:05 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I don't necessarily understand the question. However, I normally use the triangle symbol for change, then follow the symbol by p/t (position/time). confidence assessment: 3
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20:39:42 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 --> Okay, that is how I write it--I just didn't know what it was called! Now I understand the symbolism 'd on the computer too. self critique assessment: 3
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20:43:19 `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 --> 50 meters Rate contains the idea of a change in one quantity and its relationship with another. We can use the relationship of position versus time to figure out the answer to this equation by multiplying the final average rate by the time. confidence assessment: 3
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20:46:37 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 --> We multiply the final average (5 meters per second) by 10 seconds to obtain the final result of 50 total meters. self critique assessment: 3
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20:47:45 `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 x 'dt confidence assessment: 3
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20:47:55 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. self critique assessment: 3
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20:50:29 `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 --> Average velocity is, in effect, a rate. It is how much of something happens, is involved with, etc. with regard to another quantity. In terms of velocity, these ""how muches"" per ""how muches"" involve distance and time. The formulas for rate and velocity alike can be used in numerous ways to find any of three variables, provided two parts of each equation is provided. In the current problem, the final value of velocity and time are provided. One can use the velocity formula ""backwards"" to find the distance involved with the equation. confidence assessment: 3
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20:50:38 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 --> I understand. self critique assessment: 3
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20:51:48 `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 vAve('dt)='ds The result is multiplication of final velocity by time. confidence assessment: 3
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20:51:55 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 --> self critique assessment: 3
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20:52:40 `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 --> When given any two of the three variables, one can find the third by ""maneuvering"" the formula and applying the information accordingly. confidence assessment: 3
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20:52:52 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 --> I understand. self critique assessment: 3
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20:53:48 `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 vAve('dt)='ds 'dt='ds/vAve Time change equals position change divided by velocity. confidence assessment: 3
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20:53:58 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 --> I understand. self critique assessment: 3
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20:56:32 `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 --> They are all related and can be used in conjunction to find any one of the three missing pieces of information (i.e., one could use displacement and clock time to find velocity, etc.). confidence assessment: 3
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20:56:38 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 --> I understand. self critique assessment: 3
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course Phy 201 مeեHassignment #002
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20:26:39 `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 --> 12/4 = 3 meters/second If a street is, hypothetically, 12 meters long and it takes 4 seconds to travel down it, one can commonsensi""cly"" figure out that for every second one will travel 3 meters. confidence assessment: 3
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20:26:49 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 --> self critique assessment: 3
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20:27:35 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> Rate is the change in one quantity/change in another quantity. Velocity is merely a more specific ""rate"" formula, since it incorporates the idea of distance changes divided by time changes. confidence assessment: 3
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20:27:39 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: 3
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20:30:54 `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. My rationale: x is normally the independent variable, and in this case x is time; so, y is dependent on x, meaning distance (or position) is dependent on time. confidence assessment: 3
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20:31:03 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 --> self critique assessment: 3
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20:31:28 `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 really didn't miss any concepts. confidence assessment: 3
.................................................
......!!!!!!!!...................................
20:31:39 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 --> I understand. self critique assessment: 3
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20:34:07 `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 --> avg. velocity=change in distance/change in time =-6/3 =--2 meters/second If the idea here is a position change in relation to time, it is obvious that division is needed; hence, one can furthermore conclude that -6 units change in 3 (anything time measurement) will be -2; then, fill in the units (meters/second) and that is the answer. confidence assessment: 3
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20:34:57 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 understand the difference between speed and velocity; velocity is sensitive to direction, whereas speed is not (hence the reason velocity can be negative--a direction--whereas general speed cannot). self critique assessment: 2
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20:37:10 `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 --> 'ds/'dt=vAve confidence assessment: 3
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20:37:15 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 --> self critique assessment: 3
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20:38:05 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I don't necessarily understand the question. However, I normally use the triangle symbol for change, then follow the symbol by p/t (position/time). confidence assessment: 3
.................................................
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20:39:42 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.
......!!!!!!!!...................................
RESPONSE --> Okay, that is how I write it--I just didn't know what it was called! Now I understand the symbolism 'd on the computer too. self critique assessment: 3
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20:43:19 `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 --> 50 meters Rate contains the idea of a change in one quantity and its relationship with another. We can use the relationship of position versus time to figure out the answer to this equation by multiplying the final average rate by the time. confidence assessment: 3
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20:46:37 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 --> We multiply the final average (5 meters per second) by 10 seconds to obtain the final result of 50 total meters. self critique assessment: 3
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20:47:45 `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 x 'dt confidence assessment: 3
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20:47:55 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. self critique assessment: 3
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20:50:29 `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.
......!!!!!!!!...................................
RESPONSE --> Average velocity is, in effect, a rate. It is how much of something happens, is involved with, etc. with regard to another quantity. In terms of velocity, these ""how muches"" per ""how muches"" involve distance and time. The formulas for rate and velocity alike can be used in numerous ways to find any of three variables, provided two parts of each equation is provided. In the current problem, the final value of velocity and time are provided. One can use the velocity formula ""backwards"" to find the distance involved with the equation. confidence assessment: 3
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20:50:38 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 --> I understand. self critique assessment: 3
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20:51:48 `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 vAve('dt)='ds The result is multiplication of final velocity by time. confidence assessment: 3
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20:51:55 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 --> self critique assessment: 3
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20:52:40 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
......!!!!!!!!...................................
RESPONSE --> When given any two of the three variables, one can find the third by ""maneuvering"" the formula and applying the information accordingly. confidence assessment: 3
.................................................
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20:52:52 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 --> I understand. self critique assessment: 3
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20:53:48 `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 vAve('dt)='ds 'dt='ds/vAve Time change equals position change divided by velocity. confidence assessment: 3
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20:53:58 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 --> I understand. self critique assessment: 3
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20:56:32 `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 --> They are all related and can be used in conjunction to find any one of the three missing pieces of information (i.e., one could use displacement and clock time to find velocity, etc.). confidence assessment: 3
.................................................
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20:56:38 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 --> I understand. self critique assessment: 3
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