course phy 231 b{عݡ|Åassignment #002
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19:03:57 `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 --> average rate would be 3. distance/ time gives you the rate confidence assessment: 3
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19:07:39 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> it represents the relationship between distance and time confidence assessment: 3
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19:19:13 `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: 2
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19:24:15 `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 --> whether the object is moving or not, time would still continue and is not dependent on position confidence assessment: 3
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19:29:28 `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 --> -2 would be the average speed. distance over time. average velocity would be the same. confidence assessment: 2
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19:30:37 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 --> didnt know that in determining average speed, distance cannot be negative. self critique assessment: 2
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21:15:21 `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 --> dv confidence assessment: 3
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21:16:36 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 --> misunderstood the question. i thought it was asking what average velocity would be in the formula self critique assessment: 2
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21:18:20 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> v=ds/dt confidence assessment: 3
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21:19:56 `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 --> again it shows the relationship between distance and time. 5/10 =2 which is the average rate confidence assessment: 3
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21:24:53 `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=dt/ds confidence assessment: 2
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21:25:34 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 --> was unsure with this one self critique assessment: 1
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21:28:21 `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 between 2 times. the distance of an object and the elapsed time. confidence assessment: 2
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21:37:25 `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*dt=ds confidence assessment: 2
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21:41:16 `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 average velocity moves through the displacement and then through the clocked time. confidence assessment: 2
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21:42:38 `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 --> ds/vAve=dt confidence assessment: 2
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21:43:28 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 need to remember to write out the steps in any problem . not just the answer self critique assessment: 2
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21:47:45 `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 --> displacement of an object through the velocity of the object gives the clocked time of the object confidence assessment: 2
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