course Phy 121 õ£«ÈÛû·ŽU†äµoÑ_üÒ®ä³ùÎþassignment #002
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20:41:56 `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 so the object is moving at 3m/s. confidence assessment: 3
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20:42:16 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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20:45:22 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> The definition of a rate is a certain quantity or amount of one thing considered in relation to a unit of another thing and used as a standard or measure. So the distance moved 3m is considered in relation to the time it took it to move that distance 1 second to give us a measure of velocity. confidence assessment: 2
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20:48:44 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 --> So to be a rate the two quantities have to be divided and the second would be dependant(in this case seconds). self critique assessment: 2
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20:50:19 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> Time is dependant on object position. confidence assessment: 2
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20:52:23 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 --> So object position is dependent on time because it is independent of object position and any other variables. self critique assessment: 2
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20:55:39 `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 --> Position of the object is dependant on time because the clock is always running. confidence assessment: 2
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20:56:29 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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21:02:44 `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 --> 6m/3s=2 so its average velocity is 2m/s. Average velocity is distance covered divided by time I don't know what commonsense images and ideas are needed to find it? confidence assessment: 2
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21:04:12 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 --> My answer should have been -2m/s because velocity the average rate at which position changes so it can be positive or negative. self critique assessment: 2
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21:05:40 `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 --> 'va confidence assessment: 0
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21:06:59 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 --> vAve is the figure that stands for average velocity. Total equation is vAve='ds/'dt. self critique assessment: 2
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21:08:30 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> Like they appear in the question. confidence assessment: 0
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21:09:32 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 --> 'd stands for the greek symbol delta. self critique assessment: 2
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21:15:04 `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 --> 5m/s*10s=50 so the object moved 50 meters. Not sure how to explain the second half of the question. confidence assessment: 0
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21:22:14 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 --> The only way I can think to put what you are asking for in words is if you were to visualize the distance 5 meters then multiply that by 10 so you would have 10 segments of 5 meters with each having a time value of 1 second. self critique assessment: 2
Remember that average velocity is an average rate of change of position with respect to clock time and
average acceleration is an average rate of change of velocity with respect to clock time. &#
Remember that average velocity is an average rate of change of position with respect to clock time and
average acceleration is an average rate of change of velocity with respect to clock time.
The example below defines the term 'average rate', connects the definition to the slope of a graph, and applies that definition to a situation involving temperature and clock time. &#
ave rate = (change in A) / (change in B).
From this it follows that
change in A = ave rate * change in B
and also that
change in B = change in A / ave rate
where 'ave rate' here is an abbreviation for the specific term 'average rate of change of A with respect to B.
On a graph of A vs. B, the quantity A is represented relative to the vertical axis and the quantity B relative to the horizontal. Between two graph points, therefore, the change in A is the 'rise' from one point to the other, and the change in B is the 'run' from the same first point to the second. The average rate of change of A with respect to B is then
ave rate = (change in A) / (change in B) = rise / run.
Since rise / run between two points is the slope of the straight line segment between those points, we can identify an average rate of change of A with respect to B as the slope between two points on the graph of A vs. B.
Since slope = rise / run, we see that
rise = slope * run and
run = rise / slope
Interpreting slope as ave rate, rise as change in A and run as change in B, this again tells us that change in A = ave rate * change in B, and change in B = change in A / ave rate. &#
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21:24:17 `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 is the distance traveled in the time given. confidence assessment: 1
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21:28:45 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 --> The expression for the change of `ds `ds = vAve * `dt. self critique assessment: 2
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21:33:40 `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 obtained by dividing the change in a quantity by the change in another quantity on which is dependent and is used as a standard or measure. So 'dt and 'ds are the quantitys and 'ds is dependant on 'dt. When 'ds is divided by 'dt we get vAve which is the measurment in this case average velocity. confidence assessment: 2
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21:34:14 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 --> self critique assessment: 3
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21:45:35 `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 --> Multiply both sides by 'dt so now you have canceled out 'dt from the right side and moved it from the left side so the equation looks like this 'dt*vAve='ds. Now 'ds='dt*vAve. confidence assessment: 1
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21:46:29 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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21:48:37 `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 --> ? confidence assessment: 0
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21:51:32 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 --> Because we know when time is multiplied by average velocity we get distance displaced. self critique assessment: 2
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21:52:18 `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 both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt. confidence assessment: 3
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21:55:22 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 didn't pay close attention and thought this question was a repeat but I understand this 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. is how to solve for 'ds. self critique assessment: 2
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21:59:05 `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 --> Because we know that distance traveled divided by average velocity gives us the time it took to travel that distance. confidence assessment: 1
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21:59:29 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 --> self critique assessment: 3
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