course Phy 201
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RESPONSE --> The object is moving at an average rate of 3 meters/second. This conclusion was obtained by: 12 meters/ 4 seconds = 3 m/s confidence assessment: 3
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20:32:50 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 --> The commonsense image is very helpful and I should have also added that example into my answer. self critique assessment: 2
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20:35:18 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> A rate is a quanity measured with respect to another measured quantity. Therefore, the measured quantity of 12 meters with respect to 4 seconds produced the rate of 3 m/s. confidence assessment: 3
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20:37:12 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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20:38:48 `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. Time is constantly changing so the position of the object is dependent on where it is at a particular time. confidence assessment: 2
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20:42:27 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 --> Exactly!! self critique assessment: 3
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21:25:01 `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 think my explanation was pretty thorough. The only thing I didn't address directly is that clock time is independent of anything. I simply stated that position change is dependent on time. confidence assessment: 3
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21:25:26 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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21:36:31 `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 --> The average speed and the average velocity of the object is 2 m/s to the left. The result comes out to be -2 so we can say that is moving 2 m/s in the left direction. We can visualize this by thinking about a meter stick and when dividing the meters up into 3 second intervals, the object is moving 2 m/s for each second. confidence assessment: 2
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21:38:14 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 didn't realize that velocity could be positive or negative. I should have figured that out from the question but this information will be very helpful to know! self critique assessment: 2
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21:40:24 `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 stands for the average velocity confidence assessment: 2
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21:41:42 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 --> I completely misunderstood the question! Now that I realize what it was asking I completely understand the answer! self critique assessment: 3
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21:43:17 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> `ds is change in position and `dt is the time interval of an object. You can write it out in words this way or by writing delta s and delta t on the paper. Delta, in this case, means change in something. confidence assessment: 2
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21:43:36 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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21:46:58 `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 --> The object moves 50 meters. (5 meters/second) * (10 seconds) = 50 meters The problem is related to the concept of rate because rate is defined as the change in a quantity divided by change in another quantity which is dependent. In this case the distance the object is moving (5 m/s) is dependent on the time (10 seconds). From the rate the object is moving we can find the actual distance it is moving. confidence assessment: 3
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21:50:33 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 --> In this problem we multiplied the 2 quantities to obtain 50 meters. self critique assessment: 2
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21:52:45 `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 represents the change in position of the object. `ds = vAve * `dt confidence assessment: 3
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21:53:22 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 --> ok self critique assessment: 3
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21:57:46 `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 --> Rate is defined as the change in a quantity in relation to change in another quantity which is dependent. vAve, `dt, `ds all rely on one another to find the other. Therefore, a change in one is dependent on the change of another; the rate at which an object moves is dependent on the time interval of an object which helps us to solve average velocity. confidence assessment: 2
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21:58:08 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 --> ok self critique assessment: e
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21:59:59 `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 Thus, we get (`dt) (vAve) = `ds or `ds = (vAve) (`dt) confidence assessment: 3
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22:00:21 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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22:01:36 `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 --> This result simply shows that all are dependent on the other and as a flow diagram shows you can find the result of all three with relation to one another. confidence assessment: 2
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22:02:36 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 completely understand. self critique assessment: 3
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22:11:42 `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 Multiply both sides by `dt. (`dt) * (vAve) = `ds Then divide both sides by vAve. `dt = `ds * vAve We now see that the time interval equals change in position multiplied by average velocity. confidence assessment: 3
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22:12:03 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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22:13:47 `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 --> The result shows us that the time interval is equal to the change in position multiplied by the average velocity. All three can be moved around to find the result of the other. confidence assessment: 3
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22:22:06 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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