course PHY 201 9/10 10:45 pmvvvv
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Given Solution: 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 span 3 meters, corresponding to the distance moved in 1 second, on the average. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: It is related to the concept of rate because velocity is the change in distance divided by the change in clock time. If an object moved 30 meters in 10 seconds, we divide the 30 by the 10 and get the rate of 3m/s. confidence rating: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. More specifically The rate of change of A with respect to B is defined to be the quantity (change in A) / (change in B). An object which moves 12 meters in 3 seconds changes its position by 12 meter during a change in clock time of 3 seconds. So the question implies Change in position = 12 meters Change in clock time = 3 seconds When we divide the 12 meters by the 3 seconds we are therefore dividing (change in position) by (change in clock time). In terms of the definition of rate of change: the change in position is the change in A, so position is the A quantity. the change in clock time is the change in B, so clock time is the B quantity. So (12 meters) / (3 seconds) is (change in position) / (change in clock time) which is the same as average rate of change of position with respect to clock time. Thus average velocity is average rate of change of position with respect to clock time. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My answer wasnt as large but I feel it got the point across. ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q003. Is object position dependent on time or is time dependent on object position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The object position is dependent on time because the object can only move so far in a set amount of time. confidence rating: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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 (this might not be so at the most fundamental level, but for the moment, unless you have good reason to do otherwise, this should be your convention). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not believe I missed any concepts. confidence rating: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Be sure you have reviewed all the definitions and concepts associated with velocity. If theres anything you dont understand, be sure to address it in your self-critique. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The object that is displaced -6m in 3 seconds cannot have a speed. Speed is the measurement of distance with respect to time. Since distance cannot be negative, the speed of this cannot be computed. Velocity on the other hand, can be negative, because velocity can go in either direction. confidence rating: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Speed is the average rate at which distance changes with respect to clock time. Distance cannot be negative and the clock runs forward. Therefore speed cannot be negative. Velocity is the average rate at which position changes with respect to clock time, and since position changes can be positive or negative, so can velocity. In general distance has no direction, while velocity does have direction. Putting it loosely, position is just how fast something is moving; velocity is how fast and in what direction. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: vAve= `ds/`dt confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Average velocity is rate of change of position with respect to clock time. Change in position is `ds and change in clock time is `dt, so average velocity is expressed in symbols as vAve = `ds / `dt. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I use the symbol delta, which is a triangle. This stands for the change of a quantity. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You use the Greek capital Delta when writing on paper or when communicating outside the context of this course; this is the symbol that looks like a triangle. See Introductory Problem Set 1. `d is used for typewritten communication because the symbol for Delta is not interpreted correctly by some Internet forms and text editors. You should get in the habit of thinking and writing the Delta symbol when you see `d. You may use either `d or Delta when submitting work and answering questions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The object moved 50m. This problem is related to the concept of rate, because we start out with a rate and want to see the distance traveled in a specific time interval. The formula would be Avg rate = `ds/ `dt. 5m/s = `ds/10s `ds = 5m/s * 10s `ds = 50 m confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. The definition of rate of change states that the rate of change of A with respect to B is (change in A) / (change in B), which we abbreviate as `dA / `dB. `dA stands for the change in the A quantity and `dB for the change in the B quantity. For the present problem we are given the rate at which position changes with respect to clock time. The definition of rate of change is stated in terms of the rate of change of A with respect to B. So we identify the position as the A quantity, clock time as the B quantity. The basic relationship ave rate = `dA / `dB can be algebraically rearranged in either of two ways: `dA = ave rate * `dB or `dB = `dA / (ave rate) Using position for A and clock time for B the above relationships are ave rate of change of position with respect to clock time = change in position / change in clock time change in position = ave rate * change in clock time change in clock time = change in position / ave rate. In the present situation we are given the average rate of change of position with respect to clock time, which is 5 meters / second, and the change in clock time, which is 10 seconds. Thus we find change in position = ave rate * change in clock time = 5 cm/sec * 10 sec = 50 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q009. If vAve stands for the rate at which the position of the object changes with respect to clock time (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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: we start with the formula vAve= `ds/ `dt Now just set the equation to `ds on one side `ds = vAve*`dt confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: To find the change in a quantity we multiply the rate by the time interval during which the change occurs. The velocity is the rate, so we 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: We know what it means to multiply pay rate by time interval (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour). When we multiply vAve, for example in in units of cm / sec or meters / sec, by `dt in seconds, we get displacement in cm or meters. Similar reasoning applies if we use different measures of distance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The definition of rate is Rate= change in quantity a with respect to quantity b. So to apply this to velocity we get vAvg = change in position with respect to time. vAvg= `ds/ `dt confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve for `ds we multiply both sides by `dt which gives us `ds = vAve* `dt confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q012. How is the preceding result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: When I think of velocity I think of the speedometer in my car. To get the distance traveled in a certain time, we take the vAvg and multiply it by the time. This gives us the displacement of the object. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For most of us 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok ********************************************* Question: `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To solve the `dt we will first multiply both sides by `dt. This will give us `dt * vAve = `ds. Next we will divide both sides by vAve to give us `dt = `ds/vAve confidence rating: ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ********************************************* Question: `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If we wanted to figure out how long it took for an object to travel a certain distance we divide the distance by the avg velocity. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 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. This is equivalent to the calculation `dt = `ds / vAve. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval. When dealing with displacement, velocity and time interval, we can always check our thinking by making the analogy with a simple example involving miles, hours and miles/hour. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating: ok
course MTh 173 9/10 11:48 pm 002. vvvv
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Given Solution: `aThe three points on the graph are (3, 5000), (7, 5300) and (12, 5500). The rise between the first point and the second is from 5000 to 5300, or 300, and the run is from 3 to 7, or 4, so the slope is 300 / 4 = 75. Note that the 300 represents $300 and the 4 represents 4 months, so the slope represents $300 / (4 months) = $75 / month, which is the average rate of change during the first time interval. The rise between the second point and the third is from 5300 to 5500, or 200, and the run from 7 to 12 is 5, so the slope is 200 / 5 = 40. This slope represents the $40/month average rate of change during the second time interval. Click on 'Next Picture' to see the graph. `routine graph2 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q002. Look at your results for the slopes, and look the results for the average rates of change. What do you notice? In what way then does the graph represent the average rate of change? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I noticed that the slope is the avg rate of change. The slope between the first and second point was 75. This is actually the rate of change. Rate of change is defined as the change in quantity a / change in quantity b. This is what slope is. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWe see from this example that the slope of a graph of value vs. clock time represents the rate at which value is changing with respect to clock time. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok ********************************************* Question: `q003. To what extent do you think your graph, consisting of 3 points with straight line segments between them, accurately depicts the detailed behavior of the stocks over the 9-month period? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Our graph is very vague compared to what one would look like with weekly or day to day points. Stocks fluctuate greatly and just using 3 points for a 9 month period is definitely not specific enough. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStocks can do just about anything from day to day-they can go up or down more in a single day than their net change in a month or even a year. So based on the values several months apart we can't say anything about what happens from day to day or even from month to month. We can only say that on the average, from one time to another, the stocks changed at a certain rate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q004. From the given information, do you think you can accurately infer the detailed behavior of the stock values over the nine-month period? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The only thing we could accurately infer is the trend between March to July and July to December. Anything more specific than this we cannot accurately say for sure. confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aNot on a day-to-day basis, and not even on a month-to-month basis. All we can see from the given information is what might be an average trend. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: ok