course Phy 121 6/28 8 PM 002. Velocity*********************************************
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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): Self-critique rating: ********************************************* Question: `q002. How is the preceding problem related to the concept of a rate? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The preceding problem is related to the concept of a rate because it is the change in one variable (distance) compared to another variable (time). Confidence Assessment:
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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): Self-critique rating: ********************************************* Question: `q003. Is object position dependent on time or is time dependent on object position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Object position is dependent on time. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Self-critique rating: ********************************************* 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 did not explain the concept of rate because distance and time are such fundamental concepts. I should have explained that a rate is defined as the average change of A with respect to B or change in A divided by change in B. Confidence Assessment: 3
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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): Self-critique rating: ********************************************* 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: -6 meters / 3 seconds = -2 meters / second Velocity = -2 meters/second Speed = 2 meters/second. I picture the object falling six feet, which means that its position is being reduced but it is moving so speed is positive. I dont think there is such thing as negative speed. Confidence Assessment: 3
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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) : Self-critique rating: ********************************************* 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 Assessment: 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): Self-critique rating: ********************************************* Question: `q007. How do you write the expressions `ds and `dt on your paper? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I do not know how to write these on paper. Confidence Assessment: 0
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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): `d is written using the Greek delta symbol (triangle). Using typewriter notation, it is written using either `d or Delta. Self-critique rating: 3 ********************************************* 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: vAve = `ds / `dt 5 meters/second = 10 seconds / `dt (5 meters/second) * `dt = 10 seconds `dt = 10 seconds / 5 meters/second = 2 meters This problem is related to the concept of rate because it deals with the change in one variable compared to the change in another. Confidence Assessment: 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. Given the rate at which quantity A changes with respect to quantity B, and the change in quantity B, we have The given rate, which is `dA / `dB The change in B, which is `dB We can therefore multiply `dA / `dB by `dB to get `dA / `dB * `dB = `dA. That is, you can get the change in the first quantity (the A quantity) if you know the average rate (`dA / `dB) and the change in the second quantity (the B quantity). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I do not understand this problem and why `dA and `dB come into play. Is my solution incorrect?
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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): Self-critique rating: ********************************************* 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: These quantities are related by the definition of a rate because vAve is the change in position divided by the change in time. A rate is a change in one variable divided by the change in another. Confidence Assessment: 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): Self-critique rating: ********************************************* 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: In order to solve the equation vAve = `ds / `dt for `ds, you multiply both sides by `dt to get vAve * `dt = `ds. Confidence Assessment: 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): Self-critique rating: ********************************************* 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: The preceding result is related to our intuition about the meanings of the terms average velocity, displacement and clock time because average velocity can be thought of as how far an object travels in a given amount of time. If we are multiplying the velocity by time, we come up with the distance that is traveled. Confidence Assessment: 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): Self-critique rating: ********************************************* 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: vAve = `ds / `dt vAve * `dt = `ds `dt = `ds / vAve Confidence Assessment: 3
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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): 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: When planning a trip, we typically know the distance that needs to be traveled. We also know that, on the highway, the average speed is going to be around 60 miles per hour. We can therefore divide the number of miles by 60 miles per hour in order to calculate how long it will take us to arrive at our destination. Confidence Assessment: 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): Self-critique rating: You should submit the above questions, along with your answers and self-critiques. You may then begin work on the Questions, Problem and Exercises, as instructed below. "