course Phy 121 £«ñΫ“™ü„R͉µÃ…w˶Ɇœ´×Ó¢¥Õassignment #005
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23:48:38 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
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RESPONSE --> What I know: v0, accel., 'dt What I need to find: vf and 'ds aAve = 'dv / 'dt OR aAve = (v2 - v1) / (t2 - t1) Since I know aAve and 'dt use a version of this formula to find vf (or as above, v2). vAve = 'ds / 'dt Use the above stuff to find vf I can plug into vAve = (v0 + vf) /2 to find vAve. Once I know vAve, I already know 'dt so I can plug into vAve = 'ds / 'dt and solve for 'ds. confidence assessment: 1
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23:48:49 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
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RESPONSE --> I think I was right. self critique assessment: 3
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23:54:29 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> Okay, knowing v0 and vf you can find 'dv (vf - v0). Also know initial and final velocities you can find vAve (vAve = (vf + v0) / 2). Because you also know 'dt and you now know 'dv you can find acceleration (a = 'dv / 'dt). Because you know average velocity and 'dt you can find 'ds (vAve * 'dt). confidence assessment: 1
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23:54:47 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **
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RESPONSE --> My answer was a least partially correct. self critique assessment: 3
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00:09:55 Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.
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RESPONSE --> What I know: v0 = 0 km / hr vf = 10 km / hr vAve = 5 km / hr What I need to find: 'dt 'ds = 3937.0 km (according to Google) My estimate is 394 hours. confidence assessment: 1
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00:11:19 It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.
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RESPONSE --> My estimate was not exactly the same as this one, but I was thinking very closely to the way I needed to in order to get this answer. I also understand how the conversion factors work. self critique assessment: 2
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00:19:33 All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?
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RESPONSE --> First of all no one knows how long a human life will last, but I am going to say the average life expectancy for a human being is 68. On average the heart beats 120 times per minute. If there are sixty seconds in one minute, 60 minutes in one hour, 24 hours in one day and 365 days in one year, and the human is expected to live for 68 years that comes out to 2.57 * 10^11 heartbeats. confidence assessment: 1
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00:20:01 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
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RESPONSE --> I think my answer was pretty close to what it should have been. self critique assessment: 2
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00:20:14 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?
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RESPONSE --> Not applicable. confidence assessment: 0
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00:20:24 ** For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **
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RESPONSE --> Not applicable. self critique assessment: 3
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00:21:07 Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I'm still having a little trouble connecting how to find a lot of information from so few numbers. self critique assessment: 2
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00:21:19 ** I had to get a little help from a friend on vectors, but now I think I understand them. They are not as difficult to deal with as I thought. **
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RESPONSE --> self critique assessment: 3
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ädÃÛ«¿»áÝI„Ÿ²wò`ìýÖîýÜ\zwŠ¥ assignment #005 005. Uniformly Accelerated Motion Physics I 06-17-2007
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22:45:01 `q001. Note that there are 9 questions in this assignment. If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Suppose that an object increases its velocity at a uniform rate, from an initial velocity of 5 m/s to a final velocity of 25 m/s during a time interval of 4 seconds. By how much does the velocity of the object change? What is the average acceleration of the object? What is the average velocity of the object?
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RESPONSE --> Concerning the velocity of the object, if this question is asking just for the change in velocity then you would use: 'dv = v2 - v1 = 25 - 5 = 20 m/s If it is looking for average velocity then you would use: vAve = (vf + v0)/2 = (25+5)/2 = 15 m/s Concerning acceleration, you would use the following formula: aAve = 'dv/'dt = (25-5)/4 = 5 m/s confidence assessment: 1
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22:45:38 The velocity of the object changes from 5 meters/second to 25 meters/second so the change in velocity is 20 meters/second. The average acceleration is therefore (20 meters/second) / (4 seconds) = 5 m / s^2. The average velocity of the object is the average of its initial and final velocities, as asserted above, and is therefore equal to (5 meters/second + 25 meters/second) / 2 = 15 meters/second (note that two numbers are averaged by adding them and dividing by 2).
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RESPONSE --> I think I missed the units concerning average acceleration, but all of the numbers were right. self critique assessment: 2
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22:48:19 `q002. How far does the object of the preceding problem travel in the 4 seconds?
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RESPONSE --> I will use the formula vAve = 'ds/'dt and solve for the variable 'ds. From the last problem, vAve = 15 m/s. 15 m/s = 'ds/4sec Multiply both sides by 4: 60 m = 'ds confidence assessment: 1
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22:48:33 The displacement `ds of the object is the product vAve `dt of its average velocity and the time interval, so this object travels 15 m/s * 4 s = 60 meters during the 4-second time interval.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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22:54:32 `q003. Explain in commonsense terms how we determine the acceleration and distance traveled if we know the initial velocity v0, and final velocity vf and the time interval `dt.
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RESPONSE --> The formula for determining average acceleration is: aAve = 'dv/'dt I would find distance by using the formula: vAve = 'ds/'dt and solve for 'ds. Another formula for vAve is: vAve = (vf + v0)/2. Since we know vf, v0 and dt, I would use vAve = (vf + v0)/2 to find vAve. Then I would plug in these known variables (vAve and 'dt) into the formula vAve = 'ds/'dt and solve for 'ds. To find acceleration, because you know initial and final velocites as well as the time interval I would plug the values into an above mentioned formula to find average acceleration. confidence assessment: 1
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22:54:49 In commonsense terms, we find the change in velocity since we know the initial and final velocities, and we know the time interval, so we can easily calculate the acceleration. Again since we know initial and final velocities we can easily calculate the average velocity, and since we know the time interval we can now determine the distance traveled.
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RESPONSE --> I think my answer was at least mostly correct. self critique assessment: 3
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23:10:41 `q004. Symbolize the situation by first giving the expression for the acceleration in terms of v0, vf and `dt, then by giving the expression for vAve in terms of v0 and vf, and finally by giving the expression for the displacement in terms of v0, vf and `dt.
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RESPONSE --> Acceleration: aAve = 'dv / 'dt or a = (vf - v0) / 'dt Average velocity: vAve = (vf + v0) / 2 Displacement: vAve = 'ds / 'dt Multiply both sides by 'dt: 'dt * vAve = 'ds 'ds = change in position or displacement confidence assessment: 1
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23:10:57 The acceleration is equal to the change in velocity divided by the time interval; since the change in velocity is vf - v0 we see that the acceleration is a = ( vf - v0 ) / `dt.
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RESPONSE --> I think I have the correct answer. self critique assessment: 2
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23:12:45 `q005. The average velocity is the average of the initial and final velocities, which is expressed as (vf + v0) / 2.
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RESPONSE --> I am not sure what the question is here, but I do know that this statement is true and correct. The formulas for finding average velocity are: vAve = (vf + v0) /2 or vAve = 'ds/'dt confidence assessment: 1
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23:14:32 When this average velocity is multiplied by `dt we get the displacement, which is `ds = (v0 + vf) / 2 * `dt.
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RESPONSE --> Okay, I did not get that this is what I was supposed to find by reading the previous question. In fact, I did not see anything posed as a question. What I read was a statement concerning average velocity, but I see here another way in which to find displacement. self critique assessment: 2
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23:25:41 `q006. This situation is identical to the previous, and the conditions implied by uniformly accelerated motion are repeated here for your review: If the acceleration of an object is uniform, then the following statements apply: 1. A graph of velocity vs. clock time forms a straight line, either level or increasing at a constant rate or decreasing at a constant rate. 2. The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity). 3. The velocity of the object changes at a constant rate (this third statement being obvious since the rate at which the velocity changes is the acceleration, which is assumed here to be constant). 4. The acceleration of the object at every instant is equal to the average acceleration of the object. Describe a graph of velocity vs. clock time, assuming that the initial velocity occurs at clock time t = 0. At what clock time is the final velocity then attained? What are the coordinates of the point on the graph corresponding to the initial velocity (hint: the t coordinate is 0, as specified here; what is the v coordinate at this clock time? i.e., what is the velocity when t = 0?). What are the coordinates of the point corresponding to the final velocity?
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RESPONSE --> The graph has a linear relationship and is increasing at a constant rate. However, I am not sure how to proceed about answering the question from here because according to the four statements given in the question, ""The average velocity of the object over any time interval is equal to the average of its velocity at the beginning of the time interval (called its initial velocity) and its velocity at the end of the time interval (called its final velocity)."" I am not given an initial velocity, so I am assuming the initial velocity is zero. If the initial velocity is zero and the initial time is zero, then how am I supposed to know how far the object moved and in what time interval it moved? I just don't see how I can find the final velocity with so little information. confidence assessment: 0
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23:26:44 The initial velocity of 5 m/s occurs at t = 0 s so the corresponding graph point is (0 s, 5 m/s). The final velocity of 25 meters/second occurs after a time interval of `dt = 4 seconds; since the time interval began at t = 0 sec it ends at at t = 4 seconds and the corresponding graph point is ( 4 s, 25 m/s).
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RESPONSE --> Where did the initial velocity of 5 m/s come from because I didn't see it in the question unless I missed it somewhere? The rest of the solution makes sense. self critique assessment: 2
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23:27:57 `q007. Is the graph increasing, decreasing or level between the two points, and if increasing or decreasing is the increase or decrease at a constant, increasing or decreasing rate?
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RESPONSE --> This answer will come from my answer to the previous question which I am confused about, so based on the example in the class notes I am going to say that there is a constant increase between the two points. confidence assessment: 0
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23:28:31 Since the acceleration is uniform, the graph is a straight line. The graph therefore increases at a constant rate from the point (0, 5 m/s) to the point (4 s, 25 m/s).
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RESPONSE --> It looks like I made a good guess, but this solution makes sense to me. self critique assessment: 2
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23:31:16 `q008. What is the slope of the graph between the two given points, and what is the meaning of this slope?
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RESPONSE --> Slope is defined as rise over run. But, this answer is also determined from answers to one of the earlier questions I was confused about. Based solely on the example in the class notes, I am going to say the rise is 2 and the run is 1.7 for a slope of 1.18. confidence assessment: 0
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23:32:14 The rise of the graph is from 5 m/s to 25 m/s and is therefore 20 meters/second, which represents the change in the velocity of the object. The run of the graph is from 0 seconds to 4 seconds, and is therefore 4 seconds, which represents the time interval during which the velocity changes. The slope of the graph is rise / run = ( 20 m/s ) / (4 s) = 5 m/s^2, which represents the change `dv in the velocity divided by the change `dt in the clock time and therefore represents the acceleration of the object.
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RESPONSE --> I know how to find the slope and I would have gotten the correct answer, had I known the correct final velocity and time coordinates. self critique assessment: 2
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23:41:53 `q009. The graph forms a trapezoid, starting from the point (0,0), rising to the point (0,5 m/s), then sloping upward to (4 s, 25 m/s), then descending to the point (4 s, 0) and returning to the origin (0,0). This trapezoid has two altitudes, 5 m/s on the left and 25 m/s on the right, and a base which represents a width of 4 seconds. What is the average altitude of the trapezoid and what does it represent, and what is the area of the trapezoid and what does it represent?
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RESPONSE --> To calculate the area of the trapezoid under the graph we can multiply the average height times its width. The average height of a trapezoid is just the average of its heights which are 5 m/s and 25 m/s. So the average height of the trapezoid is 15 m/s. Multiplying average height by width we multiply the 15 m/s average height, which represents average velocity, bythe 4-second width, which represents time interval, we get area (15 m/s) * (4 sec) = 60 meters, which represents the position change. confidence assessment: 1
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23:42:14 The two altitudes are 5 meters/second and 25 meters/second, and their average is 15 meters/second. This represents the average velocity of the object on the time interval. The area of the trapezoid is equal to the product of the average altitude and the base, which is 15 m/s * 4 s = 60 meters. This represents the product of the average velocity and the time interval, which is the displacement during the time interval.
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RESPONSE --> I think my answer is at least partially correct. self critique assessment: 2
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