#$&* course phy 121 9/17 around 3:00 006. `query 6*********************************************
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Given Solution: The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions: • the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities justification in terms of definitions: if acceleration is uniform, then the v vs t graph is linear so that the average velocity is equal to the average of initial and final velocities • multiplying the 15 m/s average velocity by the 10 s time interval interval we get displacement 150 m justification in terms of definitions: ave velocity is ave rate of change of position with respect to clock time, so vAve = `ds / `dt, from which it follows that `ds = vAve * `dt • the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s; so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2 justification in terms of definitions: ave acceleration is ave rate of change of velocity with respect to clock time, so aAve = `dv / `dt. in this situation acceleration is uniform, so we can if we wish use just plain a instead of aAve The student's solution is not consistent with the given information, which specified acceleration 2 m/s^2 and displacement 125 meters. A solution to the problem: Using the fourth equation of motion with the given information (`ds, a and vf) we have vf^2 = v0^2 + 2 a `ds , which we solve for v0 to get v0 = +- sqrt( vf^2 - 2 a `ds) = +- sqrt( (30 m/s)^2 - 2 * (2 m/s^2) * (125 m) ) = +- sqrt( 900 m^2 / s^2 - 500 m^2 / s^2) = +- sqrt( 400 m^2 / s^2) = +- 20 m/s. If vf = 20 m/s then we could directly reason out the rest (vAve would be 25 m/s, so it would take 5 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (20 m/s + 30 m/s) = 5 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - 20 m/s) / (2 m/s^2) = 10 m/s / (2 m/s^2) = 5 s. If vf = -20 m/s then we could directly reason out the rest (vAve would be (-20 m/s + 30 m/s) / 2 = 5 m/s, so it would take 25 s to go 125 m), or we could use the first or second equation of motion to find `dt The first equation `ds = (vf + v0) / 2 * `dt gives us `dt = 2 * `ds / (vf + v0) = 2 * 125 m / (-20 m/s + 30 m/s) = 25 s. Alternatively the second equation vf = v0 + a `dt gives us `dt = (vf - v0) / a = (30 m/s - (-20 m/s) ) / (2 m/s^2) = 50 m/s / (2 m/s^2) = 25 s. STUDENT COMMENT Umm, evidently I did NOT understand the problem.. even looking back, I’m still not sure how everything in the question exactly relates to the answer. I understand the given answer and what it means, but the original questions are very confusing to me! INSTRUCTOR RESPONSE The point is that the student's solutions are inconsistent. Using the initial velocity obtained by the student, the change in velocity would be 30 m/s and the acceleration would be 3 m/s^2. This of course contradicts the given acceleration, which was 2 m/s^2. So the student's solution contradicts the given information. STUDENT COMMENT I do not understand this problem at all. The information given all makes sense and I know that I am looking for some way to make it all fit together to verify the students results. I understand it a little better and understand the path I was supposed to take after reading through the explanation, but am going to have to work on problems like these. INSTRUCTOR RESPONSE It would be really beneficial for you to go through the given solution one phrase at a time, and tell me exactly what you do and do not understand. For example in the first few lines: The student obtained initial velocity zero and time interval 10 seconds. Along with the given 30 m/s final velocity this would lead to the following conclusions: • the average velocity is 15 m/s, which since acceleration is uniform is simply the average of the initial and final velocities **** do you understand how the given information leads to the conclusion that the average velocity is 15 m/s? **** **** what do you and do you not understand about the meaning of the statement 'since acceleration is uniform is simply the average of the initial and final velocities'? **** the change in velocity would be 30 m/s, since the velocity changes on this interval from 0 m/s to 30 m/s **** what do you and do you not understand about the details of this statement and its overall meaning? **** so the acceleration would be aAve = `dv / `dt = 30 m/s / (10 s) = 3 m/s^2 **** what do you and do you not understand about the meaning of the statement acceleration is `dv / `dt? **** **** what do you and do you not understand about why in this case `dv is 30 m/s and `dt is 10 s? **** **** what do you and do you not understand about the calculation 30 m/s / (10 s) = 3 m/s^2? **** You should deconstruct the entire solution, one phrase at a time, and tell me what you do and do not understand about each. With that information I can help you address the things you don't understand. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: An automobile coasts at high speed down a hill, speeding up as it goes. It encounters significant air resistance, which causes it to speed up less quickly than it would in the absence of air resistance. If the positive direction is down the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the air resistance positive or negative? If the positive direction is up the hill, then • Is the direction of the automobile's velocity positive or negative? • Is the direction of the automobile's acceleration positive or negative? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since the automobile is speeding up its acceleration is in its direction of motion. The velocity is down the hill. The direction of the velocity and acceleration are both down the hill, and the direction of the air resistance is up the hill. If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If you are riding in the car your perception is that the forward direction is the one in which you are moving. You can stick your hand out the window and feel that the air resistance is in the 'backward' direction. Since the automobile is speeding up its acceleration is in its direction of motion. The velocity is down the hill. Thus the direction of the velocity and acceleration are both down the hill, and the direction of the air resistance is up the hill. Therefore • If the direction down the hill is positive then the velocity is positive, the acceleration is positive and air resistance is negative. • If the direction down the hill is negative then the velocity is negative, the acceleration is negative and air resistance is positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: (gen and univ phy) At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40000 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: You should know how the milliliter, liter and cubic meter, three common measure of volume, are related: • a milliliter is the volume of a cube 1 cm on a side • a liter is the volume of a cube 10 cm on a side • a cubic meter is the volume of a cube 1 meter on a side so that • 1 liter = (10 cm)^3 = 1000 cm^3 or 1000 milliliters • 1 cubic meter = (100 cm)^2 = 1 000 000 milliliters • 1 cubic meter = 1 000 000 milliliters / (1000 milliliters / liter) = 1000 liters You should also understand the following images, which will allow you to visualize and thereby reason out these and similar relationships • It takes 10 sides of length 10 cm to make a 1 meter side, so to fill a cubic meter with 10 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A cubic meter is therefore 10 * 10 * 10 liters, or 1000 liters. • It takes 10 sides of length 1 cm to make a 10 cm side, so to fill a liter (a 10 cm cube) with 1 cm cubes we would need 10 layers, each with 10 rows and each row with 10 cubes, for a total of 10 * 10 * 10 cubes. A liter is therefore 10 * 10 * 10 milliliters, or 1000 milliliters. It is also helpful to visualize the relationship between a cubic meter and a cubic kilometer: • A kilometer is 1000 meters. • A cubic kilometer is therefore (1000 meters) ^ 3 = 1 000 000 000 m^3, or 10^9 m^3, or a billion m^3. • A cubic kilometer is visualized as 1000 layers each with 1000 rows each made up of 1000 one-meter cubes, for a total of 1000 * 1000 * 1000 = 1 000 000 000 one-meter cubes. Finally you should understand what a cylinder is and how we find its area. • The volume of a cylinder is equal to the area of its cross-section multiplied its altitude, as you saw in the q_a_initial_problems. • The links below explain prisms and cylinders, and their volumes, in elementary terms: http://www.mathsisfun.com/geometry/prisms.html http://www.mathsisfun.com/geometry/cylinder.html The solution: A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube. A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2. 1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day. • Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year. Visualize the surface of the lake as the base of a large large irregular cylinder. The volume of a cylinder is the product of the area of its base and its altitude. • The volume of water corresponding to a depth change `dy is therefore `dy * A, where A is the area of the lake. • It might be helpful to think of a layer of ice several centimeters thick on top of the lake. The cross-sections of this layer all have very nearly the same size and shape, so it can be viewed as a cylinder with cross-sectional area equal to the area of the lake, and a thickness of several centimeters. The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2. `dy * A = Volume so • `dy = Volume / A = 4.3 * 10^6 m^3 / (5 * 10^7 m^2) = .086 m or 8.6 cm. This estimate is based on 4 people per family. A different assumption would change this estimate. STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract. INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth. Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side. You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares. If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation. ANOTHER INSTRUCTOR COMMENT: The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model. COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2 INSTRUCTOR COMMENT: Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares. 1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. ** STUDENT QUESTION Can you explain this part of the problem: The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2. Where did 10^6 m^2 / km^2 come from? I think I understand the rest of the problem. INSTRUCTOR RESPONSE 1 km = 1000 m, or 10^3 m. So (1 km)^2 = (10^3 m)^2, or 1 km^2 = 10^6 m^2. Thus 10^6 m^2 / (1 km^2) is a division of a quantity by an equal quantity. It follows that 10^6 m^2 / km^2 = 1 So the product 50 km^2 * 10^6 m^2 / km^2 is just 50 km^2 * 1, or 50 km^2. Of course 50 km^2 * 10^6 m^2 / km^2 is also equal to 5 * 10^7 m^2. Thus 50 km^2 = 5 * 10^7 m^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: univ 1.74 (11th edition 1.70) univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** Letting the three vectors be A, B and C (C unknown) and the x axis point east we have A at 0 degrees, B at 315 degrees and the resultant at 0 degrees. Ax = 2, Ay = 0 (A is toward the East, along the x axis). Bx = B cos(315 deg) = 2.47, By = B sin(315 deg) = -2.47. Rx = 5.8, Ry = 0. Since Ax + Bx + Cx = Rx we have Cx = Rx - Ax - Bx = 5.8 - 2 - 2.47 = 1.33. Since Ay + By + Cy = Ry we have Cy = Ry - Ay - By = 0 - 0 - (-2.47) = 2.47. C has magnitude sqrt( Cx^2 + Cy^2) = sqrt ( 1.33^2 + 2.47^2) = sqrt(7.9) = 2.8, representing 2.8 km. C has angle arctan(Cy / Cx) = arctan(2.47 / 1.33) = 61 deg. ** STUDENT QUESTION Why is it 315 deg and not 45? Would you write it 315 deg northwest of east since it goes in that direction? INSTRUCTOR RESPONSE With the x and y axes in standard position, with the x axis pointing east and the y axis north, the southeasterly direction lies in the fourth quadrant, at 315 degrees as measured counterclockwise from the positive x axis. If you measure your vectors from anywhere else you can't use the simple relationships Ax = A cos(theta) and Ay = A sin(theta). With this convention you can, and the signs of the trig functions automatically take care of the + and - signs of the components. You should of course also be able to use triangle trigonometry, but the circular trigonometry of this solution will be used in most of the solutions you will see in the queries and qa's. STUDENT COMMENT/QUESTION: I had this totally wrong. I am sooooo confused as the student above was why the B vector would be 315 degrees? How come it is in the 4th quadrant? I put my vector going in the SE direction. Why is this incorrect? INSTRUCTOR RESPONSE You could orient your x-y coordinate system in any way you wish, as long as the positive y axis is at 90 degrees counterclockwise relative to the positive x axis. You could orient the system so that the easterly direction is the y direction, which would be consistent with your 90 degree angle for the 2 km vector. If so, the x direction would have to be toward the south. This would put the southeasterly direction in the first quadrant. The southeasterly direction would be at 45 degrees with this orientation. The given solution assumes the x axis to be pointing east, so the y axis points north, and the southeasterly direction is 'below' the x axis, in the 4th quadrant, at 315 degrees. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q**** query univ 1.86 (11th edition 1.82) (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT: A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz 3.6 * 2.4 * cos (140 deg) = -6.62 To check for consistency we can calculate the components of A and B: Ax = 3.6 * cos(70 deg) = 1.23 Ay = 3.6 * sin(70 deg) = 3.38 Bx = 2.4 * cos (210 deg) = -2.08 By = 2.4 * sin(210 deg) = -1.2 dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough. ---------------------- Cross product: | A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554. Finding the components we have (Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k = ((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k = 0 i + 0 j + 5.55 k, or just 5.55 k, along the positive z axis ('upward' from the plane). INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward. The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* "
#$&* course phy 121 9/17 around 3:20 007. Acceleration of Gravity
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Given Solution: We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I went more with equation way to solve the problem so I did not have as much explanation. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q002. What are the ramp slopes associated with these accelerations? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: For the 5-second trial, where acceleration was 4 cm/s^2, the 'rise' of the ramp was .5 cm and the 'run' was nearly equal to the 50-cm length of the ramp so the slope was very close to .5 cm / (50 cm) = .01. For the 3-second trial, where acceleration was 11 cm/s^2, the 'rise' of the ramp was 1 cm and the 'run' was very close to the 50-cm length, so the slope was very close to 1 cm / (50 cm) = .02. For the 2-second trial, where the acceleration was 25 cm/s^2, the slope is similarly found to be very close to 1.5 cm / (50 cm) = .03. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the best-fit straight line (i.e., the straight line that comes as close as possible, on the average, to the three points). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). • The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. • If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. • Alternatively the graph could that acceleration vs. ramp slope is increasing at an increasing rate. STUDENT COMMENT I am a little unclear about why the lines could be considered scattered from an actual linear behavior, unless it is because the line is not exactly straight and seems to have a curve. INSTRUCTOR RESPONSE The graph points don't line along a perfectly straight line. Based on these three points, you could possibly infer that the graph curves. A curved graph would be consistent with the data. If experimental uncertainties are large enough, the graph could also be consistent with a best-fit straight line. • The best-fit straight line for this data wouldn't go through any of the graph points; some points will lie above the line and some below. • So the points would to an extent be scattered about the line. • As long as the degree of scattering can be explained by the uncertainties in our measurements, then it is possible that the straight-line, or linear, model is consistent with the data. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I didn’t explain why the graph was not clear. ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (0, -6 cm/s^2) and (.05, 42 cm/s^2). the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity. STUDENT QUESTION Once I graphed the points using the correct coordinates I was able to see the correct graph, and its slope of 48cm/s^2/ .05 =960cm/s^2. ??????????My question here is what if on my graph the points were at around 42cm/s^2 instead of the 48cm/s^2, and my answer was 840 or so. Is there room for the variation in how you plot the graph, if I started with different number or something, or my graph is not as precise.????????????????????????? ?????????????Also I am still a little unclear, seeing how I had my point graphed wrong in the first place, about why we would use the slope of the graph .5/ 50cm to get .01....I really cannot seem to rationalize this. ????????????????????????? INSTRUCTOR RESPONSE Your slope is based on two points; it can't be calculated based on a single point because you need two points to get the rise and run. If the two points on your straight line were, say, (0, -10 cm/s^2) and (.05, 48 cm/s^2) then the rise would be 48 cm/s^2 - 10 cm/s^2 = 58 cm/s^2, your run would be .05 and your slope would be 58 cm/s^2 / .05 = 1060 cm/s^2. If the rise between the two points on your straight line was 42 cm/s^2, with the run still .05, then your slope would be 42 cm/s^2 / .05 = 840 cm/s^2, as you say. There is in fact only one best-fit line for the data, but since we're estimating the best-fit line it is expected that our estimates of its slope and location will differ somewhat. Certainly it would be possible for one person to get an estimate of 840 cm/s^2, while another might estimate 960 cm/s^2 and another 1060 cm/s^2. If all three people used the same three data points, and actually found the unique line that best fits those three points, then all three would get identical results. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 100 cycles of a pendulum of this length should require approximately 115 seconds. 115 sec / (100 cycles) = 1.15 sec / cycle confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: T = 2 `pi / `sqrt(g) * `sqrt(L) multiply both sides by `sqrt(g) T * `sqrt(g) = 2 `pi `sqrt(L) Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T Squaring both sides we finally obtain g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain • T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain • `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain • g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain • g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated. STUDENT QUESTION I do not understand how if you divide both sides by the sqrt (g), the you end up with t* sqrt(g) = 2' pi ' sqrt (L), when sqrt(L) was divided by 2 'pi.......why wouldnt it be t * sqrt (g) = 2'pi / sqrt(L)....what happens to the division sign there????? INSTRUCTOR RESPONSE The division sign is between pi and sqrt(g). It doesn't apply to sqrt(L). Multiplications and divisions are done in order, as stated in the order of operations. This means that you multiply 2 by pi, then divide the result by sqrt(g), then multiply the result by sqrt(L). If the sqrt(L) was to be part of the denominator, the expression would have to be written 2 pi / ( sqrt(g) * sqrt(L) ). In that case you would begin by multiplying sqrt(g) by sqrt(L), then you would do multiplications and divisions in order, multiplying 2 by pi then dividing by the product (sqrt(g) sqrt(L)). The correct expression: The expression 2 pi / sqrt(g) * sqrt(L) is represented in a computer algebra system as This representation is consistent with the order of operations. The common misinterpretation of the expression: The expression 2 pi / sqrt(g) * sqrt(L) is often misinterpreted as However the above represents 2 pi / (sqrt(g) * sqrt(L) ), not our original expression 2 pi / sqrt(g) * sqrt(L) &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ 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. Questions, Problems and Exercises You should answer the questions and work the problems as given below, as appropriate to your course. Your work should normally be handwritten, should include diagrams and sketches where appropriate, and should go into your Physics notebook (not into your lab notebook). If the course is not specified for a problem, then students in all physics courses should do that problem. Principles of Physics students need not do the questions or problems that are specified for General College Physics or University Physics. General College Physics students need not do questions or problems specified for University Physics. University Physics students should do all questions and problems. Principles of Physics students may if they wish do some of the questions and problems specified for General College Physics, but this is neither expected nor required. These problems are accessible to Principles of Physics students, but are generally more challenging that what is expected of students in this course. (Some University Physics problems will also be accessible to Principles of Physics students, though some will not.) General College Physics students who wish to do so are welcome to work some or all of the University Physics questions and problems, though this is neither expected nor required. Many of the University Physics questions and problems are more challenging than those expected of General College Physics students, and a number of the problems require the use of calculus, which is not expected of General College Physics students. You are not expected to submit these questions and problems. It would take too much of your time to key in all the answers and solutions. You should solve these problems and answer these questions in your notebook, in a form you can later reference and, if you later desire, revise. The Query at the end of the assignment will ask you selected questions, which you will at that time be expected to answer based on the work you have done in your notebook. Remember that you are always welcome to ask questions at any time. Any question about a problem should include a copy of the problem and a summary of what you do and do not understand about it. Questions related to q_a_ 1. If a cart is found to have acceleration 30 cm/s^2 when it rolls freely down an incline having slope .05, and acceleration 70 cm/s^2 on an incline having slope .08, then • At what average rate is its acceleration changing with respect to ramp slope? • What estimate does this result give us of the acceleration of gravity? 2. If a pendulum of length 30 cm goes through 54 cycles in a minute, then what is the corresponding estimate of the acceleration of gravity? If there is a 1% uncertainty in the count, then what is the uncertainty in the estimate? Questions related to Introductory Problem Sets These questions all relate to the figure below: 1. According to the figure, from what two quantities do we determine vAve? • How is it that vAve is determined from these quantities? That is, if we are given the values of these two quantities, how do we find the value of vAve? Why is this just common sense? According to the figure, from what two quantities do we determine `dv? • How is it that `dv is determined from these quantities? That is, if we are given the values of these two quantities, how do we find the value of `dv? Why is this just common sense? 2. If we were to sketch a line from vAve to `dt, then the figure would contain a triangle with its vertices labeled vAve, `ds and `dt. This would indicate that there is a relationship among these three quantities which is dictated by at one or more of the following: • common sense, • the definition of average velocity, • the definition of average acceleration, • fact that for uniform acceleration the average velocity on an interval is equal to the average of the initial and final velocities. Answer the following: • What is the relationship among vAve, `ds and `dt? • How is this relationship the result of one or more of the four things in the list? 3. If we were to sketch a line from `dv to `dt, then the figure would contain a triangle with its vertices labeled `dv, `ds and `dt. This would indicate that there is a relationship among these three quantities which is dictated by at one or more of the following: • common sense, • the definition of average velocity, • the definition of average acceleration, • fact that for uniform acceleration the average velocity on an interval is equal to the average of the initial and final velocities. Answer the following: • What is the relationship among `dv, `ds and `dt? • How is this relationship the result of one or more of the four things in the list? Questions/problems for Principles of Physics Students The graph below depicts a trapezoid on a graph of v vs. t for some unspecified object. The questions that follow are related to this graph. 1. Sketch and label four trapezoids of your own: • Construct the first trapezoid so that vf is four times as great as v0. • Construct the second trapezoid so that vAve and `dt are the same as for the first, but vf is half as great as v0. • Construct the third trapezoid so that vf and v0 are the same as on the first, but `dt is half as great. • Construct the fourth trapezoid so that `dv is the same as for the first, but vAve is twice as great. Suppose that each trapezoid represents a runner. Describe what the runner is doing in each case, and how it differs from what he or she is doing in each of the other cases. 2. List the four trapezoids you sketched in the preceding problem, in each of the orders indicated below: • order the trapezoids from the one with the least `dv to the one with the greatest. • order the trapezoids from the one with the least slope to the one with the greatest. • order the trapezoids from the one with the least area to the one with the greatest. 3. Identify the line segment in the depicted trapezoid whose length correspond to each of the following quantities: • v0 • vf • `dv • vAve • `dt 4. In terms of the picture of the trapezoid, how do we construct the line segment for `dv? In terms of the picture, how do we construct the line segment for vAve? In terms of the picture, how do we construct a rectangle whose area is equal to that of the trapezoid? In terms of the picture, how do we construct a triangle whose legs are `dv and `dt? 5. What are the height and width of the rectangle whose area is equal to that of the depicted trapezoid? What is the meaning of the height? What is the meaning of the width? How are the height and width of a rectangle used to find its area? • What therefore is the meaning of the trapezoid's area for the motion of the object? • What is the expression for the area of the trapezoid? 6. What is the rise of the 'red' triangle in the figure (the 'red' triangle is the triangle having two red legs; its hypotenuse is blue)? What is the run of that triangle? What therefore is the slope of the triangle? What does the rise of the triangle mean in terms of the motion of the object? What does the run ise of the triangle mean in terms of the motion of the object? What therefore does the slope of the triangle mean in terms of the motion of the object? 7. The slope of the trapezoid is (vf - v0) / (tf - t0), which could also have been written (vf - v0) / `dt. • Explain how this gives us the equation a = (vf - v0) / `dt. • Solve this equation for vf. • If you solved correctly, your result will be identical to one of the equations of uniformly accelerated motion. Which equation is that? 8. The area of the trapezoid is (vf + v0) / 2 * ( tf - t0) = (vf + v0) / 2 * `dt. This represents the product of average velocity and time interval, i.e., the displacement. • If you set `ds equal to this expression, you get one of the equations of uniformly accelerated motion. Which is equation do you get? ________________________________________ Be prepared at any time to explain the meaning of the statement • The first two equations of uniformly accelerated motion can be obtained by finding the slope and the area associated with a v vs. t trapezoid where the initial velocity is v0, the final velocity is vf and the time interval is `dt. ________________________________________ Questions/problems for General College Physics Students The figure shown under the heading 'Questions/problems for Principles of Physics Students' applies here as well. 1. What aspects of the figure are related to the definition of the average rate of change of position with respect to clock time, and how are they related? 2. What aspects of the figure are related to the definition of the average rate of change of velocity with respect to clock time, and how are they related? 3. How is it that we can assert that, for this graph, the average velocity is the average of the initial and final velocities? 4. Sketch an example of a velocity vs. clock time graph for a time interval, with the property that the average velocity on the interval is less than the average of the initial and final velocities. Be prepared to describe your sketch. 5. Derive the third and fourth equations of motion from the first two. Questions/problems for University Physics Students 1. If the function s(t) = .3 m/s^3 * t^3 - 2 m/s^2 * t^2 + 5 m/s * t + 12 m, then: • show s(t) is in units of meters if if t is clock time in seconds (to show this you need to first show that every term of the s(t) function has units of meters) • the corresponding velocity function v(t) is the derivative with respect to clock time t of the s(t) function; find this velocity function and show that v(t) has units of m/s when t has clock time seconds • the corresponding acceleration function a(t) is the derivative with respect to clock time t of the velocity function, making it the second derivative with respect to t of the position function; find this function • According to your position function, what are the positions at t = 2 seconds and t = 2.1 seconds? What therefore is the average rate of change of position with respect to clock time during this interval? How does this compare with the velocity at t = 2 seconds? 2. Using the a(t) function you obtained in #1, integrate to find the general form of the v(t) and s(t) functions (you will obtain the general form if you include the integration constant with each integration; note that if you get two different integration constants you can't call them both 'c'; you might, for example, choose to call them c1 and c2). Find the values of the integration constants required to fulfill the conditions v(0) = 4 m/s and s(0) = 20 m. If one object moves according to the position function given in the first problem, and another moves according to the position function just obtained, how will the motions of the two objects compare? • Will they ever have the same acceleration? • Will they ever be at the same position? • Will they ever have the same velocity? 3. If the acceleration of an object is constant, its initial velocity is v0 and its initial position is s0 we can find its velocity and position functions, in terms of a, v0, s0 and the clock time t. Using integration, find the velocity function v(t) and the position function s(t): • Start with the acceleration function a(t) = a, where the right-hand side consists of the constant number a. • Use integration to find the general velocity functions v(t) and s(t), in terms of the integration constants obtained for each integration (you might choose to call these constants c1 and c2 to distinguish them). • Initial velocity v0 means that the velocity at the t = 0 instant is v0, and initial position s0 means that the velocity at the t = 0 instant is s0. Use this information to express your two integration constants in terms of v0 and "