#$&* course PHY 201 006. `query 6 submitted 10 Feb 11 around 7:56 AM. 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, Im 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): 2, I needed to refer to the given solution to find the correct direction of the solution. ------------------------------------------------ Self-critique Rating: 2 ********************************************* 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: If the positive direction is down the hill, then the direction of the automobiles velocity down the hill is positive also. The direction of the air resistance is negative because its in the opposite direction. If the positive direction is up the hill, then the direction of the automobiles velocity up the hill is positive also. The direction of the air resistance is negative because its in the opposite direction. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: 3 ********************************************* 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 #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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
You should of course always attempt a solution and detail your thinking about what the problem means and how you might solve it. Even an incorrect attempt forms a basis for correction and subsequent understanding. &#
The given solution reads as follows: '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. ' If you take this phrase by phrase and let me know what you do and do not understand, I can help clarify. If you are sure you understand, of course, this isn't necessary.*@ ------------------------------------------------ Self-critique Rating: 0 I had to rely totally on the given solution/Instructors response. I had to read it several times to get an understanding. ********************************************* 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^.............................................
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: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!