course Phy 231 _퍭{KǣVassignment #006
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15:36:31 ** 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. **
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RESPONSE --> Ok self critique assessment: 3
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15:48:39 **** query univ 1.82 (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product
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RESPONSE --> The plane would be traveling 3.6 km at 70 degree of north of east, and 2.4 km at 210 degree from east or 30 south from west. To locate this plane from origin it would be found by combing the scalar distances. First we find the 3.6 * sin 70 = 3.38, and 3.6 * cos 70 = 1.23. Then find 2.4 * sin 210 = -1.2 and 2.4 * cos 210 = -2.08; therefore, the plane will be located 3.38 - 1.2 = 2.18 km north, and 1.23 - 2.08 = - .85 km east or simply .85 km west of the airport. This would be givin as a straight line distance of 5.48 km at 140 degrees of east. confidence assessment: 3
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16:03:59 ** 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. **
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RESPONSE --> I understood what the student was doing until they reach the point of dot product. They obtained a value different from their final answer and from my own that I found by using a seperation of the motions as he did with the format of Ax,Ay, etc...When he used this dot product the value of -6.61 was found and does not match the answer given. So how is this relevant to the problem? How is there a z axis since the plane was only being concerned with it 2 motions of direction along the x and y axis? I was going with the format of the question 1.66 that involves an airplane in my 11th edition text. I think they were discrepancies in problems between 10th and 11th edition in this instance. self critique assessment: 1
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