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sol2final

# sol2final - HW Solutions 2 8.01 MIT Prof Kowalski Topics...

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HW Solutions 2 - 8.01 MIT - Prof. Kowalski Topics: Vectors and Two dimensional motion 1) 1.38 Please refer to figure 1.26 p.35. To add the several vector displacements, we use components, refer- ring to east and north instead of x and y. The net northward dis- placement is 2.6 + (3.1) (sin 45) =+ 4.8 km, and the net eastward displacement is (4.0)+ (3.1) (cos 45) = + 6.2 km. The magnitude of the net displacement is (4 . 8) 2 + (6 . 2) 2 = 7 . 8 km, using Pythago- ras’ theroem. To find the direction, note that the ratio of northward to eastward displacement is the tangent of the angle with the east axis: arctan( 4 . 8 6 . 2 ) = 38 noth of east. * The total displacement is a vector starting from the starting point towards the end. By sketching the vectors to scale you’ll see the agreement with the method of components. 2) 1.70 Please refer to figure 1.32 p.37. The displacement vectors of the three legs must sum up to give the total observed displacement. This vector equality provides TWO equations, one for each component - the net x displacement should be 5.80 km and the net y displacement should be 0: The third leg must take the sailor east the distance: (5.80) - (3.50) cos 45 - (2.00) = 1.33 km and a distance north (3.5) sin 45 = 2.47 km. magnitude of the displacement is (1 . 33) 2 + (2 . 47) 2 = 2 . 81 km and the direction arctan( 2 . 47 1 . 33 ) = 62north of east. 1

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3) 1.50 Please refer to figure 1.28 p.35. Method : (Product of magnitudes × cos θ ) The angles must be found from the figure. A.B = AB cos θ AB = (12 × 15) cos 93 = 9 . 4 m 2 B.C = BC cos θ BC = (15 × 6) cos 80 = +15 . 6 m 2 C.A = CA cos θ CA = (12 × 6) cos 173 = 71 . 5 m 2 4) 1.52 The problem as initially posed was ambiguous as to which figure was referred to.” Either of these solutions will be graded as correct.
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