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Chapter03

# Chapter03 - 1 A vector a can be represented in the...

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1. A vector a G can be represented in the magnitude-angle notation ( a , θ ), where 2 2 x y a a a = + is the magnitude and 1 tan y x a a θ § · = ¨ ¸ © ¹ is the angle a G makes with the positive x axis. (a) Given A x = 25.0 m and A y = 40.0 m, 2 2 ( 25.0 m) (40.0 m) 47.2 m A = + = (b) Recalling that tan θ = tan ( θ + 180°), tan –1 [(40.0 m)/ (– 25.0 m)] = – 58° or 122°. Noting that the vector is in the third quadrant (by the signs of its x and y components) we see that 122° is the correct answer. The graphical calculator “shortcuts” mentioned above are designed to correctly choose the right possibility.

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3. The x and the y components of a vector G a lying on the xy plane are given by cos , sin x y a a a a θ θ = = where | | a a = G is the magnitude and θ is the angle between G a and the positive x axis. (a) The x component of G a is given by a x = 7.3 cos 250° = – 2.5 m. (b) and the y component is given by a y = 7.3 sin 250° = – 6.9 m. In considering the variety of ways to compute these, we note that the vector is 70° below the – x axis, so the components could also have been found from a x = – 7.3 cos 70° and a y = – 7.3 sin 70°. In a similar vein, we note that the vector is 20° to the left from the – y axis, so one could use a x = – 7.3 sin 20° and a y = – 7.3 cos 20° to achieve the same results.

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4. (a) The height is h = d sin θ , where d = 12.5 m and θ = 20.0°. Therefore, h = 4.28 m. (b) The horizontal distance is d cos θ = 11.7 m.
5. The vector sum of the displacements G d storm and K d new must give the same result as its originally intended displacement o ˆ (120 km)j d = G where east is # i , north is # j. Thus, we write storm new ˆ ˆ ˆ (100 km)i, i j. d d A B = = + G G (a) The equation storm new o d d d + = G G G readily yields A = –100 km and B = 120 km. The magnitude of G d new is therefore equal to 2 2 new | | 156 km d A B = + = G . (b) The direction is tan –1 ( B / A ) = –50.2° or 180° + ( –50.2°) = 129.8°. We choose the latter value since it indicates a vector pointing in the second quadrant, which is what we expect here. The answer can be phrased several equivalent ways: 129.8° counterclockwise from east, or 39.8° west from north, or 50.2° north from west.

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6. (a) With r = 15 m and θ = 30°, the x component of G r is given by r x = r cos θ = (15 m) cos 30° = 13 m. (b) Similarly, the y component is given by r y = r sin θ = (15 m) sin 30° = 7.5 m.
(a) We compute the distance from one corner to the diametrically opposite corner: 2 2 2 (3.00 m) (3.70 m) (4.30 m) + + . (b) The displacement vector is along the straight line from the beginning to the end point of the trip. Since a straight line is the shortest distance between two points, the length of the path cannot be less than the magnitude of the displacement.

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Chapter03 - 1 A vector a can be represented in the...

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