ch03 - Chapter 3 1 The x and the y components of a vector a...

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73 Chapter 3 1. The x and the y components of a vector a lying on the xy plane are given by cos , sin x y a a a a where | | a a is the magnitude and is the angle between a and the positive x axis. (a) The x component of a is given by cos (7.3 m)cos250 2.50 m x a a    . (b) Similarly, the y component is given by sin (7.3 m)sin 250 6.86 m 6.9 m. y a a      The results are depicted in the figure below: 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 (7.3 m)cos70 2.50 m, (7.3 m)sin 70 6.86 m. x y a a           Similarly, we note that the vector is 20° to the left from the – y axis, so one could also achieve the same results by using (7.3 m)sin 20 2.50 m, (7.3 m)cos20 6.86 m. x y a a           As a consistency check, we note that 2 2 2 2 ( 2.50 m) ( 6.86 m) 7.3 m x y a a  
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CHAPTER 3 74 and 1 1 tan / tan [( 6.86 m) /( 2.50 m)] 250 y x a a , which are indeed the values given in the problem statement. 2. (a) With r = 15 m and = 30°, the x component of r is given by r x = r cos = (12 m) cos 30° = 10 m. (b) Similarly, the y component is given by r y = r sin = (12 m) sin 30° = 6.0 m. 3. A vector a can be represented in the magnitude-angle notation ( a , ), where 2 2 x y a a a is the magnitude and 1 tan y x a a is the angle a 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 = tan ( + 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. The results are depicted in the figure below: We can check our answers by noting that the x - and the y - components of A can be written as cos , sin x y A A A A
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75 Substituting the results calculated above, we obtain (47.2 m)cos122 25.0 m, (47.2 m)sin122 40.0 m x y A A       which indeed are the values given in the problem statement. 4. The angle described by a full circle is 360° = 2 rad, which is the basis of our conversion factor. (a) 2 rad 20.0 20.0 0.349 rad 360   . (b) 2 rad 50.0 50.0 0.873 rad 360   . (c) 2 rad 100 100 1.75 rad 360   . (d) 360 0.330 rad = 0.330 rad 18.9 2 rad . (e) 0 0 360 2.30 rad 2.30 rad 132 2 rad . (f) 360 7.70 rad = 7.70 rad 441 2 rad . 5. The vector sum of the displacements d storm and d new must give the same result as its originally intended displacement o ˆ (120 km)j d where east is i , north is j . Thus, we write storm new ? ?
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