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# Chap 3 - MOTION IN TWO AND THREE DIMENSIONS 3 EXERCISES...

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3.1 MOTION IN TWO AND THREE DIMENSIONS E XERCISES Section 3.1 Vectors 17. I NTERPRET We interpret this as a problem involving finding the magnitude and direction of a (displacement) vector in two dimensions. D EVELOP In two dimensions, a displacement vector can generally be written as, in unit vector notation, r r i r j x y = + ˆ ˆ , where r x and r y are the x - and y -components of the displacements, respectively. The magnitude of r is r r r x y = + ( ) ( ) , 2 2 and the angle r makes with the + x axis is θ = tan 1 r r y x We choose + x direction to correspond to east and + y for north. E VALUATE With the coordinate system established above, the components of the displacements are r x = − 220 m (220 m, x ) and r y = + 150 m (150 m, + y ). Therefore, the magnitude of the displacement r is r r r x y = + = + = ( ) ( ) ( ) ( ) 2 2 2 2 220 150 266 m m m The direction of r is θ = = = tan tan 1 1 150 220 14 r r y x m m 5 7 . ° A SSESS The displacement vector r lies in the second quadrant. It makes an angle of 145 7 . ° with the + x axis. Alternatively, the direction of r can be specified as 34.3 ° N of W, or 55.7 ° W of N, or by the azimuth 304.3 ° (CW from N), etc. 18. (a) The length of the semicircle is 1 2 2 15 2 47 8 ( ) ( . ) . π π π r r = = = cm cm. (b) The magnitude of the displacement vector, from the start of the semicircle to its end, is just a diameter, or 2 15 2 30 4 ( . ) . cm cm. = 19. I NTERPRET We interpret this as a problem involving the addition of two displacement vectors in two dimensions and finding the magnitude and direction of the resultant vector. The object of interest is the migrating whale. D EVELOP Using Equation 3.1, we see that in two dimensions, a vector A can be written as, in unit vector notation, A A i A j A i j x y A A = + = + ˆ ˆ (cos ˆ sin ˆ ) θ θ where A A A x y = + 2 2 and θ A y x A A = tan ( / ). 1 Similarly, we express a second vector B as B B i B j x y = + = ˆ ˆ B i j B B (cos ˆ sin ˆ ). θ θ + The resultant vector C is C A B A B i A B j A B x x y y A B = + = + + + = + ( ) ˆ ( ) ˆ ( cos cos θ θ ) ˆ ( sin sin ) ˆ ˆ ˆ i A B j C i C j A B x y + + = + θ θ 3

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3.2 Chapter 3 E VALUATE From the problem statement, the first segment of the travel can be written in unit-vector notation as (with A = 360 km and θ A = ° 135 ) r A A i j i A A = + = ° + (cos ˆ sin ˆ ) ( )(cos ˆ sin θ θ 360 135 km 135 254 6 254 6 ° = − + ˆ ) ( . ) ˆ ( . ) ˆ j i j km km Similarly, the second segment of the travel can be expressed as (with B = 400 km and θ B = ° 90 ) r B B i j j B B = + = (cos ˆ sin ˆ ) ( ) ˆ θ θ 400 km Thus, the resultant displacement vector is r r r C A B C i C j i x y = + = + = − + + ˆ ˆ ( . ) ˆ [( . ) 254 6 254 6 km km ( )] ˆ ( . ) ˆ ( . ) ˆ 400 254 6 654 6 km km km j i j = − + The magnitude of r C is C C C x y = + = + = 2 2 2 2 254 6 654 6 702 4 ( . ) ( . ) . km km km and its direction is θ = = tan tan .
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Chap 3 - MOTION IN TWO AND THREE DIMENSIONS 3 EXERCISES...

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