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Unformatted text preview: 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, ˆ ˆ , x y r r i r j ∆ = ∆ + ∆ r where x r ∆ and y r ∆ are the x and ycomponents of the displacements, respectively. The magnitude of r ∆ r is 2 2 ( ) ( ) , x y r r r ∆ = ∆ + ∆ and the angle r ∆ r makes with the + x axis is 1 tan y x r r θ ∆ = ∆ 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 220 m x r ∆ =  (220 m,  x ) and 150 m y r ∆ = + (150 m, + y ). Therefore, the magnitude of the displacement r ∆ r is 2 2 2 2 ( ) ( ) ( 220 m) (150 m) 266 m x y r r r ∆ = ∆ + ∆ = + = The direction of r ∆ r is 1 1 150 m tan tan 145.7 220 m y x r r θ ∆ = = = ° ∆ A SSESS The displacement vector r ∆ r lies in the second quadrant. It makes an angle of 145.7 ° with the + x axis. Alternatively, the direction of r ∆ 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 cm) 47.8 cm. r r π π π = = = (b) The magnitude of the displacement vector, from the start of the semicircle to its end, is just a diameter, or 2(15.2 cm) 30.4 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 r can be written as, in unit vector notation, ˆ ˆ ˆ ˆ (cos sin ) x y A A A A i A j A i j θ θ = + = + r where 2 2 x y A A A = + and 1 tan ( / ). A y x A A θ = Similarly, we express a second vector B r as ˆ ˆ x y B B i B j = + = r ˆ ˆ (cos sin ). B B B i j θ θ + The resultant vector C r is ˆ ˆ ˆ ˆ ˆ ˆ ( ) ( ) ( cos cos ) ( sin sin ) x x y y A B A B x y C A B A B i A B j A B i A B j C i C j θ θ θ θ = + = + + + = + + + = + r r r E VALUATE From the problem statement, the first segment of the travel can be written in unitvector notation as (with 360 km A = and 135 ) A θ = ° ˆ ˆ ˆ ˆ ˆ ˆ (cos sin ) (360 km)(cos135 sin135 ) ( 254.6 km) (254.6 km) A A A A i j i j i j θ θ = + = ° + ° =  + r Similarly, the second segment of the travel can be expressed as (with 400 km B = and 90 ) B θ = ° 3.1 3 3.2 Chapter 3 ˆ ˆ ˆ (cos sin ) (400 km) B B B B i j j θ θ = + = r Thus, the resultant displacement vector is ˆ ˆ ˆ ˆ ˆ ˆ ( 254.6 km) [(254.6 km) (400 km)] ( 254.6 km) (654.6 km) x y C A B C i C j i j i j = + = + =  + + =  + r r r The magnitude of C r is 2 2 2 2 ( 254.6 km)( 254....
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This note was uploaded on 09/12/2009 for the course PH PH101 taught by Professor Prof.kim during the Spring '09 term at Korea Advanced Institute of Science and Technology.
 Spring '09
 Prof.Kim
 Physics

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