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

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3.1 MOTION IN TWO AND THREE DIMENSIONS EXERCISES Section 3.1 Vectors 17. INTERPRET We interpret this as a problem involving finding the magnitude and direction of a (displacement) vector in two dimensions. DEVELOP In two dimensions, a displacement vector can generally be written as, in unit vector notation, ∆∆ r rr ir j xy =+ ˆˆ , where r x and r y are the x - and y -components of the displacements, respectively. The magnitude of r r is r ()() , 22 and the angle r 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. EVALUATE 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 r is r = −+ = ( )( ) 2 2 220 150 266 mmm The direction of r r is = = = −− 11 150 220 14 r r y x m m 57 . ° ASSESS 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 47 8 () ( . ) . ππ π == = 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. = 19. INTERPRET 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. DEVELOP Using Equation 3.1, we see that in two dimensions, a vector r A can be written as, in unit vector notation, r AA iA jA i j A A =+= + (cos ˆ sin ˆ ) θθ where A and Ay x = tan ( / ). 1 Similarly, we express a second vector r B as r BB iB j Bi j ˆ sin ˆ + The resultant vector r C is r CAB A B i A B j A B xx yy A B =+= + + + = + ˆ ˆ ( cos cos ) ˆ (s i n s i n ) ˆˆˆ B jC iC j AB x y ++ = + 3
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3.2 Chapter 3 EVALUATE 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 AA i j i =+ = ° + (cos ˆ sin ˆ ) ( )(cos ˆ sin θθ 360 135 km 135 254 6 254 6 °=− + ˆ )( . ) ˆ (. ) ˆ ji j km km Similarly, the second segment of the travel can be expressed as (with B = 400 km and B 90 ) r BB i j j = ˆ sin ˆ ) ˆ 400 km Thus, the resultant displacement vector is rr r CABC iC j i xy =+= + = + + ˆˆ (.) ˆ [( . ) 254 6 254 6 km km ( ) ] ˆ ˆ ˆ 400 254 6 654 6 km km km j =− + The magnitude of r C is CC C = + = 22 2 2 254 6 654 6 702 4 ( .) . km km km and its direction is = = −− tan . . 11 654 6 254 6 C C y x km km ° ° 68 75 111 ., o r We choose the latter solution () 111 ° since the vector (with <> 00 and ) lies in the second quadrant. ASSESS As depicted in the figure, the resultant displacement vector r C lies in the second quadrant. The direction of r C can be specified as 111 ° CCW from the x -axis (east), or 45 23 7 68 7 °+ °= ° ..
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Chap 3 - MOTION IN TWO AND THREE DIMENSIONS 3 EXERCISES...

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