HW_3_solutions

HW_3_solutions - P112 Spring 2007 Homework#3 Solutions...

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P112 Spring 2007 Homework #3 Solutions 1. (not boxed) Let the x + direction be east (E), and the y + direction be north (N). Subscript definitions: p stands for plane, g stands for ground, and a stands for air. i v ˆ ) km/h 220 ( a p, ! = r (i.e., 220 km/h west (W)). At h 500 . 0 = ! t , j i r ˆ ) km 20 ( ˆ ) km 120 ( g p, ! + ! = " r . Therefore, j i j i r v ˆ ) km/h 40 ( ˆ ) km/h 240 ( h 500 . 0 ˆ ) km 20 ( ˆ ) km 120 ( g p, g p, ! + ! = ! + ! = " " = t r r . (a) This part is asking for the velocity of the air (wind) with respect to the ground, g a, v r . j i i v v v v v ˆ ) km/h 40 ( ˆ ) km/h 240 ( ˆ ) km/h 220 ( g p, a p, g p, p a, g a, ! + ! + ! ! = + ! = + = r r r r r . j i v ˆ ) km/h 40 ( ˆ ) km/h 20 ( g a, ! + ! = r . km/h 45 ) km/h 40 ( ) km/h 20 ( | | 2 2 g a, = ! + ! = v r . ° = ! " # $ % = 63 20 40 tan 1 g a, v r ( . Since both x v g, a, and y v g, a, are negative, g a, v r points 63 ° below the x ! direction or 63 ° south of west. b) For this part of the problem, j v ˆ ) km/h 40 ( g a, ! = r (i.e., 40 km/h due south). Again, km/h 220 a p, = v . Here we want to know what direction of a p, v r will result in the plane traveling due west. g a, g p, a g, g p, a p, v v v v v r r r r r ! = + = . In order for the plane to travel due west, require 0 g, p, = y v . So, working with the y component of the velocity equation yields km/h 40 ) km/h 40 ( 0 g, a, g, p, , a p, = ! ! = ! = y y y v v v . ° = ! " # $ % = ! ! " # $ $ % = 10 220 40 sin sin 1 a p, a, p, 1 a p, v v y v r . Also notice that since 0 g, a, = x v , 0 g, p, a, p, < = x x v v (negative because the plane should travel due west). Thus, a p, v r points 10 ° above the x ! direction or 10 ° north of west.
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Aside: km/h 216 ) km/h 40 ( ) km/h 220 ( 2 2 2 a, p, 2 a p, a, p, ! = ! ! = ! ! = y x
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HW_3_solutions - P112 Spring 2007 Homework#3 Solutions...

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