CHAPTER 3 - 19 - 62. If v...

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v BW v PW v PB θ 45° x y 62. If  v PA  is the velocity of the airplane with respect to the air,  v PG  the velocity of the airplane with respect to the ground, and  v AG  the velocity of the air(wind) with respect to the ground, then  v PG   v PA  +  v AG   , as shown in the diagram.  ( a ) From the diagram we find the two components of  v PG : v PGE  =  v AG  cos 45° = (90.0 km/h) cos 45° = 63.6 km/h; v PGS  =  v PA  –  v AG  sin 45° = 550 km/h – (90.0 km/h) sin 45° = 486 km/h. For the magnitude we have v PG  = ( v PGE 2  +  v PGS 2 ) 1/2  = [(63.6 km/h) 2  + (486 km/h) 2 ] 1/2  =         490 km/h . We find the angle from tan   =  v PGE / v PGS  = (63.6 km/h)/(486 km/h) = 0.131, which gives          = 7.45° east of south . ( b ) Because the pilot is expecting to move south, we find the easterly distance  from this line from
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CHAPTER 3 - 19 - 62. If v...

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