Suppose that you wish to fly directly from Sacramento to KlamathFalls, which is

Question

Suppose that you wish to fly directly from Sacramento to KlamathFalls, which is 245 miles due North of Sacramento. a) What ground speed (in mph) is necessary tomake the trip directly form Sacramento to Klamath Falls in 1.94hours?
b) If there is a steady 17.0 mph wind blowingtoward a heading of phi = 60.0 degrees east of south (asshown below), then what airspeed and apparent heading are requiredto make the trim to Klamath Falls in the desired 1.94 hours? Youmay write your answer in standard polar form if you prefer. please answer both parts a) What ground speed (in mph) is necessary tomake the trip directly form Sacramento to Klamath Falls in 1.94hours?
b) If there is a steady 17.0 mph wind blowingtoward a heading of phi = 60.0 degrees east of south (asshown below), then what airspeed and apparent heading are requiredto make the trim to Klamath Falls in the desired 1.94 hours? Youmay write your answer in standard polar form if you prefer. please answer both parts

Explanation / Answer

Distance between Sacramento to Klamath Falls is S= 245 miles (a).time t= 1.94 hours So, ground speed v = S / t = 126.288 mph i.e., grounded velcoity v = 126.288 j   Since it is in north direction Speed in polar coordinates = ( 126.288 miles , 90degrees ) (b). wind speed = 17 mph angle = 60 degrees velocity of wind v ' = 17 sin 60 i - 17 cos60 j                              = 14.722 i - 8.5 j here i is an unit vector in east directionand j is an unit vector in north direction relative velocity V = v - v '                             = 126.288 j -14.722i +8.5 j                             = -14.722 i + 134.788 j Magnitude of velocity V = [ -14.722 ^ 2 + 134.788^ 2 ]                                      = 135.58 mph If the vector makes an angle with negative x-axis thentan = 134.788/14.722                                                                                          = 83.76 degrees So, angle made with positive x-axis ' = 180 -                                                             =96.233 degrees So, velocity in polar coordiantes = ( 135.58 mph ,96.233 degrees )

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