Question
An airplane has an airspeed of 450 km/hr bearing N45E. The wind velocity is 30 km/hr in the direction N30W. What is the ground speed and what is its direction?.
I have gotten this far:
A= 450cos(45), 450sin(45) = (318.20, 318.20)
W= 30cos(120), 30sin(120) = (-15, 25.98)
A+W= 303.2 +344.18
|A+W|= 458.7 km.hr
tan-1(344.18/303.2)= 48.62
How do I change 48.62 to degrees east of north, which is the actual direction if the aircraft relative to due north (round to the nearest tenth degree).
I have gotten this far:
A= 450cos(45), 450sin(45) = (318.20, 318.20)
W= 30cos(120), 30sin(120) = (-15, 25.98)
A+W= 303.2 +344.18
|A+W|= 458.7 km.hr
tan-1(344.18/303.2)= 48.62
How do I change 48.62 to degrees east of north, which is the actual direction if the aircraft relative to due north (round to the nearest tenth degree).
Answers
Henry
Our answers are identical:
458.7km/hr @ 48.62 deg. CCW.
90 - 48.62 = 41.38 deg,E of N.
You did a good job on this problem!
458.7km/hr @ 48.62 deg. CCW.
90 - 48.62 = 41.38 deg,E of N.
You did a good job on this problem!