Im Am SO lost please help....vectors :(?

Question

hi. Im am having so much trouble with this airplane vector problem, which our teacher never showed us how to do. The answers for this problem is 325.5mph 30 degrees South of East. The question is: An airplane is going 300mph 20 degrees south of east. The wind is blowing 60mph north. If the wind stops blowing, and the pilot makes no changes, what will be the new speed and direction of the plane relative to the ground. Thank you so much :)))))

Answer 1

First draw a picture – it will be a triangle representing the airplane speed and direction (heading flown), the wind speed and direction and finally, the resultant ground speed and direction.  Look at the angles and sides of the triangle that you know and can calculate.  THEN, use the law of cosines to find the speed you don’t know – airspeed.

Law of cosines:  c^2=a^2 + b^2 – 2abcosC where in this case, a, b, and c are speeds and A, B, C are the angles opposite of the sides.  If c is the airspeed of the airplane, you know that the angle between the ground track and wind is 110 degrees … then plug in the numbers.

Airspeed = sqrt( 300^2 + 60^2 – 2*300*60*cos(110))

You get the 325 for airspeed.

THEN, you can calculate the remaining angles in the triangle that you drew.  Use the law of sines: a/sinA = b/sinB = c/sinC.  A, b and c are the speeds, airspeed, groundspeed and wind speed.  You know the one angle – 110 degrees, the rest you can calculate.

Answer 2

If you are a student pilot, you need to be instructed on the use of the E6b flight computer (sorta like a slide rule). If you are just a academic student,I am not going to do your homework for you.

Instructions on the use of the E6b can be found at:

http://www.gleim.com/aviation/computerin...