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lamerali
Aug20-08, 12:24 PM
A pilot wishes to fly form city A to city B, a distance of 720 km on a bearing of 70 degrees. The speed of the plane is 700 km/h. An 60 km/h wind is blowing on a bearing of 110 degrees. What heading should the pilot take to reach his or her destination? How long will the trip take?

I have answered the question but i am not sure if it correct. if anyone can just check it for me I would be very grateful.

my answer is below...

Let Vd represent the direction and distance between the two cities
Let Vw represent the velocity of the wind
Let VP represent the velocity of the plane.

Find the x and y coordinates of Vd
\theta = 90 - 70 = 20 degrees
Vd = [720cos20, 720sin20] = [676.6, 246.3]

Now find the x and y coordinates of Vw
\theta = 110 - 90 = 20 degrees
Vd = [60cos20o, - 60sin20o] = [56.4, -20.5]

Now find Vp

|Vp| = \sqrt{733^2 + 255.8^2}
= 767 km

directional angel;
\theta = tan^-1 \frac{255.8}{733}
= 17 degrees

90 - 17 = 73 degrees

speed = distance / time
time = distance / speed
time = 767 km / 700 km per hour
= 1.1 hours

Therefore, the pilot should fly at a heading of 73 degrees and it will take 1.1 hours to get from city A to city B.

Any help is appreciated
THANKS!

cellotim
Aug20-08, 02:45 PM
I get 67 degrees of bearing and 744 km.

OwenMc
Aug20-08, 04:07 PM
As a pilot myself (little planes, no way I'd make it to 700km/h!) I do these sort of calculations in my sleep!
You're both right, depending how you define the wind. The standard definition is the bearing the wind blows FROM. If you use this you get lamerali's answers. Use the incorrect approach (which some people seem to find easier, but just confuses me!) of being the direction to wind is blowing TO, and you get cellotim's answers.
I've attached a spreadsheet which I use for flightplanning, with the figures for both of these cases in. Hope this helps!

lamerali
Aug21-08, 03:16 AM
Great! thanks guys! :D