# Did I do this vector question right?

• Jun 8th 2010, 10:39 AM
kmjt
Did I do this vector question right?
A plane is heading S 70degrees W with a groundspeed of 625km/h. If the pilot is steering west at an airspeed of 665km/h, what must be the windspeed and wind direction?

This is the diagram I drew, forgive me if the scaling is off:

http://img686.imageshack.us/img686/9...rquestion1.png

The magnitude of the wind I found was:

w^2 = a^2 + g^2 -2(a)(g)(cos110degrees)

Is this wrong? I want to know if I have the right answer for the magnitude before attempting to find the direction of the wind. (Smirk)
• Jun 8th 2010, 11:02 AM
skeeter
$\displaystyle w^2 = a^2 + g^2 - 2ag \cos(20)$
• Jun 8th 2010, 11:32 AM
Soroban
Hello, kmjt!

Your diagram is incorrect . . .

Quote:

A plane is heading S 70° W with a groundspeed of 625 km/h.
If the pilot is steering west at an airspeed of 665 km/h,
. . what must be the windspeed and wind direction?

Code:

                665     A o - - - - - - - - - - - o O       \              20° *         \              *         \          *           \      *  625           \  *             o             B

The plane is flying from O to A at 665 km/hr.
. . $\displaystyle \angle AOB = 20^o$

The wind is blowing from A to B.

The plane's resultant flight is: .$\displaystyle OB = 625$ km/hr.

Law of Cosines: . $\displaystyle AB^2 \:=\:665^2 + 625^2 - 2(665)(625)\cos20^o \;=\;51,\!730.50897$

. . Hence: .$\displaystyle AB \;\approx\;227.4$

Then: .$\displaystyle \cos A \;=\;\frac{665^2 + 227.4^2 - 625^2}{2(665)(227.4)} \;=\;0.341588668$

. . Hnce: .$\displaystyle \angle A \;\approx\;70^o$

Therefore, the windspeed is 227.4 km/hr and its direction is: .S 20° E.

Interesting!
Triangle ABO is virtually a right triangle . . .
.
• Jun 8th 2010, 12:06 PM
kmjt
Doh thats what I got the first time I just thought it was wrong because usually we have to calculate something to find what angle the wind is blowing at.. Thanks :D