• Feb 14th 2010, 01:12 PM
sam314159265
Hi, just wondering if anybody could help me with the following vector word problem.

An airplane is flying at 550 km/h on a heading of 080 degrees. The wind is blowing at 60 km/h from a bearing of 120 degrees.
1) Draw a vector diagram
2) Find the ground velocity of the airplane

I'm not sure if I should be using true bearings or quadrant bearings for the diagram, and every time I work out the velocity/bearing I get a different answer than the one provided at the back of my textbook.

Would appreciate any help, thanks
• Feb 14th 2010, 01:55 PM
Soroban
Hello, sam314159265!

Quote:

An airplane is flying at 550 km/hr on a bearing of 80°.
The wind is blowing at 60 km/hr from a bearing of 120°.

1) Draw a vector diagram

2) Find the ground velocity of the airplane

Code:

              C        Q       P      o        *       *      *  *  60  :       :    *    *    :       :    *        *60°:       :  *          * :       :  *        40°  o B       : *        *  80°:       :*    *  550      :     A o                *                         R

The plane flies from $\displaystyle A$ to $\displaystyle B\!:\;AB \,=\, 550,\;\angle PAB \,=\, 80^o \,=\,\angle ABR$

The wind is blowing from the southeast: .$\displaystyle BC \,=\,60,\;\angle QBC \,=\,60^o$

Hence: .$\displaystyle \angle ABC \,=\,40^o$

Law of Cosines: .$\displaystyle AC^2 \;=\;550^2 + 60^2 - 2(550)(60)\cos40^o \;=\;255,\!541,0668$

. . Hence: .$\displaystyle AC \;=\;505.6510699 \;\approx\;505.5$ km/hr.

Law of Cosines: .$\displaystyle \cos(\angle CAB) \;=\;\frac{550^2 + 505.5^2 - 60^2}{2(550)(505.5)} \;=\;0.997087043$

. . Hence: .$\displaystyle \angle CAB \;=\;4.374216288^o \;\approx\;4.4^o$

Then: .$\displaystyle \angle PAC \;=\;\angle PAB - \angle CAB \;=\;80^o - 4.4^o \;=\;75.6^o$

Therefore, the plane is flying at 505.5 km/hr at a bearing of 75.6°.

• Feb 14th 2010, 04:43 PM
sam314159265