A plane is flying 650 mph in a direction of 12 degrees south of west. Meanwhile, a wind is blowing form the north at 30 mph. Find the resulting speed and direction of the plane. Help please, thanks
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A plane is flying 650 mph in a direction of 12 degrees south of west. Meanwhile, a wind is blowing form the north at 30 mph. Find the resulting speed and direction of the plane. Help please, thanks
Attachment 26076Please consult the colorcoded diagram I made for you.
Note that there are two velocity vectors in play here  the airplane velocity vector is in RED and the wind velocity vector is in GREEN.
Further, note that I have defined West as the positive X direction and South as the positive Y direction.
The first step is to determine the X and Y coordinates for each velocity vector. Let's start with the RED airplane velocity vector. Note that its angle is 12 degrees SOUTH OF WEST. That is, find the West direction, and move toward the South direction by 12 degrees. My diagram illustrates this. The Xcomponent of the airplane velocity is adjacent to the 12 degree angle, and the Ycomponent of the airplane velocity is opposite the 12 degree angle. Thus, , and . Succinctly, .
The velocity components for the wind vector (green) are simple  the Xcomponent is 0 and the Ycomponent is 30 mph. That is, .
Summing our forces, we get
The resultant speed is simply then
The resultant direction is SOUTH OF WEST. In other words, .
EDIT: The diagram was formerly upsidedown. Fixed now.
Hello, limar96!
Do you have to use vectors?
Quote:
A plane is flying 650 mph in a direction of 12 degrees south of west.
Meanwhile, a wind is blowing form the north at 30 mph.
Find the resulting speed and direction of the plane.
Note: represents
The plane is flying from toCode:
O
W  +         o    
: 12o * * 
: * θ * 
A o 650 * 
102o * 
 * x 
30  * 
 * 
 * 
B o 

. .
The wind is blowing from to
. .
Let
Law of Cosines:
. .
Hence: .
Law of Cosines:
. .
Hence: .
Therefore: .