Pilot Direction Due North West Given 250 km/h Vp & 50 km/h Vw

[BigCountry]

A plane with a still air speed of 250 km/h is flying due NorthWest. At the same time a wind is blowing toward the South at 50 km/h. In what direction should the pilot head to continue traveling due North West? (The plane must continue with a 250 km/h velocity)

My Answer:

Velocity of plane (Vp) = 250
Velocity of wind (Vw) = 50

sin y = y/x = 50/250
y = 11.5 degrees

90 degrees - 11.5 degrees = 78.5 degrees

The plane must head 78.5 degrees N of W to continue in a 45 degree
N of W line.

Is this correct? Any help is much appreciated.

 

[quantumdude]

Your method looks wrong. You need to write down a vector for the velocity of the plane with respect to the air (call it \vec{v}_{PA}) and a vector for the velocity of the air with respect to the Eart (call it \vec{v}_{AE}. Then you add them up to get the velocity of the plane with respect to the Earth (call it v_{PE}).

\vec{v}_{PE}=\vec{v}_{PA}+\vec{v}_{AE}

From there, find your angle.

 

[lightgrav]

You draw the vectors in the right directions, added tail-to-tip!
your triangle doesn't have a 90-degree in it, but you can use
the law of cosines since you do know two legs and the 135 angle.

 

[abhijitlohiya]

i think use the formula for resultsnt
R=sq.root(p*p+Q*Q+2pqcos135)
find the resultant.
for direction use tan x=p cos 135/p+qsin135

 

[andrevdh]

The resulting motion, \vec r, of the plane is the vector sum of is still air speed, \vec s, and the wind speed, \vec w. The plane has to fly in the direction of the \vec s vector in order to have a resultant motion in the direction of the \vec r vector. Since the \vec r vector is making an angle of 45_o with the "x-axis" its x- and y-components have the same magnitude. Assuming that the \vec s vector makes an angle \theta with the x-axis we can therefore say that:
r_x\ =\ r_y
which gives
s\cos(\theta)\ =\ s\sin(\theta)\ -\ 50

 

[andrevdh]

If my previous relation is a bit too challenging try solving for the angle between \vec r and \vec s, say angle x, via the sine rule:
\frac{\sin(x)}{w}=\frac{\sin(135^o)}{s}