# Vectors Question

1. Jan 18, 2006

### 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 travelling due North West? (The plane must continue with a 250 km/h velocity)

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.

2. Jan 20, 2006

### Tom Mattson

Staff Emeritus
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 $v_{PE}$).

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

3. Jan 20, 2006

### 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.

4. Jan 21, 2006

### 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

5. Jan 21, 2006

### 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
$$s\cos(\theta)\ =\ s\sin(\theta)\ -\ 50$$

Last edited: Nov 29, 2006
6. Jan 22, 2006

### 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:
$$\frac{\sin(x)}{w}=\frac{\sin(135^o)}{s}$$