# Math Help - Vector airplane question

1. ## Vector airplane question

An airplane is flying at 500km/hr in a wind blowing 60 km/hr toward the SE. In what direction should the plane head to end up going due east? What is the plane's speed relative to the ground?

I assume that the plane is heading due east to start and the wind would add to the i component of speed. I got that the wind was 42.4 i and - 42.4 j so tangent of angle is 42.4 over 542.4 so angle would be arctan of this which is 4.4 degrees. Therefore I have that the plane would need to head north of east by 4.4 degrees. The answer is 4.87 degrees north of east I must be missing another part of the calculation???? The plane's speed is 540.63 km/hr but by my calculations it should be 542.4. Can someone tell me how i am going wrong this time, please.

2. the plane will have to steer a heading north of east to maintain an easterly track over the ground.

basic vector equation ...

air vector + wind vector = ground vector

let $\theta$ = angle of the air vector relative to east.

east/west components ...

$500\cos{\theta} + 60\cos(-45) = G$ , where G is the groundspeed east

north/south components ...

$500\sin{\theta} + 60\sin(-45) = 0$

solve for $\theta$ in the second equation, then use that angle in the first equation to find G.

3. Hello, Frostking!

An airplane is flying at 500km/hr in a wind blowing 60 km/hr toward the SE.
In what direction should the plane head to end up going due east?
What is the plane's speed relative to the ground?
Code:
                              B
*
*   | *
500    *       |   * 60
*           |h    *
*               |       *
*  θ                |     45° *
A *   *   *   *   *   *   *   *   *   * C
: - - - - - - - - b - - - - - - - - :

The plane flies from $A$ to $B$ at 500 km/hr.
The wind is blowing from $B$ to $C$ at 60 m/hr.
The result flight is $AC$, directly east.

In the right triangle at the right: . $\sin45^o \:=\:\frac{h}{60}\quad\Rightarrow\quad h \:=\:60\sin45^o$ [1]

In the right triangle at the left: . $\sin\theta \:=\:\frac{h}{500}$ [2]

Substitute [1] into [2]: . $\sin\theta \:=\:\frac{60\sin45^o}{500} \:=\:0.984852814$

Hence: . $\theta \;=\;4.867561141^o$

The direction should be about: . $4.868^o$ north of east.

We have: . $\angle B \:=\:180^o - 45^o - 4.868^o \:=\:130.132^o$

Law of Sines: . $\frac{b}{\sin130.132^o} \:=\:\frac{500}{\sin45^o} \quad\Rightarrow\quad b \:=\:540.6266461$

Its ground speed is about: . $540.63$ km/hr.