1. ## airpalne vector question

A plane is traveling between two cities that are 2000 miles apart. The destination city is north of the departure city. The plane is leaving at 10 am and is to arrive at 2 pm. There is a NNE wind (N 22.5 degrees NE) at 20 mph.

I know the normal speed of the plane will 500 mph. The plane must also correct for the wind aiming the nose to the west because of the wind. But will the plane need to increase speed because of the wind? I am a little stuck on how to set up the problem.

2. Originally Posted by hotdogking
A plane is traveling between two cities that are 2000 miles apart. The destination city is north of the departure city. The plane is leaving at 10 am and is to arrive at 2 pm. There is a NNE wind (N 22.5 degrees NE) at 20 mph.

I know the normal speed of the plane will 500 mph. The plane must also correct for the wind aiming the nose to the west because of the wind. But will the plane need to increase speed because of the wind? I am a little stuck on how to set up the problem.
very rough sketch attached ... not even close to scale, but it gives you an idea what's happening.

air vector + wind vector = ground vector

A + W = G

ground vector ... G = 500 mph due North

wind vector ... W = 20 mph blowing toward 22.5 degrees West of South

need to find the air vector, A.

using the cosine law ...

$A^2 = W^2 + G^2 - 2WG\cos(157.5^\circ)$

$A \approx 519$ mph

now use the sine law to find the direction, $\theta$, the airplane should steer relative to North ...

$\frac{\sin{\theta}}{20} = \frac{\sin(157.5)}{A}$

$\sin{\theta} = \frac{20\sin(157.5)}{A}$

$\theta \approx 1^{\circ}$

3. Yes, he must increase since the wind his blowing against him at 22.5 degrees.

$500+20cos(22.5)=518.48 \;\ mph$

$x=20sin(22.5)=7.65$

So, he must adjust is heading $tan^{-1}(\frac{7.65}{518.48})=.845 \;\ \text{degrees}$.

Not much adjustment. Just a little under one degree.