An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.
An airplane takes off at a ground velocity of 200km/hr toward the east, climbing at an angle of 14 degrees. A 20km/hr wind is blowing from the north. Determine the resultant air and ground velocities and their magnitudes.
Use a coordinate system with the x-axis pointing East, the y-axis pointing North and the z-axis pointing up into the sky.
Then the North wind can be described by:
$\displaystyle \vec w=(0, -20, 0)$
and the starting airplane as
$\displaystyle \vec p=(200\cdot \cos(14^\circ)\ ,\ 0\ ,\ 200\cdot \sin(14^\circ))$
The resulting vector which described the way of the airplane is:
$\displaystyle \vec r = \vec w + \vec p = (200\cdot \cos(14^\circ)\ ,\ -20\ ,\ 200\cdot \sin(14^\circ))$
and it's magnitude is:
$\displaystyle |\vec r|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2+ (200\cdot \sin(14^\circ))^2}=\sqrt{40400} \approx 201\ \frac{km}h$
The course over ground is described by:
$\displaystyle \vec g = (0,-20)+(200\cdot \cos(14^\circ)\ ,\ 0)=(200\cdot \cos(14^\circ)\ ,\ -20)$ which yields
$\displaystyle |\vec g|=\sqrt{(200\cdot \cos(14^\circ))^2+(-20)^2}\approx 195.1\ \frac{km}h$
The back of my book says that the air velocity is [200, 20, 200 tan14] which gives a magnitude of 207.1 km/h.
Also, the ground velocity is given as[200,20,0] which gives a magnitude of 201.0 km/h.
I couldn't make sense of this answer. Should I assume that the book is wrong.
No, it's me who made the mistake. I overlooked that the ground speed of the airplane was given (I assumed that the speed through the air was 200 km/h).
So in short:
$\displaystyle \vec w=(0, -20, 0)$
$\displaystyle \vec p=(200\ ,\ 0\ ,\ 200\cdot \tan(14^\circ))$
$\displaystyle \vec r = \vec w + \vec p = (200\ ,\ -20\ ,\ 200\cdot \tan(14^\circ))$
$\displaystyle
|\vec r|=\sqrt{(200)^2+(-20)^2+ (200\cdot \tan(14^\circ))^2}=\sqrt{42886.578} \approx 207.1\ \frac{km}h$
And the speed over ground is then:
$\displaystyle
|\vec g|=\sqrt{(200)^2+(-20)^2}\approx 201\ \frac{km}h
$
Again: I'm sorry for the confusion!