Question

If an airplane is flying directly north at 300.0 km/h, and a crosswind is hitting the airplane at 50.0 km/h from the east, what is the airplane's resultant velocity?

Answer 1

Answer:

magnitude = 304.14 km/h

direction: $9.46^o$ West of North

Explanation:

The final plane's vector velocity will be the result of the vector addition of one pointing North of length 300 km/h, another one pointing West of length 50 km/h.

To find the magnitude of the final velocity vector (speed) we need to apply the Pythagorean theorem in a right angle triangle with sides: 300 and 50, and find its hypotenuse:

$|v|=\sqrt{300^2+50^2}=\sqrt{92500} = 304.14$ km/h

The actual direction of the plane is calculated using trigonometry, in particular with the arctan function, since the tangent of the angle can be written as:

$tan(\theta)=\frac{50}{300} = \frac{1}{6} \\\theta = arctan(\frac{1}{6} ) = 9.46^o$

So the resultant velocity vector of the plane has magnitude = 304.14 km/h,

and it points $9.46^o$ West of the North direction.