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

magnitude = 304.14 km/h

direction:
`9.46^o` West of North

Step-by-step 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|=√(300^2+50^2)=√(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)=(50)/(300) = (1)/(6) \\\theta = arctan((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.