Thread: Vector problem

A plane is headed due west with an air speed of 212 km/hr. It is driven from its course by a wind from due north blowing at 23.6 km/hr. Find the ground speed of the plane and the actual direction of travel.

I'm not sure how to complete this problem correctly. Here is what I have so far:
Vw=-23.6j Vp=-212i
Vtotal=-212i+-23.6j
tan(theta)=23.6/212 --> theta=6.35 degrees

The plane is heading West of South at 6.35 degree. But How do I calculate the ground speed of the plane? Please help!

It always can be useful to make a design. Have you done that? ...

Design is provided in the problem

Originally Posted by metelskiy

Design is provided in the problem

Draw a right triangle in standard position with terminal side in the third quadrant.
The "adjacent" side is 212.
The "opposite" side is 23.6
The hypotenuse is found using the Pythagorean Theorem.
So you're 1 step away.

Originally Posted by TheChaz

Draw a right triangle in standard position with terminal side in the third quadrant.
The "adjacent" side is 212.
The "opposite" side is 23.6
The hypotenuse is found using the Pythagorean Theorem.
So you're 1 step away.

Ok I found the speed to be 213.31km/hr. So will my final answer be correct if I say that the plane is traveling at speed of 213.31km/hr west of south at 6.35 degrees?

Originally Posted by metelskiy

Ok I found the speed to be 213.31km/hr. So will my final answer be correct if I say that the plane is traveling at speed of 213.31km/hr west of south at 6.35 degrees?

All your calculations are OK. But..

The plane is travelling 6.35° south of west. Better use the course over ground (COG) which is 263.65°.