# Linear Algebra - Vectors

1. Oct 30, 2009

### Whiz

1. The problem statement, all variables and given/known data

A plane flies at 400 km/h [south 30 degrees east]. The wind is blowing from the east at 100 km/h. Find the resulting speed of the plane.

2. Relevant equations

Not sure

3. The attempt at a solution

I've drawn a diagram similar to this:

|\
| \
| \
|30\
| \
| \ 400 km/h
| \
|_____\
100km/h

I thought this was just a simple question of finding the left side of the triangle, but I don't seem to get the answer. Any hints is greatly appreciated.

Last edited: Oct 30, 2009
2. Oct 30, 2009

### Staff: Mentor

You have drawn a 30-60-90 right triangle. If the hypotenuse is 400 km/hr, the base would be 200 km/hr, not 100 km/hr.

Decompose the plane's velocity into a south component and and east component.

3. Oct 30, 2009

### Whiz

Hmm so would this work?

use 30 degrees and 400 km/h to find the speed going straight south

cos (30) * (south_speed/400)
south_speed = cos(30) * 400 = 346

Then use speed going straight south and incorporate the 100km/h wind blowing west, use Pythagorean theorem and find the resultant speed(hypotenuse)?

346^2 + 100^2 = 130000^2
= 360.5

Does that seem right?

4. Oct 30, 2009

### LCKurtz

Set up a vector for the plane velocity: P = < ?, ?>
Write a vector for the wind velocity: W = <?, ?>

The resultant velocity is V = P + W and its length is the groundspeed.

5. Oct 30, 2009

### Whiz

Can you please show me how? I was only only taught how to do it when the points were given.

6. Oct 30, 2009

### LCKurtz

Think of the plane as at the origin with the x direction East and y direction North. The direction described as "South 30 degrees East" is the polar coordinate angle -60 degrees. So the plane velocity is given to you in polar coordinates, r = 400 and $\theta = -\pi/3$. Use the polar coordinate equations to get P = <x, y> for the airplane. And you should easily be able to write the wind vector since you are given its length 100 and direction "from the East".

By the way, I didn't say your answer was incorrect. I'm just trying to show you how you should set up such problems.

Last edited: Oct 30, 2009
7. Oct 31, 2009

### Staff: Mentor

Yes, this is pretty close. I get a slightly larger figure since I didn't round off what you're calling south_speed, which I get as 346.41 km/hr (approx.).