# Homework Help: Relative Velocity Airplane Problem

1. Sep 9, 2014

### lexie47

Q: The pilot of an aircraft flies due east relative to the ground in a wind blowing 20 km/h toward the south. If the speed of the aircraft in the absence of wind is 70 km/h, what is the speed of the aircraft relative to the ground?

I set up the vector of wind WRT ground pointing due south, then the vector of the plane WRT the ground extending due E from the tail of the first, so the resultant vector is the plane WRT the wind. I tried solving this my just using the pythagorean theorem bc of the presence of a right angle, but was terribly wrong. Can someone point me in the right direction, I'm just totally stumped on this one, thanks!

2. Sep 9, 2014

### SteamKing

Staff Emeritus
Why don't you post your work? You may have made a silly mistake which can be easily spotted by a fresh set of eyes.

As always, please use the HW template when posting questions to the HW forums. It is there for your benefit.

3. Sep 9, 2014

### lexie47

Sorry, this is my first post, will stick to the template next time!
I attached a picture of my breaking the vectors into components, which lead me nowhere because I ended up with V_pg = V_pg.

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4. Sep 9, 2014

### phinds

In addition to what SteamKing said, I would add, your diagram is a VERY important part of your work. Without the diagram we really can't see what you are doing.

EDIT. HA! you beat me to it

5. Sep 9, 2014

### phinds

Seriously? You're going to post the picture sideways? That gives me a kink in my neck and disincentives me to look at your problem.

6. Sep 9, 2014

### lexie47

It just uploaded that way I'm sorry, trying to attach another way.

7. Sep 10, 2014

### lexie47

Can't find another way to attach, sorry for being a pain, I'm not familiar with forums. Thanks for looking anyway.

8. Sep 10, 2014

### billy_joule

You've misunderstood the question and labeled your vectors wrong.

9. Sep 10, 2014

### lexie47

I got it, thank you, its 67 km/hr if using the theorem.

10. Sep 10, 2014

### billy_joule

Still wrong ;-)

You need to rethink how you've labeled your vectors.

11. Sep 10, 2014

### lexie47

That's what my teacher posted as the answer though, unless she's wrong?

I'm getting Vpg= √((V_pw )^2-(V_wg)^2)

Last edited: Sep 10, 2014
12. Sep 10, 2014

### lexie47

with numbers plugged in its V_pg = √(70)^2 - (20)^2 = 67 km/hr

13. Sep 10, 2014

### billy_joule

The planes speed is 70km/hr east RELATIVE TO THE AIR that's what this statement means:

"If the speed of the aircraft in the absence of wind is 70 km/h"

The air speed is 20km/hr south.

The resultant velocity is the sum of these two velocity vectors; add these two vectors, do not subtract one from the other.

14. Sep 10, 2014

### lexie47

But the velocity of the plane WRT the ground is due E and V of the wind WRT the ground is due S; since those two vectors from a right triangle, that leaves V of the plane WRT the wind as the hypotenuse, so the vectors have to be subtracted since we're solving for V of the plane WRT the ground.

15. Sep 10, 2014

### billy_joule

No it is not. This is the where you went wrong.

The question asks for the velocity WRT the ground;

you don't know it yet, you need to calculate it. It is not due east. It is not 70km/hr. The southern wind blows the plane off course and alters it's velocity WRT ground.

This is the airspeed of the plane. It is the speed of the plane relative to the air. in the absence of any wind it is also the speed relative to the ground.

Once there is wind (say 20km/hr to the south for example..) this no longer applies (hence boldface).

Consider this analogy;

You are standing still on a large treadmill that is heading east at 7m/s.
Your velocity relative to the ground is 7m/s. Your velocity relative to the treadmill is 0 m/s

You then start walking across the treadmill south a 2m/s.

Now you are traveling 7 m/s south AND ALSO 2m/s east. Do you think you velocity relative to the ground will be greater than 7 m/s? What would the vectors look like?