# Relative velocity

1. Sep 25, 2012

### izelkay

1. The problem statement, all variables and given/known data
I would like to check my answers for #3 on this practice exam:
http://cyclotron.tamu.edu/dhy/sample_exam1_phys218.pdf [Broken]

2. Relevant equations

3. The attempt at a solution
b) v(plane-wind)x = 30cos30
v(plane-wind)y= 30sin30 + 160

c) 8.44° East of North
d) 1.4 hours

Is this correct or wrong?

Last edited by a moderator: May 6, 2017
2. Sep 25, 2012

### Simon Bridge

We get this sort of question a lot.
The core problem is that you don't know how to tell if your answers are correct or not.
You need to figure it out because you are supposedly training to tackle problems where nobody knows the answer: so there is nobody to ask. At best you'll find out that someone else ends up with the same figures as you - bet they could have made the same mistakes.

How do you tell?

One way is to see if the physical ideas match the end result.

The problem has a pilot tasked with travelling due north (from CLL to 76F) - a wopping 176 nanometers (I'm guessing "nm" means "nautical miles" here - should be M, NM, or nmi). Airspeed is 160knots. The 30knot wind comes in from 30deg west of North.

If the aircraft were to point due North, then the wind will tend to blow it east and south. To correct the pilot will want to turn into the wind ... towards the west. Make sense?

3. Sep 25, 2012

### izelkay

Yeah, that's how I approached the problem. (The wind is approaching from East of North though, not West of North). I feel like I did it right, but I just don't know. I'll just have to go to my professor's open office hours, I guess. Thank you for replying.

4. Sep 26, 2012

### Simon Bridge

<rereads> Oh yes - so it is :/ well done spotting the deliberate mistake ... erm.
note: sin(30)=1/2 =0.5, cos(30)=√3/2 ≈0.87

Wind vector components would be:
30sin(30) = 15knots (W)
30cos(30) = 25.98knots (S)