# Relative Motion: Vectors

1. Sep 1, 2008

### RedBarchetta

1. The problem statement, all variables and given/known data
A jetliner with an airspeed of 1000 km/h sets out on a 1500-km flight due south. To maintain a southward direction, however, the plane must be pointed 15 degrees west of south. If the flight takes 100 min, what is the wind velocity?

3. The attempt at a solution
This one has been giving me issues. What do you do here? Does 'airspeed' mean that the plane is traveling 1000 km/h with respect to the air? Here is what I've tried:

First of all, I find the components of the jetliner. I take -y to be south and +x to be east.
$$$\overrightarrow {V_{pa} } = 1000 \cdot \cos (15^ \circ )i + 1000 \cdot \sin (15^ \circ )j$$$

So there is the velocity vector with respect to the air. Now from this equation, I can multiply by the time 100min or (5/3)h, to get the displacement vector:

$$$\overrightarrow {R_{net} } = \frac{5} {3}1000 \cdot \cos (15^ \circ )i + \frac{5} {3}1000 \cdot \sin (15^ \circ )j$$$

I'm not quite sure what to do after this....How do you find the wind velocity from this?

My book says the answer is:
$$$(259km/h)i + (65.9km/h)j$$$

2. Sep 1, 2008

### LowlyPion

I haven't worked the numbers, but ...

Perhaps if you separated the X and Y components of vectors for both wind and Jet, and added them separately such that they arrive at the desired coordinate, you could find the wind direction and magnitude more readily?

3. Sep 1, 2008

### RedBarchetta

^^ I'm not sure. The question is, where does the plane need to stop? The question states its traveling on a trip 1500 km south. I'm assuming you want to end up 1500 km due south exactly. So from the point the airplane starts, its displacement needs to be exactly 1500 south. No change in horizontal position can happen. I'm still lost....but I'll try this:

So......

(Vx_plane)=965.926 km/h
(Vy_plane)=258.819 km/h

V_net=(Vx_plane+Vx_wind)i+(Vy_plane+Vy_wind)j

(Vx_plane+Vx_wind) * (5/3h) = 0

(Vy_plane+Vy_wind) * (5/3h) = 1500 km

Plane's X-displacement - (Vx_plane) * (5/3h) = 1609.88 km

Plane's Y-displacement (Vy_plane) * (5/3h) = 431.365 km

So:

X wind:

(965.926 km/h + Vx_wind) * (5/3h) = 0

1609.88 km + (5/3)*Vx_wind = 0

1609.88 km = -(5/3)*Vx_wind

(-3/5)*1609.88 =Vx_wind = -965.926

Y wind:

(258.819 km/h+Vy_wind) * (5/3h) = 1500 km

431.365 + (5/3)*Vy_wind = 1500km

(5/3)*Vy_wind=1500-431.365 km

Vy_wind=1068.64*(3/5) = 641.181

V_wind vector magnitude = 1159.36 so now I can get components:

1159.36 * cos(15)i + 1159.36 * sin(15)j = V_wind

1119.86 i + 300j = V_wind

.........Wrong. This problem is ridiculously hard.

4. Sep 1, 2008

### LowlyPion

This would be my understanding of the problem as well.

You also know that time was 100 minutes = 1.6 hrs to get there.

The next step would be to identify the components of the wind velocity V.

It's angle θ will allow us to identify Vx as V Cosθ and Vy as V Sinθ.
(We will arbitrarily define θ to be relative to positive X axis.)

Now add these components into the Sin15/Cos15 terms of the planes velocity. This gives you for X components

0 = (-1000*Sin15 + V Cosθ ) * 1.6 hours
(Plane is going negative x)
(You multiply by 1.6 hours because V*t = displacement and you want to calculate things in x,y space. Not V space.)

And for the Y components

-1500 = (-1000*Cos 15 +V Sinθ) * 1.6 hours
(Plane is going negative y)

This should be a bit easier to calculate no?

5. Sep 1, 2008

### RedBarchetta

Thanks for your help Pion! So let me confirm this: We are saying that the angle of the wind velocity is arbitrary and what we really solve for is simple the magnitude of the components to end up with our answer.

I solved those formulas and left out the 15 degrees for the Wind's "V cos(theta)" and "V sin(theta)".

I would think that you would plug 15 degrees into every theta, but apparently that's wrong since we actually don't know the angles of the wind velocity.

6. Sep 1, 2008

### LowlyPion

There are 2 equations and 2 unknowns. You don't know V and you don't know θ. That's what you solve for. V is magnitude and θ is direction.

What is arbitrary is what definition you make for θ - that is what angle θ is referenced to. It's value of course needs to be calculated.

Good luck.