# Airplane and vectors

## Homework Statement

A plane flies at an airspeed of 250 km/h. A wind is blowing at 80km/h in the direction 60 degrees east of north. In what direction should the head in order to fly due north.

## Homework Equations

I turned the wind into a vector assuming that the wind is blowing 30 degrees from the x axis

<69.3,40>

I then thought that because the final direction of the plane would be pointing up the Y axis that the i component should be zero, and to make the wind's i component and the airplane's heading i component equal to zero then the air plane would have to head in the following direction of this vector <-69.3, ? >. so I don't know the j component.

I think my main problem is that I don't know how to apply the 250 km/h airspeed to the problem because I want the speed relative to the ground, not to the air.

Can anyone help me out on converting that 250 airspeed to ground speed? or how to apply it in a useful way to the problem?

## The Attempt at a Solution

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tiny-tim
Homework Helper
Hi EV33! Simplest way is to draw a vector triangle …

you know that one side is due north, one side is length 80 at 60º to north, and the third side has length 250 …

put arrows on them to make sure they're the right way round, and use either trig or components to find the unknown angle. Thank you.

That gave me the right answers.

I still have one problem though. I don't understand how I can add two vectors together that are from different reference points. The 250 is in reference to the moving air. The 80 is in reference to the ground. Can someone explain to me how this works?

Maybe I have misunderstood what airspeed means. I take airspeed as meaning the speed relative to the air but does it just really mean its speed in the air relative to the ground?

tiny-tim
Homework Helper
Hi EV33! (just got up :zzz: …)
I still have one problem though. I don't understand how I can add two vectors together that are from different reference points. The 250 is in reference to the moving air. The 80 is in reference to the ground. Can someone explain to me how this works?
All vectors are relative.

For a position vector, that's obvious … the vector is from one position to another.

But it's also true for a velocity vector … it's from one velocity to another!

So just as you can write a position vector as AP, and get equations like AP + PG = AG,

you can write a velocity vector as AP, and get equations like AP + PG = AG,

except perhaps it's clearer if you emphasise that they're velocities by writing VAP + VPG = VAG.

In this case, A P and G are (the velocities of) the ground the air and the plane.

If you always write it this way, you won't go wrong. Maybe I have misunderstood what airspeed means. I take airspeed as meaning the speed relative to the air but does it just really mean its speed in the air relative to the ground?
Yes, airspeed is VAP, the velocity of the plane relative to the air.

So the velocity of the plane relative to the ground is VGP = VGA + VAP = windspeed + airspeed.

(silly name isn't it? … "windspeed" is the correct term for the speed of the wind, but "airspeed" isn't the correct term for the speed of the air! )​