How Does Wind Affect the Velocity of an Aircraft?

[oaky6]

A jet airliner moving initially at 300 mi/h due east enters a region where the wind is blowing at 100 mi/h in a direction 27 degress north of east. What is the new velocity of the aircraft to the ground?

This is how i did it: Vag=Vaw + Vwg

Vag=?
Vaw(aircraft relative to wind)= 300
Vwg(wind relative to ground) = 100


drawing out the vectors i have Vag as my hypotonuse, Vaw as my base and Vwg as my (oppositte angle of 27 degrees)...solving for Vag I get 316
but that's not the right answer. I also need to find at what degrees the aircraft is going (north of east) but i dk how to do that?

any suggestions?
i was wondering if someone could draw the vectors out for me if possible

 

[Astronuc]

Consider north up (+j) and east to the right (+i),

So the aircraft is traveling 300 mi/h i, when it encounters a wind which has a velocity of 100 mi/h in the direction which is 27° N from east.

Now the angle of 27° is between the axis pointing east (+i) and the wind vector.

To obtain the new aircraft vector with respect to ground, one adds the aircraft vector and the wind speed vector (with respect to ground), which then gives a resultant, which is the new aircraft speed with respect to ground.

So 300 i + W_i is the new V_i, and the Northward component is V_j = W_j since the northward wind component carries the plane northward.

W = W_i i + W_j j

Then use the Pythagorean theorem to calculate the magnitude of resultant vector, and the angle is the arctangent of the V_j/V_i.

 

[kyle8921]

I would approach this problem by breaking down the velocities into their components. The initial velocity of the aircraft is 300 mi/h due east, which can be represented as 300 mi/h in the x-direction and 0 mi/h in the y-direction. The wind velocity can also be broken down into its components, with a magnitude of 100 mi/h at an angle of 27 degrees north of east. This can be represented as 100*cos(27) mi/h in the x-direction and 100*sin(27) mi/h in the y-direction.

To find the new velocity of the aircraft to the ground, we can add the x-components and y-components separately. The x-component will be 300 mi/h + 100*cos(27) mi/h = 356.7 mi/h. The y-component will be 0 mi/h + 100*sin(27) mi/h = 45.6 mi/h. This gives us a new velocity of 356.7 mi/h at an angle of tan^-1(45.6/356.7) = 7.4 degrees north of east.

In summary, the new velocity of the aircraft to the ground is 356.7 mi/h at an angle of 7.4 degrees north of east. This calculation takes into account the relative velocities of the aircraft and the wind, and can be visualized by drawing the vectors as you described.