How Does Wind Affect a Plane's Velocity?

AI Thread Summary
A plane traveling at 220 m/s northeast encounters a headwind of 45 m/s, affecting its resultant velocity. The northerly and easterly components of the plane's velocity are calculated using trigonometric functions. The east-west component is adjusted by subtracting the wind's velocity, while the north-south component remains unchanged. The resultant velocity is determined using the Pythagorean theorem, with the components forming a right triangle. Clarifications were made regarding the direction of the wind, confirming it was indeed a headwind.
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Homework Statement


A plane heads out to L.A with a velocity of 220m/s in a NE direction, relative to the ground, and encounters a wind blowing head-on at 45m/s, what is the resultant velocity of the plane, relative to the ground.

Homework Equations



Pythagorean Theorem ?
Simple subtraction

The Attempt at a Solution


Here is an Image I made in paint to help visualize the solution of the problem... I believe I'm not solving this correctly.http://img443.imageshack.us/img443/9841/physics1.jpg

Uploaded with ImageShack.us

I'm pretty much grabbing in the dark here since I cannot understand everything. Sorry for wasting your time in a way.
 
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NE is at a bearing of 45 degrees so that that "northerly" component of the plane's velocity is (220 m/s)cos(45 deg) and the "easterly" component is (220 m/s)sin(45 deg). Now, the westward wind doesn't affect the velocity along the N-S axis, but it does affect the velocity along the E-W axis. Taking east to be the positive direction and west to be the negative direction along the E-W axis, the new velocity along this axis is:

(220 m/s)sin(45 deg) - 45 m/s = v_ew

(where I've given it a name, to remove clutter).

The resultant total velocity is easy to find. The N-S and E-W components still form a right triangle, so that the sum of their squares is equal to the square of the total velocity (in magnitude):

(v_total)^2 = [(220 m/s)cos(45)]^2 + (v_ew)^2

EDIT: you can also get the angle of the resultant from this same triangle.
 
Doesn't the problem say a head on wind? where are you getting westerly?
 
Zula110100100 said:
Doesn't the problem say a head on wind? where are you getting westerly?

Good call, I got it from the OP's diagram, but I see now that the OP was just wrong.

To the OP: if the two vectors lie along the same line, you can add them by simply adding their magnitudes (or subtracting them if they are in opposite directions).
 
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