How Does Wind Affect a Plane's Velocity?

[vaironl]

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.

 

[cepheid]

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.

 

[Zula110100100]

Doesn't the problem say a head on wind? where are you getting westerly?

 

[cepheid]

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).

 

[asteriatic]

I can confirm that your approach is correct. The Pythagorean Theorem can be used to calculate the resultant velocity of the plane, as it involves the addition of two vectors (the plane's velocity and the wind's velocity). The resultant velocity can also be found by subtracting the wind's velocity from the plane's velocity. Both methods will give the same answer.

To solve the problem, we can use the formula c^2 = a^2 + b^2, where c represents the resultant velocity, and a and b represent the two vectors being added together. In this case, a = 220m/s and b = 45m/s.

Plugging these values into the formula, we get c^2 = (220m/s)^2 + (45m/s)^2 = 48400m^2/s^2 + 2025m^2/s^2 = 50425m^2/s^2.

To find the resultant velocity, we take the square root of both sides, c = √50425m^2/s^2 = 224.47m/s.

So the resultant velocity of the plane, relative to the ground, is 224.47m/s.

Alternatively, we can also find the resultant velocity by subtracting the wind's velocity from the plane's velocity. In this case, the resultant velocity would be 220m/s - 45m/s = 175m/s.

Both methods give us the same answer, which confirms that your approach is correct. Good job!