- #1

- 1

- 0

A plane is flying due west at 235 km/h and encounters a wind from the north at 45 km/h. What is the plane's new velocity with respect to ground in standard location?

Equations:

Plane Equation (magnitude) (COS Angle) = x and (magnitude) (SIN Angle) =y

Wind Equation (magnitude) (COS ANgle) =x and (magnitude) (SIN Angle) =y

adding your x's and y's

Angle tan-1 y/x

plane angle + wind angle = adjusted angle

adjusted angle - new angle = current angle

Magnitude = square root of x squared and y squared

My attempt:

Before I start I just want to apologise for spelling all my equations out because I have no idea what I am doing here. This is my first time taking a physics course and I am completely lost

THe first issue that I was presented in this problem is finding out the angles. Which I am still unsure about. I know for example, northeast with no angle is a 45 degree angle, but if it is due north, due east, due south and due west i really have no idea. If someone can clarify that for me that would be great.

Plane:

(235 km/h) (COS 0) =235 (235km/h) (SIN 0) =0

* I picked 0 for the angle (due west) because flying in a horizontial line would not have an angle?!*

Wind:

(45 km/ h) (COS 90) = 0 (45 km/h) (SIN 90) =45

* I picked 90 for the angle (due north) because flying in a vertical line would create a 90 degree?!*

Current Angle:

x=235+0=235

y=0+45=45

angle tan-1 = 235/45 =79.11

180+90=270-79.11=190.89

Current Magnitude:

square root (235) squared + (45) squared = 239.27

New velocity to respect of ground 239.27 km/h < 190.89 WN

I tried to make an educated guess on my vector angles, but am unsure if I chose wisely. Besides checking my work can someone clarify that due north and south will always be 90 degrees and due east and west will always be 0 degrees (unless otherwise specified).

Thank you,

Ilene