# Chapter 3 - Kinematics in Two Dimensions; Vectors - General Problems: 54

65km/h

#### Work Step by Step

Call east the positive x direction and north the positive y direction. Let P denote the Plane, A the Air, and G the Ground. The pair “PG”, for example, represents the plane’s motion relative to the ground. $$\vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}}$$ The speed of the plane relative to the air is given as 185 km/h, and the speed of the plane relative to the ground is 135 km/h. The plane’s velocity relative to the ground is at 285 degrees. Equate the y components of the velocity vectors. $(135 km/h)(sin 285 ^{\circ}) = -185 km/h + v_{AG, y}$ The wind’s velocity relative to the ground has a y-component of 54.6 km/h. Now consider the x components of the relative velocity equation. $(135 km/h)(cos 285 ^{\circ}) = 0 km/h + v_{AG, x}$ The wind’s velocity relative to the ground has an x-component of 34.9 km/h. The magnitude of the wind's velocity is $v = \sqrt {34.9^2 + 54.6^2} = 65km/h$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.