Chapter 3 - Kinematics in Two Dimensions; Vectors - Problems: 44

a. 628km/h, 5.82$^o$ east of south. b. 16.5km

Work Step by Step

Call east the positive x direction and north the positive y direction. a. 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. $$\vec{v_{PG}} = \vec{v_{PA}} + \vec{v_{AG}}$$ $$= (0, -688) km/h + (90.0 cos 45 ^{\circ}, 90.0 sin 45 ^{\circ}) km/h$$ $$= (63.6, -624) km/h$$ $$\vec{v_{PG}} = (63.6, -624) km/h$$ Find the magnitude using the Pythagorean Theorem and the direction using the definition of the tangent. $$v_{PG} = 628 km/h$$ The angle is at -84.2$^o$, or 5.82$^o$ east of south. b. The distance the plane is “off” is just the wind speed multiplied by the time. $$(90 \frac{km}{h})(\frac{11}{60} hour) = 16.5 km$$

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