5/20/2018 Solution guide to problem 4.1

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Set up the coordinate axes so that north is the positive y-direction, and

east is the positive x-direction. The wind is blowing at 30 km/h from thedirection N45W, so that its velocity vector is 30 km/h S45E, which canbe written as vwind =30(cos (45) i sin (45)j). With respect to the stillair, the velocity vector of the plane is 200 km/h N60E, or equivalentlyvplane=200(cos (30) i+sin(30)j). The velocity of the plane relative to theground isv = vwind+ vplane

= (30 cos (45) + 200 cos (30)) i+ (30 sin (45) + 200 sin (30))j=

15

2 +100

3i+

100 152 j 194.4 i+78.8j.

The ground speed is

|v

|

(194.4)2 + (78.8)2

209.8 km/h. The angle the

velocity vector makes with thex-axis is arctan 78.8194.4 22.1. Therefore,the true course of the plane is about N(90 22.1)E = N 67.9E.

5/20/2018 Solution guide to problem 4.1

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