Question

The speed of a plane in the absence of wind is 500 km/hr. In the presence of a wind, in order to fly due north at a speed of 450 km/hr with respect to the ground, the pilot must point the plane in a direction 35^o east of due north. Which of the following vectors (in unit vector notation) describes the velocity of the wind with respect to the ground?

Answer 1

Consider that the wind has a velocity V, in a direction of ^o west of due north.
Consider that north is along y-axis and east is along x-axis
Component of wind's velocity due north = V cos
Component of plane's velocity due north = 500 cos35
The net velocity of the plane, due north = 450
V cos + 500 cos (35) = 450
V cos + 500 cos(35) = 450
V cos = 450 - 500 cos(35)    ...(1)

Horizontal component of wind's velocity = V sin
Horizontal component of plane's velocity = 500 sin(35) (-)
Net component of plane's horizontal velocity = 0
V sin - 500 sin(35) = 0
V sin = 500 sin(35) ...(2)

(2) / (1) gives
tan = (500 sin(35)] / [450 - 500 cos(35)]
= 7.094
= tan^-1(7.094)
= 81.98^o

V = 500 sin(35) / sin(81.98)
= 289.62 m/s

V = - V sin + V cos
= - 289.62 sin(81.98) + 289.62 cos(81.98)
= - 286.79 + 40.42