Question

The captain of a plane wishes to proceed due west. The cruising speed of the plane is 256 m/s relative to the air. A weather report indicates that a 44.3-m/s wind is blowing from the south to the north. In what direction, measured with respect to due west, should the pilot head the plane?

Answer 1

Answer:

Angle = 9.965° South of West.

Explanation:

In the question,

The speed of the Plane w.r.t to Wind = 256 m/s

Speed of the Wind w.r.t to Ground = 44.3 m/s

Direction of Wind = North from South

Direction Plane wishes to go = West

So,

Using the vectors,

Speed of Wind w.r.t Ground is given by,

$v_{w,g}=44.3j$

Resultant speed of Plane w.r.t Ground is given by,

$v_{p,g}=-A(i)$

So,

From the Vector's Triangle Sum property,

$v_{p,w}=v_{p,g}-v_{w,g}\\v_{p,w}=-Ai-44.3j$

Therefore, the plane has to move in the direction of 'III quadrant' or 'South-West' direction.

Now,

In the triangle, using Pythagoras Theorem,

$(256)^{2}=(44.3)^{2}+(v_{p,g})^{2}\\v_{p,g}=\sqrt{63573.51}\\v_{p,g}=252.13\,m/s$

The speed of the plane = 252.13 m/s

Now,

$sin\theta=\frac{v_{w,g}}{v_{p,w}}=\frac{44.3}{256}=0.173\\\theta=sin^{-1}(0.173)\\\theta=9.965$

Therefore, the angle at which the plane should fly to go West is = 9.965° South of West.