Canada geese migrate essentially along a north–south direction for well over a thousand kilometers in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north-south direction.)

The given data can be listed below.

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

The term "relative velocity" describes how a moving item might appear to one observer in his own frame.

The diagram for the motion of the bird is given below.

The angle at which the bird is flying is given by,

$\sin \theta =\frac{v_{GA}}{v_{BA}}$

Here, v_GA is the velocity of air and v_BA is the velocity of bird.

Substitute all the values in the above,

$\begin{array}{rcl}\theta & =& {\sin }^{-1}\left(\frac{40}{100}\right)\\ & =& {\sin }^{-1}\left(0.4\right)\\ & =& 23.6^o\end{array}$

Thus, the bird should fly at an angle of 23.6^o in southwest direction.

The velocity component of bird is given by,

$\begin{array}{rcl}{v}_{BG}& =& v_{BA}\cos \varphi \\ & =& 100\times \cos \left(23.6^o\right)\\ & =& 100\times 0.9164\\ & =& 91.64m/s\end{array}$

The time taken by bird is given by,

$t=\frac{D}{v_{BG}}$

Substitute all the values in the above expression.

$\begin{array}{rcl}t& =& \frac{500km}{91.64km/h}\\ & =& 5.45h\end{array}$

Hence, the time taken by the bird to cover a ground distance from north to south is 5.45 h.