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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 40-km/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? (Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)

Respuesta :

Answer

given,

taking x-direction at west  and y-direction as north

Wind is moving from east to west

  x = 40 Km/h, y = 0 Km/h

bird has to fly opposite to the wind

speed of wind, x = -40 Km/h

a) angle of bird

[tex]\theta = sin^{-1}(\dfrac{40}{100})[/tex]

       θ = 23.58° South of West

b) Speed of the bird = 100 Km/h

Speed of bird north to south = 100 cos 23.58°

                                                = 91.65 Km/h

time taken to travel 500 Km

[tex]t = \dfrac{500}{91.65}[/tex]

    t = 5.45 hr

time to travel 500 Km from North to south is equal to 5.45 hr.

The velocity of the bird can be directed such that the resultant velocity of

the bird and the wind will be in the southward direction.

(a) To travel directly southward relative to the ground, the bird should head

to an angle direction of approximately South 23.58° East.

(b) The time it would take the bird to travel 500 km from North to South is

approximately 5.46 hours.

Reason:

component of the velocity are;

The magnitude of the velocity of the bird = 100 km/h

Direction of the bird = North-South direction

Magnitude of the velocity of the wind = 40-km/h

Direction of the wind = East-west direction

(a) The component vector of the two velocities are;

Velocity of bird, v[tex]_{bird}[/tex] = -100·j

Velocity of wind, v[tex]_{wind}[/tex] = 40·i

The direction of the velocity of the bird will be southward if it has a

component of velocity to accommodate the velocity of the wind.

Let θ represent the angle of the direction of the bird relative to the

negative x-axis, we have;

-100 × cos(θ) = -40·i

  • [tex]\theta = arcos \left(\dfrac{-40}{-100} \right) \approx 66.42^{\circ}[/tex]

θ ≈ 66.42° in the the third quadrat, given that both the y, and x, values are

negative.

In the South-West direction, the angle is therefore; 90 - 66.42 ≈ 23.58°

Therefore, the bird should fly in the direction south 23.58° east to fly

directly southward relative to the ground

(b) The component of the birds velocity southward when moving in the

new direction, [tex]v_y[/tex], is given as follows;

[tex]v_y[/tex] = 100 × sin(θ)

Therefore;

Southward velocity, [tex]v_y[/tex] = 100 × sin(66.42°)

The time, t, it takes to travel 500 km, is given as follows;

[tex]t = \dfrac{500 \, km}{100 \, km/hr \times sin(66.42^{\circ})} \approx 5.46 \, hr[/tex]

The time it would take the bird to travel 500 km from North to South, t ≈

5.46 hours

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