Question

A wildlife researcher is tracking a flock of geese. The geese fly 4.0km due west, then turn toward the north by 40 degrees and fly another 3.5km.

Part A: How far west are they of their initial position?
Part B: What is the magnitude of their displacement?

Answer 1

Answer:

(a). The distance from the initial position is 6.68 km.

(b).  The magnitude of their displacement is 7.05 km.

Explanation:

Given that,

Geese fly 4.0 km due west, then turn to north by 40° and fly another 3.5 km.

(a). We need to calculate the distance from the initial position

Using formula of distance

$D=AB+BC\cos\theta$

Put the value into the formula

$D=4.0+3.5\cos40$

$D=6.68\ km$

(b). We need to calculate the magnitude of their displacement

Using formula of displacement

$AC=\sqrt{CD^2+AD^2}$

$AC=\sqrt{(3.5\sin40)^2+(6.68)^2}$

$AC=7.05\ Km$

Hence, (a). The distance from the initial position is 6.68 km.

(b).  The magnitude of their displacement is 7.05 km.

Answer 2

Answer:

a.6.68 km

b.7.01 km

Explanation:

The gees fly due to west=4.0 km

$\theta=40^{\circ}$

Geese fly another distance due to north=3.5 km

a.We have to find the west distance from their initial position

From below diagram

The x-component of the total distance that geese flies which is the west distance from the initial position

$D_x=4.0+3.5cos40^{\circ}$

$D_x=4.0+2.68=6.68 km$

B. The vertical component of total distance that gees flies from initial position

$D_y=3.5sin40^{\circ}=2.25km$

Magnitude of displacement ,D=$\sqrt{D^2_x+D^2_y}$

Substitute the values then we get

Magnitude of their displacement,D=$\sqrt{(6.68)^2+(2.25)^2}=7.01 km$

Magnitude of their displacement,D=7.01 km