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A person walks in the following pattern: 3.4 km north, then 2.2 km west, and finally 4.9 km south. (a) How far and (b) at what angle (measured counterclockwise from east) would a bird fly in a straight line from the same starting point to the same final point?

Respuesta :

Answer:

[tex]FO=2.663\ km[/tex]               [tex]\theta=214.284^{\circ}[/tex]

Explanation:

When a person walks according to the given instruction in the following directions:

  • 3.4 km North
  • then, 2.2 km West
  • then finally, 4.9 km South

The person stands in the South-west direction from the initial point taken at the origin.

Now we observe that the person has covered a distance form the initial point to the final point F.

[tex]Distance=3.4+2.2+4.9[/tex]

[tex]Distance=10.5\ km[/tex]

  • When a bird directly flies from the initial point to the final point then it covers the displacement between the two points which is the shortest distance between the two points.

In triangle OAF:

FA=1.5 km

OA=2.2 km

Here the displacement is given by FO.

Using Pythagoras theorem:

[tex]FO=\sqrt{(OA)^2+(FA)^2}[/tex]

[tex]FO=\sqrt{(2.2)^2+(1.5)^2}[/tex]

[tex]FO=2.663\ km[/tex]

Now the angle AOF:

[tex]\tan\angle AOF=\frac{FA}{OA}[/tex]

[tex]\tan\angle AOF=\frac{1.5}{2.2}[/tex]

[tex]\angle AOF=34.287^{\circ}[/tex]

Therefore the direction of the bird from the east direction:

[tex]\theta=180+34.287[/tex]

[tex]\theta=214.284^{\circ}[/tex]

Ver imagen creamydhaka
lublana

Answer with Explanation:

We are given that

Person walks in north direction=3.4 km

Person walks in west direction=2.2 km

Person walks in south direction=4.9 km

a.We have to find the distance flawn by bird in a straight line from the same starting point to the same final point.

x-Component of resultant vector=[tex]R_x=[/tex]-2.2 km

y-Component of resultant vector=[tex]R_y=-(4.9-3.4)=-1.5km[/tex]

Magnitude of resultant vector =[tex]R=\sqrt{R^2_x+R^2_y}[/tex]

By using the formula of magnitude of vector

r=xi+yj

[tex]\mid r\mid=\sqrt{x^2+y^2}[/tex]

Substitute the values

[tex]R=\sqrt{(-2.2)^2+(-1.5)^2}=2.7 km[/tex]

b.We have to find the direction .

[tex]Direction=\theta=tan^{-1}(\frac{y}{x})[/tex]

By using the formula

[tex]\theta=tan^{-1}(\frac{-1.5}{-2.2})=34.2^{\circ}[/tex]S of W

Resultant vector lies in III quadrant therefore, in III quadrant the angle

[tex]\theta=180+\theta=180+34.2=214.2^{\circ}[/tex] Nof E

Hence, the angle measured from the east in counterclockwise direction=214.2 degree

Ver imagen lublana