Question

person walks in the following pattern: 2.3 km north, then 2.5 km west, and finally 5.4 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

Answer 1

a.3.86 Km

b.Direction=233.3 degree

Step-by-step explanation:

We are given that

Person wale in North direction,DE=2.3 Km

Person walk in West direction==BD=2.5 Km

Person walk in South direction,EC=5.4 Km

a.We have to find the distance between initial and final position.

Vector AB=-2.5 Km

BD=-EA=-2.3 Km

CA=y=-(EC-EA)=-(5.4-2.3))=-3.1 m

AB=-DE=x=-2.5 Km

Magnitude of resultant vector=
`√(AB^2+CA^2)=√((-2.5)^2+(-3.1)^2)`

Magnitude of resultant vector=3.86 Km

Hence, the distance between initial and final position=3.86 Km

b.Direction=
`\theta=tan^(-1)((y)/(x))`

Direction=
`\theta=tan^(-1)((-3.1)/(-2.3))\approx 53.3^(\circ)` S of W

x-coordinate and y-coordinate are negative therefore, the angle lies in third quadrant.

When we measured from East then

`\theta=53.3+180=233.3^(\circ)`

Therefore, the direction=233.3 degree

Answer 2

given,

Distance travel = 2.3 km north; 2.5 km west; 5.4 km south

distance travel by the person = 2.3 + 2.5 + 5.4

= 10.2 Km

angle at which bird fly from the starting point

`tan \theta = (y)/(x)`

`tan \theta = (-3.1)/(-2.5)`

`tan \theta = 1.24`

`\theta =tan^(-1)(1.24)`

`\theta =51.12^0`

Angle from the original position is 51.12° S of W