Answer
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
[tex]tan \theta = \dfrac{y}{x}[/tex]
[tex]tan \theta = \dfrac{-3.1}{-2.5}[/tex]
[tex]tan \theta = 1.24[/tex]
[tex]\theta =tan^{-1}(1.24)[/tex]
[tex]\theta =51.12^0[/tex]
Angle from the original position is 51.12° S of W
Answer:
a.3.86 Km
b.Direction=233.3 degree
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=[tex]\sqrt{AB^2+CA^2}=\sqrt{(-2.5)^2+(-3.1)^2}[/tex]
Magnitude of resultant vector=3.86 Km
Hence, the distance between initial and final position=3.86 Km
b.Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
Direction=[tex]\theta=tan^{-1}(\frac{-3.1}{-2.3})\approx 53.3^{\circ}[/tex] S of W
x-coordinate and y-coordinate are negative therefore, the angle lies in third quadrant.
When we measured from East then
[tex]\theta=53.3+180=233.3^{\circ}[/tex]
Therefore, the direction=233.3 degree
