Firstly it is important to explain what movement is and the difference between displacement and trajectory:
Movement is the change of position of a body at a certain time.
Now, during this movement, the trajectory is the path followed by the body (is a scalar magnitude) and the displacement is the distance in a straight line between the initial and final position (is a vector magnitude).
For example, in the case of the bird flight here (1st figure attached), in a Cartesian coordinate plane where the origin is the initial position of the bird, the green arrows represent the trajectory of the bird, while the red arrow represents the resultant vector of the displacement of the bird.
This resultant vector (325,1750) is the result of a vector sum of the vectors (0,1750) and (325,0)
How do we calculate the module of the displacement?
Well, after drawing the displacement vector, we are able to see that a Right Triangle (with a 90 degree angle) is formed (see 2nd figure attached) with two known sides and the third unknown.
This unknown side value (we will call it [tex]h[/tex]) is the module of the displacement vector we want to find.
According to the Pythagorean Theorem:
In every right triangle, the square of the length of the hypotenuse [tex]h[/tex] is equal to the sum of the squares of the respective lengths of the legs (the other two sides of the triangle).
As shown in equation (1):
[tex]h^{2}={1750m}^{2}+{325m}^{2} [/tex] (1)
Squaring both sides of the equation we have:
[tex]h=\sqrt{{1750m}^{2}+{325m}^{2}}[/tex]
Then:
[tex]h=1779.9227m[/tex]
Which is the module of the displacement vector of the bird flight.
Now, if we want to find the direction of this displacement vector we have to use the sine trigonometric function:
[tex]sin(a)=\frac{OP}{h}[/tex] (2)
Where [tex]OP[/tex] is the Opposite leg to the angle [tex]a[/tex], and [tex]h[/tex] is the hypotenuse.
[tex]sin(a)=\frac{1750m}{1779.9227m}[/tex]
[tex]sin(a)=0.983[/tex]
[tex]a=arcsin(0.983)[/tex]
[tex]a=79.48\º[/tex] (3)
In addition, we know the angles [tex]a[/tex] and [tex]c[/tex] sum to [tex]90\º[/tex], so if we want to find the direction of the vector from the North, we have to find the value of [tex]c[/tex] as follows:
If [tex]90\º=a+c[/tex]
Then:
[tex]c=90\º-a[/tex] (4)
We already know the value of angle [tex]a[/tex] from equation (3). Let’s substitute it on (4):
[tex]c=90\º-79.48\º[/tex]
[tex]c=10.52\º[/tex]
Finally, the resultant displacement of the bird is 1779.9227 m with a direction of [tex]10.52\º[/tex] East of North.
