(a) 46.6 cm
To calculate the total displacement, we need to resolve each vector along the x- and y- direction.
In the following, we take:
- East as positive x-direction, West as negative x-direction
- North as positive y-direction, South as negative y-direction
Vector A:
[tex]A_x = -31.0 cm\\A_y = 0[/tex]
Vector B:
[tex]B_x = -(32.0 cm) cos 34.0^{\circ}=-25.6 cm\\B_y = -(32.0 cm) sin 34.0^{\circ} = -17.9 cm[/tex]
Vector C:
[tex]C_x = (16.0 cm)cos 57.0^{\circ} =8.7 cm\\C_y = -(16.0 cm)sin 57.0^{\circ} =-13.4 cm[/tex]
Vector D:
[tex]D_x = (16.0 cm) cos 75.0^{\circ} =4.1 cm\\D_y = (16.0 cm) sin 75.0^{\circ} =15.5 cm[/tex]
So the components of the total displacement are:
[tex]R_x = -31.0 cm - 25.6 cm +8.7 cm +4.1 cm =-43.8 cm\\R_y = 0-17.9 cm -13.4 cm +15.5 cm =-15.8 cm[/tex]
So the magnitude of the total displacement is:
[tex]R=\sqrt{R_x^2 + R_y^2}=\sqrt{(-43.8 cm)^2+(-15.8 cm)^2}=46.6 cm[/tex]
(b) 19.8 degrees south of west
First of all, let's notice that:
- both components x and y of the total displacement are negative --> this means that the direction of the displacement is south of west
This means that the angle will be given by
[tex]\theta = tan^{-1} (\frac{R_y}{R_x})[/tex]
where we have
[tex]R_x = -43.8 cm\\R_y = -15.8 cm[/tex]
Substituting, we find
[tex]\theta = tan^{-1} (\frac{-15.8 cm}{43.8 cm})=19.8^{\circ}[/tex]
and this angle is south of west.
