1) Magnitude
Let's take south as positive y-direction and east as positive x-direction. Then we have to resolve both displacements into their respective components:
[tex]d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km[/tex]
[tex]d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km[/tex]
[tex]d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km[/tex]
[tex]d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km[/tex]
So, the components of the total displacement are
[tex]d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km[/tex] east (so, 1.26 km west)
[tex]d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km[/tex] south
So, the magnitude of the resultant displacement is
[tex]d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km[/tex]
2) Direction
the direction of the hiker's displacement is
[tex]\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ}[/tex] south of west.
The resultant displacement of the hiker is 2.21 km in the direction of 74⁰ South west.
The resultant displacement of the hiker is calculated as follows;
[tex]d_{x1} = - 3.5 \times cos(55) = - 2 km\\\\d_{y1} = 3.5 \times sin(55) = 2.87 \ km\\\\d_{x2} = 2.7 \times cos (16) = 2.6 \ km\\\\d_{y_2} = -2.7 \times sin(16) = -0.74 \ km[/tex]
[tex]R_x = 2.6 \ km \ - \ 2\ km = 0.6 \ km\\\\R_y = 2.87 \ km - 0.74 \ km = 2.13 \ km[/tex]
[tex]d = \sqrt{R_x^2 + R_y^2} \\\\d = \sqrt{0.6^2 + 2.13^2} \\\\d = 2.21 \ km[/tex]
[tex]\theta = tan^{-1} (\frac{2.13}{0.6} )= 74 \ ^0[/tex]
Learn more about resultant displacement here: https://brainly.com/question/13309193
