Question

A hiker leaves her camp and walks 3.5 km in a direction of 55° south of west to the lake. After a short rest at the lake, she hikes 2.7 km in a direction of 16° east of south to the scenic overlook. What is the magnitude of the hiker’s resultant displacement? Round your answer to the nearest tenth.  km What is the direction of the hiker’s resultant displacement? Round your answer to nearest whole degree. °  south of west

Answer 1

The magnitude of the hiker’s resultant displacement is 5.6 km
The direction of the hiker’s resultant displacement is 77 degrees

Answer 2

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:

$d_{1x} = -(3.5 km) cos 55^{\circ}=-2.0 km$

$d_{1y} = (3.5 km) sin 55^{\circ}=2.87 km$

$d_{2x} = (2.7 km) sin 16^{\circ}=0.74 km$

$d_{2y} = (2.7 km) cos 16^{\circ}=2.60 km$

So, the components of the total displacement are

$d_x = d_{1x}+d_{2x}=-2.0 km +0.74 km=-1.26 km$ east (so, 1.26 km west)

$d_y=d_{1y}+d_{2y}=2.87 km + 2.60 km=5.47 km$ south

So, the magnitude of the resultant displacement is

$d=\sqrt{d_x^2+d_y^2}=\sqrt{(1.26)^2+(5.47)^2}=5.61 km$

2) Direction

the direction of the hiker's displacement is

$\theta= arctan(\frac{d_y}{d_x})=arctan(\frac{5.47}{1.26})=arctan(4.34)=77.0^{\circ}$ south of west.