Answer:
12.97 km
Explanation:
In order to find the resultant displacement, we have to resolve each of the 3 displacements along the x and y direction.
Taking north as positive y direction and east as positive x-direction, we have:
- Displacement 1: 2.00 km to the north
So
[tex]A_x = 0\\A_y = +2.00 km[/tex]
- Displacement 2: 60.0° south of east for 7.00 km
So
[tex]B_x=(7.00)(cos (-45^{\circ}))=4.95 km\\B_y = (7.00)(sin (-45^{\circ}))=-4.95 km[/tex]
- Displacement 3: 9.50 km 35.0° north of east
So
[tex]C_x=(9.50)(cos 35^{\circ})=7.78 km\\C_y=(9.50)(sin 35^{\circ})=5.45 km[/tex]
So the net displacement along the two directions is:
[tex]R_x=A_x+B_x+C_x=0+4.95+7.78=12.73 km\\R_y=A_y+B_y+C_y=2.00+(-4.95)+5.45=2.50 km[/tex]
So, the distance between the initial and final position is equal to the magnitude of the net displacement:
[tex]R=\sqrt{R_x^2+R_y^2}=\sqrt{12.73^2+2.50^2}=12.97 km[/tex]
