a boat sails 150km N60°E, then 100km West. What is it's displacement from its starting point?
North 150 cos 60 = 75 km N
East 150 sin 60 - 100 = 29.9 km E
distance = sqrt(75^2+29.9^2)
tan angle north of east = 75/29.9
if you want the compass angle east of north, it is 90 - that
To find the displacement of the boat from its starting point, we can use vector addition. Let's break down the boat's movements into components.
The boat first sails 150km N60°E. This means it travels northeast at an angle of 60° with respect to the north direction. To find the north and east components of this movement, we can use trigonometry.
The north component (N) is given by:
N = 150km * sin(60°)
Therefore, N = 150km * 0.866 ≈ 129.9km
The east component (E) is given by:
E = 150km * cos(60°)
Therefore, E = 150km * 0.5 = 75km
So, the boat's movement of 150km N60°E can be expressed as 129.9km North and 75km East.
Next, the boat sails 100km west. Since this movement is directly towards the west, the east component remains unchanged, and the north component becomes zero.
The west component (W) is -100km (negative since it is in the opposite direction of the positive east direction).
Finally, we can find the displacement by adding up the north and east components:
North component: 129.9km
East component: 75km
West component: -100km
Displacement = √[(East + West)² + North²]
= √[(75km + (-100km))² + (129.9km)²]
= √[(-25km)² + 16,888.01km²]
= √(625km² + 16,888.01km²)
= √17,513.01km²
≈ 132.24km
Therefore, the boat's displacement from its starting point is approximately 132.24km.