59.8k views
1 vote
An ocean liner leaves New York City and travels 18.0° north of east for 155 km. What

are the magnitudes of the components of the ship's displacement in the directions
due east and due north?

User IGian
by
7.7k points

1 Answer

1 vote

Final Answer:

The magnitude of the ship's displacement due east is 147.65 km, and the magnitude of the ship's displacement due north is 49.16 km.

Step-by-step explanation:

To find the components of the ship's displacement in the due east and due north directions, we use trigonometry. The displacement forms a right-angled triangle, with the eastward displacement as the adjacent side and the northward displacement as the opposite side.

Given that the ship travels 18.0° north of east for 155 km, we can use trigonometric functions to calculate these components. The eastward component can be found using cosine:

Eastward displacement = 155 km * cos(18.0°) ≈ 147.65 km

The northward component can be found using sine:

Northward displacement = 155 km * sin(18.0°) ≈ 49.16 km

Therefore, the magnitudes of the ship's displacement in the directions due east and due north are approximately 147.65 km and 49.16 km, respectively.

User Romper
by
7.6k points