An ocean liner leaves New York City and travels 26.9 ° north of east for 217 km. How far east and how far north has it gone? In other words, what are the magnitudes of the components of the ship's displacement vector in the directions (a) due east and (b) due north?

Vo = 217km[26.9o],

a. Xo = 217*Cos26.9 =

b. Yo = 217*sin26.9 =

X=193.52

Y=98.17

To find the magnitudes of the components of the ship's displacement vector in the east and north directions, we can use trigonometry. Let's start by understanding the given information.

The ship travels 26.9° north of east for a distance of 217 km. This means the angle between the displacement vector and the east direction is 26.9°, and the magnitude of the displacement vector is 217 km.

To find the magnitude of the component in the east direction, we can use the cosine of the angle. Similarly, to find the magnitude of the component in the north direction, we can use the sine of the angle.

Let's calculate:

(a) Magnitude of the component in the east direction:
Magnitude_east = 217 km × cos(26.9°)

(b) Magnitude of the component in the north direction:
Magnitude_north = 217 km × sin(26.9°)

To get the numerical values for the magnitudes, we need to calculate the cosine and sine of 26.9° using a scientific calculator or an online calculator. Let's assume we get the following values:

cos(26.9°) ≈ 0.897
sin(26.9°) ≈ 0.442

Substituting these values, we can find the magnitudes:

(a) Magnitude of the component in the east direction:
Magnitude_east ≈ 217 km × 0.897 ≈ 194.85 km

(b) Magnitude of the component in the north direction:
Magnitude_north ≈ 217 km × 0.442 ≈ 95.79 km

Therefore, the magnitudes of the components of the ship's displacement vector in the directions due east and due north are approximately 194.85 km and 95.79 km, respectively.