Question

a ship leaves port on a bearing of 44.0 degrees, and travels 11.4 mi. the ship turns east and travels 6.2 mi. How far is the ship from port and what is its bearing from.port

Answer 1

Based on the information provided, $\overrightarrow{r}$ = at a bearing angle of = 59.88° .

We are requested to calculate the overall displacement of the ship, both in magnitude and direction, once it departs the harbor under the stated parameters. First, let's define a bearing.

What is bearing in the context given above?

Angles are generally measured anticlockwise from the positive x-axis, whereas bearing angles are evaluated clockwise from the positive y-axis. A bearing is NOT a standard angle measuring tool.

Therefore, a bearing of 44.0° indicates that this is an angle 90.0° - 44.0° = 46°. This is the angle to be used in the calculations.

We're given that the first displacement is 11.4° at an angle of 46° (which was computed earlier).

Step 1  - Split the above into components

x1 = 11.4 Cos 46°

= 7.92m

y1 = 11.4Sin 46.0°

= 11.4 * 0.71933980033

= 8.2m

The second displacement is a simple 6.2mi due east, that is, the positive

x-direction. The components are thus:

Δx = x1 + x2

= 7.92 + 6.2

= 14.12mi

Δy = 1 + y2

= 8.2 + 0

= 8.2

r = √(x total)² + (y total)²

=√[(14.21)²+(8.2)²]

= √(201.9241 +67.24)

= √269.1641

= 16.4mi

The direction of the displacement vector is given by:

tan Ф = (Δy)/(Δx)

= arctan (8.2/14.12)

= arctan (0.5807365439)

= 30.12°

Recall that we were asked for the bearing angle. The bearing angle is what we get when we subtract 30.12° from 90°.

That is:
= 90 - 30.12

= 59.88°

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