Question

Two ships leave a harbor at the same time. One ship travels on a bearing S13°W at 18 miles per hour. The other ship travels on a bearing N75°E at 12 miles per hour. How far apart will the ships be after 3 hours?

a. The distance is approximately 52.3 miles.
b. The distance is approximately 58.6 miles.
c. The distance is approximately 65.2 miles.
d. The distance is approximately 71.5 miles.

Answer 1

The distance between the ships after 3 hours is approximately 71.5 miles.

Step-by-step explanation:

To solve this problem, we can break down the given information into components of motion for each ship.

First, let's consider the ship traveling on a bearing S13°W at 18 miles per hour. Since it is moving south of west, we can break its velocity down into its east and south components. The east component is given by 18 * cos(13°), and the south component is given by 18 * sin(13°).

Similarly, for the ship traveling on a bearing N75°E at 12 miles per hour, we can break its velocity down into its east and north components. The east component is given by 12 * cos(75°), and the north component is given by 12 * sin(75°).

We can then use these components to calculate the distance between the two ships after 3 hours using the Pythagorean theorem. The distance is given by sqrt((east_component1 - east_component2)^2 + (south_component1 - north_component2)^2).

Calculating all the values, we find that the distance between the ships after 3 hours is approximately 71.5 miles.