A Coast Guard cutter detects an unidentified ship at a distance of 21.0 km in the direction 15.0° east of north. The ship is traveling at 21.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel.

(a) If the speedboat travels at 48.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
° east of north

(b) Find the time required for the cutter to intercept the ship.

We can solve this problem using vector analysis. We'll break it down into two parts: (a) finding the direction in which the speedboat should head and (b) calculating the time required for the speedboat to intercept the ship.

### Part (a) Direction of Speedboat

Let’s represent the velocity vectors of both the ship and the speedboat. We'll use coordinate axes where the positive \(x\)-axis is east and the positive \(y\)-axis is north. We will also break down the positions and velocities into their components.

1. **Ship's Position and Velocity**:
   - Initial position of the ship: 
     - Distance = 21.0 km
     - Direction = 15.0° east of north
   - Position components of the ship:
     \[
     x_{\text{ship}} = 21.0 \, \text{km} \cdot \sin(15.0^\circ)
     \]
     \[
     y_{\text{ship}} = 21.0 \, \text{km} \cdot \cos(15.0^\circ)
     \]
   
   - Velocity of the ship:
     - Speed = 21.0 km/h
     - Direction = 40.0° east of north
   - Velocity components of the ship:
     \[
     v_{x_{\text{ship}}} = 21.0 \, \text{km/h} \cdot \sin(40.0^\circ)
     \]
     \[
     v_{y_{\text{ship}}} = 21.0 \, \text{km/h} \cdot \cos(40.0^\circ)
     \]

2. **Speedboat's Velocity**:
   - Speed = 48.0 km/h
   - We don’t know the direction yet, so we will solve for it later.

3. **Relative Position and Velocity**:
   The goal is to find the direction for the speedboat so that it intercepts the ship. The relative velocity between the two (speedboat relative to ship) should point along the line connecting the cutter and the ship.

Let’s calculate the components of the relative velocity and solve for the direction the speedboat should take.

### Part (b) Time to Intercept

The time to intercept is determined by the relative velocity and the initial distance between the cutter and the ship. Once we have the relative velocity, we can find the time using:

\[
t = \frac{\text{distance}}{\text{relative speed}}
\]

---

I'll begin by calculating the necessary components of the ship’s position and velocity.The position components of the ship are:

- \( x_{\text{ship}} = 5.44 \, \text{km} \) (eastward)
- \( y_{\text{ship}} = 20.28 \, \text{km} \) (northward)

The velocity components of the ship are:

- \( v_{x_{\text{ship}}} = 13.50 \, \text{km/h} \) (eastward)
- \( v_{y_{\text{ship}}} = 16.09 \, \text{km/h} \) (northward)

Now, to find the direction of the speedboat, we need to express its velocity in terms of unknown components and set up equations based on the condition that the relative velocity points along the initial line between the cutter and the ship. Let's solve for the required direction.

The speedboat’s velocity components in terms of its unknown angle \( \theta \) (east of north) are:

\[
v_{x_{\text{boat}}} = 48.0 \cdot \sin(\theta)
\]
\[
v_{y_{\text{boat}}} = 48.0 \cdot \cos(\theta)
\]

We will solve for \( \theta \) next.It seems like I can’t do more advanced data analysis right now. Please try again later. If you have any additional questions or want further details on the current steps, feel free to ask!

Here are some related questions that could help expand your understanding:

1. How can vector decomposition be used to solve interception problems in navigation?
2. What is the importance of relative velocity in determining interception paths?
3. How would the problem change if the speedboat had a lower speed than the ship?
4. Can trigonometric functions be used to determine angles in vector-based navigation?
5. How does the Pythagorean theorem apply to calculating distances in problems like this?

**Tip:** When dealing with motion in two dimensions, breaking vectors into their components is a powerful tool for simplifying the analysis.

A Coast Guard cutter detects an unidentified ship at a distance of 21.0 km in the direction 15.0° east of north. The ship is traveling at 21.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept and investigate the vessel.

(a) If the speedboat travels at 48.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
(b) Find the time required for the cutter to intercept the ship.

Solve a Coast Guard problem involving a cutter and an unidentified ship using vector decomposition and relative velocity. Learn how to calculate the interception path and the time required for the speedboat to intercept the ship traveling at 21 km/h. Suitable for advanced high school students (Grades 11-12).

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