A Coast Guard cutter detects an unidentified ship at a distance of 19.0 km in the direction 15.0° east of north. The ship is traveling at 29.0 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 40.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north in ° east of north
(b) Find the time required for the cutter to intercept the ship in min

We can approach this problem by breaking it down into vectors. The Coast Guard cutter and the unidentified ship each have velocities, and we need to determine the direction and time for the speedboat to intercept the ship. 

### Step-by-Step Solution:

**Part (a): Finding the Direction of the Speedboat**

1. **Define Vectors:**

   - Let the position of the ship relative to the cutter be represented by a vector \(\mathbf{r}_s\).
   - The velocity of the ship is given as 29.0 km/h at an angle of 40.0° east of north.
   - The velocity of the speedboat is 40.0 km/h, and we need to find the angle at which it should head to intercept the ship.

2. **Relative Position of the Ship:**
   - The initial distance of the ship is 19.0 km at 15.0° east of north. 
   - The position vector of the ship relative to the cutter is:
     \[
     \mathbf{r}_s = 19.0 (\cos 15^\circ \hat{j} + \sin 15^\circ \hat{i})
     \]
     Where \( \hat{i} \) represents the east direction and \( \hat{j} \) represents the north direction.

3. **Velocity Vectors:**
   - The velocity of the ship, \(\mathbf{v}_s\), can be written in component form:
     \[
     \mathbf{v}_s = 29.0 (\cos 40^\circ \hat{j} + \sin 40^\circ \hat{i})
     \]
   - The velocity of the speedboat, \(\mathbf{v}_b\), will also have an unknown angle \( \theta \) relative to north:
     \[
     \mathbf{v}_b = 40.0 (\cos \theta \hat{j} + \sin \theta \hat{i})
     \]

4. **Intercept Condition:**
   To intercept the ship, the relative velocity between the ship and the speedboat must cancel out the initial position vector of the ship. In other words, the relative velocity vector must be aligned in the same direction as the position vector:
   \[
   \mathbf{v}_b - \mathbf{v}_s = -\mathbf{r}_s
   \]
   This ensures that the speedboat is always moving in the correct direction to intercept the ship.

We can solve this equation for the angle \( \theta \), but first, we’ll calculate the components of the position and velocity vectors.

---

**Part (b): Finding the Time Required for Interception**

Once we know the direction of the speedboat, we can calculate the time required to intercept the ship by using the relative speed along the direction of motion.

The time \( t \) is given by:
\[
t = \frac{\text{initial distance}}{\text{relative speed along the intercept course}}
\]

Let me calculate the exact values for both parts.The time required for the speedboat to intercept the ship is approximately **103.6 minutes**.

Let me finalize the calculation for the direction the speedboat should head (part a).The speedboat should head in the direction of approximately **53.7° east of north** to intercept the ship.

### Summary:
- **Direction of speedboat**: 53.7° east of north.
- **Time to intercept**: 103.6 minutes.

Would you like further details or have any questions?

Here are 5 related questions to deepen understanding:
1. How does the speed of the speedboat affect the interception time?
2. What would happen if the ship changed its course during the interception?
3. How could we calculate the time to intercept if the speed of the ship increased?
4. What happens to the interception time if the distance between the cutter and ship increases?
5. How could we determine the point of interception in terms of coordinates?

**Tip**: Always check the relative velocity when working on interception problems, as it determines how quickly two moving objects can meet.

A Coast Guard cutter detects an unidentified ship at a distance of 19.0 km in the direction 15.0° east of north. The ship is traveling at 29.0 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 40.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north in ° east of north. (b) Find the time required for the cutter to intercept the ship in minutes.

This problem involves determining the direction and time for a Coast Guard speedboat to intercept a ship. Learn how to use vector addition, relative velocity, and trigonometric concepts to solve the problem. The ship is 19.0 km away and moving at 29.0 km/h, while the speedboat travels at 40.0 km/h. Suitable for Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation