A Coast Guard cutter detects an unidentified ship at a distance of 16.7 km in the direction 13.8° east of north. The ship is traveling at 24.9 km/h on a course at 39.2° east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speedboat travels 54.2 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
Well, it seems like the Coast Guard is going on a treasure hunt! Ahoy, mateys! Now, let's calculate the direction the speedboat should head.
First, let's break down the information we have. The unidentified ship is traveling at 24.9 km/h on a course 39.2° east of north. The speedboat, on the other hand, is speedy Gonzales with a speed of 54.2 km/h. Our mission is to intercept the vessel, so we need to find the direction.
To do that, we'll consider the vectors involved. The velocity of the unidentified ship is the vector sum of its speed and direction, while the velocity of the speedboat is also the vector sum of its speed and direction.
Now, let's find the direction for the speedboat. We'll use trigonometry here, so hold onto your captain's hat.
The x-component of the ship's velocity is 24.9 km/h * sin(39.2°), and the y-component is 24.9 km/h * cos(39.2°). To intercept it, the speedboat's velocity must cancel out these components to 0 when added to the ship's velocity.
With some calculations, we find that the x-component of the speedboat's velocity must be -16.7 km/h * sin(13.8°), and the y-component must be (-16.7 km/h * cos(13.8°)) - 24.9 km/h.
To calculate the direction, we'll use the arctan function. The ratio of the y-component to the x-component gives us the tangent of the angle. Plugging in the numbers, we find that the angle is approximately -83.1°.
But wait, we can't have a negative angle in a compass bearing! So, we need to add 180° to get the positive version of the angle.
Therefore, the direction the speedboat should head is approximately 96.9° east of north.
Now, go forth, valiant Coast Guard, and bring back the treasure! Just don't forget to share some gold doubloons with this clown bot. Argh!
To find the direction the speedboat should head, we need to determine the relative velocity between the ship and the speedboat. Let's break down the given information:
1. The ship's velocity:
- Speed: 24.9 km/h
- Course: 39.2° east of north
2. The Coast Guard cutter's detected position of the ship:
- Distance: 16.7 km
- Direction: 13.8° east of north
To determine the relative velocity between the ship and the Coast Guard cutter, we can use vector addition.
1. Convert the ship's velocity to its components:
- The northward component: 24.9 km/h * sin(39.2°) = 15.9 km/h
- The eastward component: 24.9 km/h * cos(39.2°) = 19.0 km/h (since it's east of north)
2. Determine the ship's position with respect to the Coast Guard cutter:
- The northward component of the distance: 16.7 km * sin(13.8°) = 3.8 km
- The eastward component of the distance: 16.7 km * cos(13.8°) = 3.6 km (since it's east of north)
3. Calculate the relative velocity between the ship and the Coast Guard cutter:
- The northward component: 15.9 km/h - 0 km/h = 15.9 km/h
- The eastward component: 19.0 km/h - 0 km/h = 19.0 km/h
Now, we need to calculate the direction and magnitude of the relative velocity between the ship and the speedboat. Let's break down the given information:
1. The speedboat's velocity:
- Speed: 54.2 km/h
From the relative velocity components we calculated earlier:
- The northward component: 15.9 km/h
- The eastward component: 19.0 km/h
Using these components, we can determine the direction of the relative velocity:
- Relative velocity's angle (θ): atan(19.0 km/h / 15.9 km/h) = atan(1.19) = 49.7°
To find the direction the speedboat should head, we need to calculate the angle between the relative velocity and the north direction:
- Direction of the speedboat = 90° - θ = 90° - 49.7° = 40.3°
Therefore, the speedboat should head in a direction of 40.3° east of north.
To solve this problem, we can use vector addition to find the resultant velocity of the speedboat in order to intercept the ship.
1. Let's start by breaking down the velocities into their horizontal and vertical components:
- The velocity of the unidentified ship:
- Horizontal component: 24.9 km/h * cos(39.2°)
- Vertical component: 24.9 km/h * sin(39.2°)
- The velocity of the speedboat:
- Horizontal component: 54.2 km/h * cos(theta)
- Vertical component: 54.2 km/h * sin(theta)
2. Now, we need to find the time it would take for the speedboat to reach the ship. We can use the distance between them, which is 16.7 km:
- Time = Distance / Relative Velocity
- Relative velocity = Velocity of the ship - Velocity of the speedboat
3. Calculate the relative velocity components:
- Relative velocity (horizontal) = Ship's velocity (horizontal) - Speedboat's velocity (horizontal)
- Relative velocity (vertical) = Ship's velocity (vertical) - Speedboat's velocity (vertical)
4. Calculate the time:
- Time = 16.7 km / sqrt((Relative velocity (horizontal))^2 + (Relative velocity (vertical))^2)
5. Now that we have the time, we can find the horizontal and vertical distances that the speedboat needs to travel relative to its starting point:
- Distance (horizontal) = Time * Speedboat's velocity (horizontal)
- Distance (vertical) = Time * Speedboat's velocity (vertical)
6. Finally, we can calculate the direction of the speedboat by finding the angle between the horizontal distance and the vertical distance relative to due north using the arctan function:
- Theta = arctan(Distance (horizontal) / Distance (vertical))
Note: To determine the direction as a compass bearing, we need to convert the angle from radians to degrees and adjust it to be with respect to due north.
7. Compute the direction as a compass bearing with respect to due north by:
- Compass bearing = 90° - theta
Using these steps, you can now calculate the direction in which the speedboat should head to intercept the unidentified ship.