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 22.0 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 60.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.
23.9 . ° east of north
(b) Find the time required for the cutter to intercept the ship.
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To solve this problem, we can break it down into two parts. First, we need to find the direction in which the speedboat should head to intercept the ship. Then, we can calculate the time required for the cutter to intercept the ship.
(a) To find the direction in which the speedboat should head, we can use vector addition. We need to find the resultant vector of the ship's velocity vector and the relative velocity vector of the speedboat with respect to the ship.
1. Convert the ship's velocity from the given direction and speed to components.
- The ship's velocity in the x-direction is 22.0 km/h * cos(40.0°) = 16.84 km/h.
- The ship's velocity in the y-direction is 22.0 km/h * sin(40.0°) = 14.13 km/h.
2. Find the relative velocity vector of the speedboat with respect to the ship.
- The relative velocity in the x-direction is (60.0 km/h - 16.84 km/h) = 43.16 km/h.
- The relative velocity in the y-direction is (0 - 14.13 km/h) = -14.13 km/h.
3. Add the ship's velocity vector and the relative velocity vector to find the resultant vector.
- The resultant velocity in the x-direction is 16.84 km/h + 43.16 km/h = 60.0 km/h (approximately).
- The resultant velocity in the y-direction is 14.13 km/h - 14.13 km/h = 0 km/h.
4. Find the magnitude and direction of the resultant velocity vector.
- The magnitude of the resultant velocity vector is √[(60.0 km/h)^2 + (0 km/h)^2] = 60.0 km/h (approximately).
- The direction of the resultant velocity vector is given by the arctan of the y-component divided by the x-component:
direction = arctan(0 km/h / 60.0 km/h) = 0°.
However, this direction is measured with respect to the x-axis. To express it as a compass bearing with respect to due north, we need to add 90° to the direction.
Therefore, the speedboat should head in a direction of 0° + 90° = 90° (east of north).
(b) To find the time required for the cutter to intercept the ship, we can use the distance between them and the cutter's speed.
1. Use the distance formula to find the distance between the cutter and the ship.
- The distance between them is given as 19.0 km.
2. Determine the time required by dividing the distance by the cutter's speed.
- Time = Distance / Speed
- Time = 19.0 km / 60.0 km/h = 0.317 hours (approximately).
Therefore, the time required for the cutter to intercept the ship is 0.317 hours.