Question 1178660: A ship is sailing south west with a speed of 16√2 km/h and a submarine is travelling with velocity of (20i + 15j) km/h.
(a) Find the velocity vector of the ship.
(b) Find the velocity of the ship relative to the submarine.
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's break down this problem step-by-step:
**a) Find the Velocity Vector of the Ship:**
1. **Direction:** "South West" means the ship is moving at a 45-degree angle from both the South and West directions.
2. **Angle from East:** Since it's Southwest, the angle measured counter-clockwise from the positive x-axis (East) is 180° + 45° = 225°.
3. **Velocity Components:**
* x-component (Westward): 16√2 * cos(225°)
* y-component (Southward): 16√2 * sin(225°)
4. **Calculate Cosine and Sine:**
* cos(225°) = -√2 / 2
* sin(225°) = -√2 / 2
5. **Velocity Components:**
* x-component: 16√2 * (-√2 / 2) = -16
* y-component: 16√2 * (-√2 / 2) = -16
6. **Velocity Vector:** The velocity vector of the ship is (-16i - 16j) km/h.
**Therefore, the velocity vector of the ship is (-16i - 16j) km/h.**
**b) Find the Velocity of the Ship Relative to the Submarine:**
1. **Velocity of the Ship (Vs):** (-16i - 16j) km/h
2. **Velocity of the Submarine (Vu):** (20i + 15j) km/h
3. **Relative Velocity (Vs_relative_to_Vu):** Vs - Vu
* Vs_relative_to_Vu = (-16i - 16j) - (20i + 15j)
* Vs_relative_to_Vu = (-16i - 20i) + (-16j - 15j)
* Vs_relative_to_Vu = -36i - 31j
**Therefore, the velocity of the ship relative to the submarine is (-36i - 31j) km/h.**
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