Question 1178662
Let's break down this problem step-by-step:

**a) Express the Velocity as a Column Vector:**

1.  **Bearing:** The ship is moving on a bearing of 150 degrees. This means the angle measured clockwise from North is 150 degrees.
2.  **Angle from East:** To express the velocity in terms of x (East) and y (North) components, we need the angle relative to the East axis. The angle from the East axis is 150 degrees - 90 degrees = 60 degrees clockwise from East.
3.  **Velocity Components:**
    * x-component (Eastward): 10 * cos(60°) = 10 * (1/2) = 5 km/h
    * y-component (Southward): 10 * sin(60°) = 10 * (√3 / 2) = 5√3 km/h
4.  **Column Vector:** Since the y-component is southward, it's negative.
    * Velocity vector = [5, -5√3]

**Therefore, the velocity vector is [5, -5√3] km/h.**

**b) Position After t Hours:**

1.  **Initial Position Vector:** The initial position is (20, 15), which can be written as a column vector [20, 15].
2.  **Displacement Vector:** The displacement after t hours is the velocity vector multiplied by t: [5t, -5√3t].
3.  **Position Vector:** The position at time t is the initial position plus the displacement: [20, 15] + [5t, -5√3t] = [20 + 5t, 15 - 5√3t].

**Therefore, the position after t hours is [20 + 5t, 15 - 5√3t].**

**c) Time When Due East of O and Distance:**

1.  **Due East Condition:** When the ship is due East of O, its y-coordinate must be equal to the y-coordinate of O, which is 0.
    * 15 - 5√3t = 0
    * 5√3t = 15
    * t = 15 / (5√3)
    * t = 3 / √3
    * t = √3 hours

2.  **Position at t = √3:**
    * x-coordinate: 20 + 5√3
    * y-coordinate: 15 - 5√3(√3) = 15 - 15 = 0
    * Position: [20 + 5√3, 0]

3.  **Distance from O:** The distance is the x-coordinate since the y-coordinate is 0.
    * Distance = 20 + 5√3 km
    * Distance ≈ 20 + 5(1.732)
    * Distance ≈ 20 + 8.66
    * Distance ≈ 28.66 km

**Therefore:**

* **Time (t):** √3 hours (approximately 1.732 hours)
* **Distance:** 20 + 5√3 km (approximately 28.66 km)