SOLUTION: A stationary observer O observes a ship S at noon, at a point whose coordinates relative to O are (20, 15); the units are kilometres. The ship is moving at a steady 10 km/h on a be

Question 1178662: A stationary observer O observes a ship S at noon, at a point whose coordinates relative to O are (20, 15); the units are kilometres. The ship is moving at a steady 10 km/h on a bearing of 150 degrees.
a) Express its velocity as a colum vector.
b) Write down in terms of t, its position after t hours.
c) Hence, find the value of t when it is due East of O. How far is it from O at this instant?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
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)

RELATED QUESTIONS
A curve has an equation of {{{y=xe^x}}}. The curve has a stationary point at P.... (answered by Alan3354)

Two ships leave the same port at 10:00 A.M. Ship A travels at the constant rate of 25 MPH (answered by lwsshak3)

The vertices of triangle OAB are the points O (0,0), A (0,2) and B (3,-1) a) the... (answered by greenestamps)

The vertices of triangle OAB are the points O (0,0), A (0,2) and B (3,-1), a) The point... (answered by solver91311)

A tower TR and an observer at O. |OR| = 84 m and the angle of elevation of the top of the (answered by ikleyn)

This is your question number 1 for today, out of 5 allowed. Here's your question for... (answered by richwmiller)

Ship A has position vector (9 i + 3 j) km. Relative to an observer on a second ship B,... (answered by KMST)

AB IS THE DIAMETER OF THE CIRCLE WHOSE CENTER IS O .IF THE COORDINATES OF O ARE (2,1) AND (answered by cngriffith)

AB is a diameter of circle whose center is O. The coordinates of point A are (-2,0). The... (answered by flame8855)