Engineering High School

A ship has a speed in still water of 10 ms¹ and is pointed in the direction N 30°E, but there is a current of speed 2 ms-¹ flowing towards the direction SE. By applying the Cosine or Sine Rule, find the resultant velocity of the ship, in terms of its speed and a bearing. Give your answers correct to one decimal place.​

Answers

Answer 1

Answer:

Hope this helped ^^

Explanation:

To find the resultant velocity of the ship, we can use vector addition by considering the ship's velocity in still water and the velocity of the current separately.

Let's denote the ship's velocity in still water as Vship = 10 m/s, and the velocity of the current as Vcurrent = 2 m/s.

To find the resultant velocity, we can break down the ship's velocity and the current's velocity into their horizontal (East/West) and vertical (North/South) components.

The horizontal component of the ship's velocity (Vship_h) can be calculated using the cosine rule:

Vship_h = Vship * cos(30°)

Vship_h = 10 * cos(30°) = 10 * (√3 / 2) ≈ 8.7 m/s

The vertical component of the ship's velocity (Vship_v) can be calculated using the sine rule:

Vship_v = Vship * sin(30°)

Vship_v = 10 * sin(30°) = 10 * (1 / 2) = 5 m/s

Now, since the current is flowing towards the direction SE (southeast), which is 45° from the East direction, we can break down the current's velocity into its horizontal and vertical components.

The horizontal component of the current's velocity (Vcurrent_h) can be calculated using the cosine rule:

Vcurrent_h = Vcurrent * cos(45°)

Vcurrent_h = 2 * cos(45°) = 2 * (√2 / 2) = √2 m/s

The vertical component of the current's velocity (Vcurrent_v) can be calculated using the sine rule:

Vcurrent_v = Vcurrent * sin(45°)

Vcurrent_v = 2 * sin(45°) = 2 * (√2 / 2) = √2 m/s

Now, we can find the resultant horizontal velocity (Vresultant_h) by adding the horizontal components of the ship's velocity and the current's velocity:

Vresultant_h = Vship_h + Vcurrent_h

Vresultant_h = 8.7 + √2 ≈ 10.1 m/s

Similarly, we can find the resultant vertical velocity (Vresultant_v) by adding the vertical components of the ship's velocity and the current's velocity:

Vresultant_v = Vship_v + Vcurrent_v

Vresultant_v = 5 + √2 ≈ 6.4 m/s

Finally, we can calculate the magnitude of the resultant velocity (Vresultant) using the Pythagorean theorem:

Vresultant = √(Vresultant_h² + Vresultant_v²)

Vresultant = √(10.1² + 6.4²) ≈ 11.9 m/s

The bearing of the resultant velocity can be calculated using the inverse tangent (tan⁻¹) of the vertical component divided by the horizontal component:

Bearing = tan⁻¹(Vresultant_v / Vresultant_h)

Bearing = tan⁻¹(6.4 / 10.1) ≈ 33.7°

Therefore, the resultant velocity of the ship, in terms of its speed and bearing, is approximately 11.9 m/s at a bearing of 33.7°.

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