Question

Please help me with this

Answer 1

Answer:

The correct option for the magnitude of the velocity of the ship relative to the port is...

(33 km/h)

Answer 2

Answer:

The most correct option for the magnitude of the velocity of the ship relative to the port is;

33 km/h

Explanation:

The given parameters of the cargo ship are;

The speed of the cargo ship = 25 km/h

The direction of the cargo ship, θ₁ = 20° North of East

The speed of the current = 10 km/h

The direction of the current, θ₂ = 15° North of East relative to the port

Therefore the direction of the ship relative to the port, 'θ', is given as the sum of θ₁ and θ₂ as follows;

θ = θ₁ + θ₂ = 20° + 15° = 35°

The component of the velocities relative to the port are given as follows;

The velocity of the ship, v₁ = 25·cos(35°)·i + 25·sin(35°)·j

The velocity of the current , v₂ = 10·cos(15°)·i + 10·sin(15°)·j

The velocity of the ship relative to the port, v = (25·cos(35°) + 10·cos(15°))·i + (25·sin(35°) + 10·sin(15°))·j

∴ v ≈ 30.14·i + 16.93·j

The magnitude of 'v' is $\left | v \right |$ = √(30.14² + 16.93²) ≈ 34.6

Therefore, the magnitude of the velocity of the ship relative to the port, $\left | v \right |$ ≈ 34.6 km/h.

Therefore, the closest correct option for the velocity of the ship relative to the port is 33 km/h