Relative bearing & separation — two ships leave the same port Two ships depart a port at the same instant. Ship 1 sails at 30 km/h on bearing N 32° E; Ship 2 sails at 20 km/h on bearing S 58° E. After 2 hours, how far apart are the two ships?

Introduction / Context:
To find the separation between two moving objects with different bearings, convert each motion into x–y components (east and north), sum displacements over equal time, then apply the distance formula between their positions.

Given Data / Assumptions:

• Ship 1 speed = 30 km/h, heading N 32° E.
• Ship 2 speed = 20 km/h, heading S 58° E.
• Time = 2 h; flat Earth approximation over short range.

Concept / Approach:
Displacement = speed * time. For heading N θ E: east component = s·sinθ, north = s·cosθ. For S θ E: east = s·sinθ, north = −s·cosθ. Compute each ship’s (x, y), subtract to get relative vector, then take its magnitude.

Step-by-Step Solution:

Verification / Alternative check:
20√13 ≈ 20 * 3.606 = 72.12 km, matching computed separation.

Why Other Options Are Wrong:
36.5 and 15√6 (~36.7) are too small; 100 km is too large for 2 hours at these speeds.

Common Pitfalls:
Swapping sin/cos for bearings or forgetting the negative north component for S 58° E.

Final Answer:
20√13 Km