Relative velocity of two ships — at an instant, ship A travels 6 km/h due east and ship B travels 8 km/h due north. Find the magnitude of the velocity of B relative to A (in km/h).

Introduction / Context:
Relative velocity is essential in navigation, collision avoidance, and kinematics. It is found by vector subtraction of velocities. For perpendicular velocity directions (east and north), the relative speed forms the hypotenuse of a right triangle with component differences as legs.

Given Data / Assumptions:

• v_A = 6 km/h east (positive x-axis).
• v_B = 8 km/h north (positive y-axis).
• Flat-earth approximation over short duration; instantaneous velocities.

Concept / Approach:

The velocity of B relative to A is v_B/A = v_B − v_A. Since the velocities are orthogonal, the magnitude is found using the Pythagorean theorem from the component differences.

Step-by-Step Solution:

Verification / Alternative check (if short method exists):

Since components are 6 and 8, recall the 6–8–10 Pythagorean triple, confirming 10 km/h immediately.

Why Other Options Are Wrong:

7 and 2 km/h arise from incorrect addition or subtraction of magnitudes; 1 km/h is unrealistic; 14 km/h is simple addition ignoring vector nature.

Common Pitfalls (misconceptions, mistakes):

Adding speeds as scalars; forgetting that relative velocity is a vector difference; mixing units (e.g., km/h with m/s) without conversion.

Final Answer:

10 km/h