A cat walks 43.2 m East, followed by 46.6 m North and further 75.8 m West. What is this cat's displacement vector taking reference from the starting position?

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We start from the initial position and track the cat's movements step by step to determine the displacement.
First leg: 43.2 m East increases the x-coordinate by +43.2 m, y remains 0.
Second leg: 46.6 m North increases the y-coordinate by +46.6 m, x stays at +43.2 m.
Third leg: 75.8 m West decreases the x-coordinate by 75.8 m, bringing x to 43.2 − 75.8 = −32.6 m, while y remains at +46.6 m.
Therefore, the displacement from the starting position is Δx = −32.6 m (west) and Δy = +46.6 m (north).
This can be expressed as 32.6 m West and 46.6 m North, or in Cartesian coordinates as (−32.6, 46.6) m.
Optional cross-check: the magnitude of the displacement is √(32.6^2 + 46.6^2) ≈ √(1062.76 + 2171.56) ≈ √(3234.32) ≈ 56.9 m, and the direction is arctan(46.6/32.6) ≈ 55° north of west. However, the core displacement components are the 32.6 m West and 46.6 m North derived from the vector sum of the movements.

MUF0121 Physics Unit 1 - Semester 2, 2025

A cat walks 43.2 m East, followed by 46.6 m North and further 75.8 m West. What is this cat's displacement vector taking reference from the starting position?

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