A student walks 14.4 km east and then 8.3 km north. After that, the student walks 2.3 km east. The magnitude of the student's displacement is

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Question restatement: A student walks 14.4 km east, then 8.3 km north, and finally 2.3 km east. The problem asks for the magnitude of the student’s displacement.

First, combine the eastward movements since they point in the same direction: 14.4 km east + 2.3 km east = 16.7 km east. The northward movement remains 8.3 km north.

So the resultant displacement vector has components: x = 16.7 km (east), y = 8.3 km (north).

To find the magnitude of the displacement (the straight-line distance from the starting point to the final position), use the Pythagorean theorem: magnitude = sqrt(x^2 + y^2).

Compute the squares: x^2 = 16.7^2 = 278.89, y^2 = 8.3^2 = 68.89.

Add the squares: 278.89 + 68.89 = 347.78.

Take the square root to obtain the magnitude: sqrt(347.78) ≈ 18.65.

Thus, the displacement magnitude is approximately 18.65 km.

MUF0121 Physics Unit 1 - Semester 2, 2025

A student walks 14.4 km east and then 8.3 km north. After that, the student walks 2.3 km east. The magnitude of the student's displacement is

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