Question

a particAl movemoves for 30 m along north then moves for 20m along east finally moves for 3 under root 2 m south west find magnitude of displacment

Answer 1

When the particle moves 30m north, vector obtained is 30j^Further When the partical walks 20m east, vector obtained is 20i^ Further the particles30√2 south west, which cannot be written either along x-axis or y-axis so we take components of this movement along x-axis and y-axis

along x-axis = 30√2×cos45°= 30i^(towards west)

along Y-axis = 30√2×sin45°=30j^(towards south)

Now net movement, 30j^+20i^-30j^-30i^ = -10i^ This shows the displacement of 10m towards west from origin

MARK AS BRAINLIEST

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