A particle has a displacement of $12$ $m$ towards east and $5$ $m$ towards the north and then $6$ $m$ vertically upward. Find the magnitude of the sum of these displacements.

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Solution

Total distance $= 23$ $m$ path $- A B C D$
Total displacement $= 13 + 6 = 19$ $m$ path $- A C D$

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A particle has a displacement of 12 m towards east and 5 m towards the north and then 6 m vertically upward. Find the magnitude of the sum of these displacements.

Total distance =23 m path −ABCDTotal displacement =13+6=19 m path −ACD

A particle has a displacement of $12$ $m$ towards east and $5$ $m$ towards the north and then $6$ $m$ vertically upward. Find the magnitude of the sum of these displacements.

Total distance $= 23$ $m$ path $- A B C D$
Total displacement $= 13 + 6 = 19$ $m$ path $- A C D$