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3 votes
A particle has a displacement of 12m towards east, then 5m towards north and then 6m vertically

upwards what is the magnitude of the sum of these displacements ?
1) 10.28m
2) 14.32m
3) 22.42m
4) 30.82m​

User AllenSH
by
8.0k points

1 Answer

5 votes

Answer:

D = 14.31 m

Step-by-step explanation:

It is given that,

Displacement due east is 12 m

Displacement due north is 5 m

Displacement in vertically upward direction is 6 m

We need to find the magnitude of the sum of these displacements. We know that, vertically upward, east and north are mutually perpendicular direction. So, the resultant is given by :


R=√(12^2+5^2+6^2) \\\\R=√(144+25+36) \\\\R=14.31\ m

So, the magnitude of the sum of these displacements is 14.31 m.

User Jordan Foreman
by
7.3k points
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