Question

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​

Answer 1

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.