A particle move `10 sqrt(3)` m towards east and then 10 m towards north. Find the magnitude and direction of its displacement .

Explanation

In this case AB = `10sqrt(3)` m, BC = 10 m. The initial and the final positions of the particle are A and C respectively.
`:.` The magnitude of displacement ,
`AC=sqrt(AB^(2)+BC^(2))`
=`sqrt(300+100)`
=20
If the angle between AC and AB is `theta`, then tan `theta= (BC)/(AB) = (10)/(10sqrt(3))=(1)/(sqrt(3)) " ""or",theta = 30^(@)`
This angle determines the direction of displacement.