On an open ground, a motorist follows a track that turns to his left by an angle of `60^(@)` after every `500 m`. Starting from a given turn, The path followed by the motorist is a regular hexagon with side `500 m`, as shown in the given figure specify the displacement of the motorist

at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.

Explanation

i) the path followed by the motorlist will be a closed hexagonal path. Suppose the motorist starts his journey from the, point O. He takes the turn at the point C.
Displacement = `vec(OC)`
Here, `OC=sqrt((OB)^(2)+(BC)^(2)) = sqrt((OF+FB)^(2)+(BC)^(2))`
`=sqrt((2 xx 500 xx sqrt(3)/2)^(2) + (500)^(2))`
`=500sqrt(4)` = 1000 m = 1km
Total path length = 500 m + 500 m + 500 m = 1500 m =1.5km
`("Magnitude of displacement")/("Total path length") = 1/1.5=2/3=0.67`
ii) The motorist will take the sixth turn at O.
Displacement will take the sixth turn at O.
Displacement is zero. So, displacement vector is a null vector.
Path length is 3000m, i.e. 3Km.
Ratio of magnitude of displacement and path length is zero.
iii) The motorist will take the 8th turn at B.
Magnitude of displacement = `2 xx 500 cos30^(@)`= `500sqrt(3)m`=`sqrt(3)/2km` = 0.866 km
Path length = `8 xx 500m = 4km`
Ratio of magnitude of displacement and path length is `(sqrt(3)//2)/4` i.e. `sqrt(3)/8` = 0.22