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 end of third turn

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Explanation

At III turn
Displacement `=vec(OA)+vec(AB)+vec(BC)=vec(OC)=500 cos 60^(@)+500+500 cos 60^(@)`
`=500xx1/2+500+500xx1/2=1000 m` from O to C
Distance `=500+500+500=1500 m`
So `("Displacement")/("Distance")=1000/1500=2/3`
At VI turn `:'` initial and final positions are same so displacement `=0`
and distance `=500xx6=3000 m`
`:. ("Displacement")/("Distance")=0/3000=0`
At VIII turn : Displacement `=2(500) cos (60^(@)/2)=1000xxcos 30^(@)=1000xxsqrt(3)/2=500sqrt(3) m`
Distance `=500xx8=4000 m :. ("Displacement")/("Distance")=(500sqrt(3))/4000=sqrt(3)/8`