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^(@)`
`" " = 500 xx (1)/(2) + 500 + 500 xx (1)/(2) = 1000` m
So |Displacement| = 1000 m from O to C
Distance = 500 + 500 + 500 = 1500 m `" "therefore (|"Displacement"|)/("Distance ") = (1000)/(1500) = (2)/(3)`
At VI turn
`because ` initial and final positions are same so |displacement| = 0 and distance = 500 `xx` 6 = 3000 m
`therefore (|"Displacement"|)/("Distance")=(0)/(3000) =0`
At VIII turn
`" "|"Displacement"|= 2(500)cos((60^(@))/(2)) = 1000 xx cos30^(@) = 1000 xx(sqrt3)/(2) = 500 sqrt3` m
Distance `= 500 xx 8 = 4000` m `" "therefore (|"Displacement"|)/("Distance")= (500sqrt3)/(4000) = (sqrt3)/(8)`