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

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)`