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Question
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 the displacement with the total path length covered by the motorist in each case.
Solution
(i) The path followed by the motorist 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((500 cos30^@ + 500 cos 30^@)^2 + (500)^2)`
`=sqrt((2xx500xxsqrt3/2)^2+(500)^2)`
`=500sqrt4 = 1000 m = 1 km`
Total path length = 500 m + 500 m = 1500 m = 1.5 km
`"Magnitude of displacement"/"Total path length" = 1/1.5 = 2/3 = 0.67`
(ii) After completing six turns, the motorist returns to the starting point, creating a closed hexagonal path.
The displacement is zero since the start and end points coincide |`vec("OC")`| = 0`
The total path length is 6 × 500 = 3,000 metres
Ratio of magnitude of displacement and path length is zero.
(iii) At the eighth turn, the motorist reaches point B. The displacement is the straight-line distance from the starting point O to point B.
Since ∠BOA = ∠OBA = 30°, by trigonometry, the horizontal and vertical components of the displacement are equal to OA cos(30°) and AB cos(30°), respectively.
The total displacement |`vec("OB")`| is twice the horizontal component ON3, because OA = AB, each being 500 m. Thus:
|`vec("OB")`| = 2 × 500 × cos(30°)
= `1000xxsqrt(3)/2 = 500sqrt(3) "m"`
= 0.866 km
The total path length after eight turns is 8 × 500 = 4000 m
`("|"vec("OB")"|")/"Path Length" = (500sqrt(3))/4000`
= `sqrt(3)/8`
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