The distance after which motorist take a turn = 500 m
As motorist takes a turn at an angle of 60° each time, therefore motorist is moving on a regular hexagonal path. Let the motorist starts from point A and reaches at point D at the end of third turn and at initial point A at the end of sixth turn and at point C at the end of eight turn.
Displacement of the motorist at the third turn = AD
= AO + OD = 500 + 500 = 1000 m
Total path length = AB + BC + CD
= 500+ 500+ 500 = 1500 m
Magnitude of Displacement / Total path length = 1000/1500 = 2/3 = 0.67
At the sixth turn motorist is at the starting point A.
Displacement of the motorist at the sixth turn = 0
Total path length = AB + BC + CD + DE + EF + FA = 500 +500 +500+500+500+500 = 3000m
Magnitude of displacement / Total path length = 0/3000 = 0
At the eight turn, the motorist is point C.
Displacement of the motorist = AC
Using triangle law of vector addition,
Displacement of the motorist at the end of eighth turn is 866 m making an angle 30 degrees with the initial direction of motion.
Total path length = 8 x 500 = 4000m
Magnitude of displacement / total path length = 0.22