Question

Question 4.10 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.

Chapter Motion In A Plane Page 86

Answer 1

If any particle is taking turn to his left at an angle 60° after every "d m ". Then the particle is moving on a regular hexagon .

(A)
Displacement of the motorist at the third turn = AD
= AO + OD = 500m + 500m
= 1000m

Distance covered by motorist at the third turn = AB + BC + CD
= 500m + 500m + 500m
= 1500m

So, displacement/distance = 1000/1500 = 2/3

(B) at the sixth turn motorist at A
Total Displacement for motorist = 0
Total distance covered by motorist = AB + BC + CD + DE + EF + FA = 500×6 = 3000 m

So, displacement/distance = 0/3000 = 0


(C) at the eight turns , the motorist at C .
Displacement for motorist = AC
Use resultant formula,
Angle between AB , BC = 60°
AC = √{AB² + BC² + 2.AB.BC.cos60°}
= √{ 500² + 500² + 2.(500)²×1/2}
= 500√3 m

Distance covered by motorist = perimeter of Hexagon + AB + BC = 3000 + 500 + 500 = 4000 m

So, displacement/distance= 500√3/4000 = √3/8

Answer 2

Answer:

Refer to the attachments for your answer.