Suppose that the motorist starts from the point O along the initial direction OX. After covering OA = 500 m, he turns to his left through 60° along AL and takes the first turn at the point A. After travellingadistance AB = 500 m along AL, he turns to his left through 60° and takes the second turn at the point B and so on. Thus, the motorist will take the third turn at the point C, the sixth turn at the point O (the starting point) and the eighth turn atthepoint B.
(i) At the third turn : The displacement of the motorist at the third turn is

–––

OC

. From the points A and B, draw AN1‌and‌B‌N2 perpendiculars to OC. Then
∣

–––

OC

∣=ON1+N1=N2+ N2 C
Now, ∠COA=∠BCO=60°
∴ON1=OA‌cos‌60°;N1N2=AB
and N2C=BC‌cos‌60°
∴∣

–––

OC

∣=OA‌cos‌60°+AB+BC‌cos‌60°
=500×

1

2

+500+500×

1

2

=1,000 m
The direction of the displacement

–––

OC

at the third turn is 60° to the left of the initial direction.
Total path length up to the third turn is given by
S1=OA+AB+BC=3×500
=1,500 m

∣ ––– OC ∣

S2

=

1,000

1,500

=

2

3

(ii) At the sixth turn : Since at the sixth turn, the motorist reaches the starting point, the displacement of the motorist is a null vector i.e. if S2 is path length upto the sixth turn, then S2=6×500=3,000 m
∣

–––▶

OC

∣=0∴

–––▶ ∣OC ∣

S2

=

0

3,000

=0
(iii) At the eighth turn : At the eighth turn, the displacement of the motorist will be

–––▶

OB

. From the point A, draw AN3 perpendicular to OB.
Then,
|

–––▶

OB

|=ON3+N3B
Now ∠BOA=<OBA=30°
It follows that ON3=OA‌cos‌30°;
N3B=AB‌cos‌30°
∴|

–––▶

OB

|=OA‌cos‌30°+AB‌cos‌30°
=500×

√3

2

+500×

√3

2

=500√3 m
The direction of the displacement at the eighth turn will be 30° to the left of the initial direction.
If S3 is the total length up to the eighth turn, then S3=8×500=4,000m
∴

| –––▶ OB |

S3

=

500√3

4,000

=

√3

8