NCERT Solutions
Class 11 Physics
Motion in a Plane

Q. 3.10
On an open ground, a motorist follows a track that turns to his left by an angle of 600 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.
The path followed by the motorist is a regular hexagon with side 500 m, as shown in the given figure:
Let the motorist start from point P.
The motorist takes the third turn at S. The displacement vector at S = PS
Therefore, Magnitude of displacement = PS = (PV + VS) = (500 + 500) = 1000 m
Total path length = (PQ + QR + RS) = (500 + 500 +500) = 1500 m
The motorist takes the sixth turn at point P, which is the starting point.
Therefore displacement vector is null vector.
Therefore, Magnitude of displacement = 500 + 500 = 1000m
Total path length = (PQ + QR + RS + ST + TU + UP)
= (500 + 500 + 500 + 500 + 500 + 500) = 3000 m
The motorist takes the eight turn at point R.
Displacement vector = RS which is represented by the diagonal of the ||gm PQRV
Magnitude of displacement = PR
= √ ((PQ) 2 + (QR) 2+2 (PQ) (QR) cos600)
=√ (500)2 + (500)2 + (2 x 500 x 500 x cos 600)
=√ (250000) + (250000) + (500000 x (1/2))
=866.03m
β = tan-1(500 sin 600)/ (500 + 500 cos600)
β =300
Therefore, the magnitude of displacement is 866.03 m at an angle of 30° with PR.
Total path length = (Circumference of the hexagon + PQ + QR)
= (6 × 500 + 500 + 500) = 4000 m
Comparison of magnitude of displacement with total path length:-
- 3rd turn = (Displacement)/ (Path length) = (1000m)/ (1500m) =0.666.
- 6th turn = (0m)/(3000m) = 0
- 8th turn = (866.03)/(4000m) 0.216