Motion in a plane NCERT Solutions Class 11 Physics - Solved Exercise Question 4.10

Question 4.10:

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.

Solution:

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.
∴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.
∴Magnitude of displacement = 0
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
∴Magnitude of displacement = PR
= ( PQ² + QR² + 2 (PQ)(QR) Cos 60⁰)^1/2
= ( 500² + 500² + 2 X 500 X 500 X Cos 60⁰)^1/2
= 866.03 m
β = tan^-1 ( 500 Sin 60⁰ / (500 + 500 Cos 60⁰) ) =  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
The magnitude of displacement and the total path length corresponding to the required turns is shown in the given table

Turn	Magnitude of displacement (m)	Total path length (m)
Third	1000	1500
Sixth	0	3000
Eighth	866.03; 30°	4000