Question

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.

Answer 1

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.
\(\therefore\) 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 \(R\)
\(\therefore\) Magnitude of displacement = \(PR\)

=\(\sqrt{PQ^2 + QR^2 + 2(PQ ).(QR) \cos 60\degree}\)

= \(\sqrt{ 500^2 + 500^2( 2 \times 500 \times 500 \times \cos 60\degree})\)

= \(\sqrt{250000 + 250000 + \bigg(500000 \times \frac{1}{2}\bigg)}\) 
=  \(866.03 \;m\)

\(\beta\) = \(tan^{-1} \bigg( \frac{500 \sin 60\degree }{ 500+ 500 \cos 60 \degree}\bigg)\) = \(30\)

Therefore, the magnitude of displacement is \(866.03 \;m\) at an angle of \(30\degree\) 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\(\degree\)	4000