On an open ground,a motorist follows a track that turns to his left by an angle of $60^{\circ}$ 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.

(N/A) The path followed by the motorist is a regular hexagon with side $500\; m$,as shown in the figure.
Let the motorist start from point $P$. The motorist takes the third turn at $S$.
$\therefore$ 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.
$\therefore$ Magnitude of displacement $= 0$.
Total path length $= PQ + QR + RS + ST + TU + UP = 6 \times 500 = 3000\; m$.
The motorist takes the eighth turn at point $R$.
$\therefore$ Magnitude of displacement $= PR = \sqrt{PQ^2 + QR^2 + 2(PQ)(QR) \cos 60^{\circ}}$.
$= \sqrt{500^2 + 500^2 + (2 \times 500 \times 500 \times 0.5)} = \sqrt{250000 + 250000 + 250000} = \sqrt{750000} \approx 866.03\; m$.
The angle $\beta$ with $PQ$ is given by $\tan \beta = \frac{500 \sin 60^{\circ}}{500 + 500 \cos 60^{\circ}} = \frac{\sqrt{3}/2}{1 + 1/2} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$,so $\beta = 30^{\circ}$.
Total path length $= 8 \times 500 = 4000\; m$.

Turn	Magnitude of displacement	Total path length
Third	$1000\; m$	$1500\; m$
Sixth	$0\; m$	$3000\; m$
Eighth	$866.03\; m$ at $30^{\circ}$	$4000\; m$