Question

On an open ground, a motorist follows a track that turns to his left by an angle of ${60}^o$ after every $500m$. 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

Third turn at S

$PS=PV+VS=500+500$

$=1000m$

$\text{total path length}=PQ+QR+RS$

$500+500+500$

$=1500m$

Sixth turn at $P-P\text{also a starting point Magnitude of displacement}=0$

$\text{Total path length}=PQ+QR+RS+ST+TU+UP$

$=500\times 6$

$=3000m$

$\text{eight turn at R}$

So

$PR=\sqrt{P{Q}^2+Q{R}^2+2\left(QR\right)\left(PQ\right)\cos {60}^{\circ }}$

$=866.03m$

$\text{So},\beta ={\tan }^{-1}\left(\frac{500\sin {60}^{\circ }}{500+500\cos {60}^{\circ }}\right)={30}^{\circ }$.