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(B) The cyclist moves along the circumference from point $P$ to point $S$. Since the path is circular, the points $P, O, S$ form a right-angled triang

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$\sqrt{8} \,km$

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(B) The cyclist moves along the circumference from point $P$ to point $S$. Since the path is circular, the points $P, O, S$ form a right-angled triangle where $O$ is the center of the circle. The displacement is the straight-line distance between the initial point $P$ and the final point $S$. Using the Pythagorean theorem in $	riangle POS$, the displacement $d$ is given by: $d = \sqrt{OP^2 + OS^2}$ Given the radius $R = 2 \,km$, we have $OP = OS = R = 2 \,km$. $d = \sqrt{R^2 + R^2} = \sqrt{2R^2} = R\sqrt{2}$ Substituting the value of $R$: $d = 2\sqrt{2} \,km = \sqrt{4 	imes 2} \,km = \sqrt{8} \,km$.

$A$ cyclist starts from the point $P$ of a circular ground of radius $2 \,km$ and travels along its circumference to the point $S$. The displacement of a cyclist is:

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