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 :

The correct answer is √8 km.

From the image, we can clearly see a circle with center O and four points marked on its circumference:

  • P — at the top (12 o'clock position)
  • Q — at the right (3 o'clock position)
  • R — at the bottom (6 o'clock position)
  • S — at the left (9 o'clock position)

The arrows in the diagram show the cyclist travels along the circumference in the direction P → Q → R → S (i.e., three-quarters of the circle in the clockwise direction).

Key Concept — Displacement vs. Distance:

  • Distance is the total path length traveled along the circumference.
  • Displacement is the straight-line (shortest) distance between the starting point and the ending point, regardless of the path taken.

So we need the straight-line distance from P to S.

Step 1: Set up coordinates.

Place the center O at the origin. The radius of the circle is r = 2 km.

  • P is at the top: coordinates = (0, 2)
  • S is at the left: coordinates = (-2, 0)

Step 2: Apply the Distance Formula.

The displacement = straight-line distance PS =

\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}

Substituting the coordinates of P(0, 2) and S(-2, 0):

\mathrm{PS}=\sqrt{{(-2-0)}^2+{(0-2)}^2}

\mathrm{PS}=\sqrt{{(-2)}^2+{(-2)}^2}=\sqrt{4+4}=\sqrt{8}

Alternative Method (using Pythagoras):

Since P is directly above O and S is directly to the left of O, the triangle OPS has:

  • OP = 2 km (radius, vertical)
  • OS = 2 km (radius, horizontal)
  • Angle POS = 90°

By the Pythagorean theorem:

\mathrm{PS}=\sqrt{{\mathrm{OP}}^2+{\mathrm{OS}}^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}\text{ km}

Therefore, the displacement of the cyclist from P to S is √8 km.