A car travels along a circular track of radius 70 m. It starts from point A and completes exactly half a revolution, reaching point B. What are the magnitudes of the total distance covered by the car and its final displacement from point A, respectively? (Use π=227\pi = \frac{22}{7})

Options

Option A is correct

220 m, 140 m

Option B

140 m, 220 m

Option C

440 m, 0 m

Option D

220 m, 0 m

Explanation

The radius of the circular track is R=70 mR = 70 \text{ m}. For half a revolution:

  1. Distance covered: This is the actual path length traveled by the car, which is half the circumference of the circle. Distance = 12×(2πR)=πR=227×70=22×10=220 m\frac{1}{2} \times (2 \pi R) = \pi R = \frac{22}{7} \times 70 = 22 \times 10 = 220 \text{ m}.
  2. Displacement: This is the shortest straight-line distance from the initial position (point A) to the final position (point B). For half a revolution on a circular path, the final position B is diametrically opposite to the initial position A. Therefore, the displacement is equal to the diameter of the circle. Displacement = 2R=2×70=140 m2R = 2 \times 70 = 140 \text{ m}. Thus, the distance covered is 220 m and the displacement is 140 m.