Question

An athlete runs along a circular track of diameter 80m. The distance travelled and the magnitude of displacement of the athlete when he covers \(\frac{3^{th}}{4}\) of the circle is ( in m)

• A. \(60\pi,40\sqrt2\)
• B. \(40\pi,60\sqrt2\)
• C. \(120\pi,80\sqrt2\)
• D. \(80\pi,120\sqrt2\)

Answer 1

The Correct Option is A

1. Calculate the Radius and Circumference: 

• Diameter \(d = 80\) m
• Radius \(r = \frac{d}{2} = \frac{80}{2} = 40\) m
• Circumference \(C = 2\pi r = 2 \pi (40) = 80\pi\) m

2. Calculate the Distance Traveled:

• Distance traveled \( = \frac{3}{4} \times C = \frac{3}{4} \times 80\pi = 60\pi\) m

3. Calculate the Displacement:

• The athlete is at the end of two perpendicular radii after completing \(\frac{3}{4}\) of the circle.
• Displacement \( = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2} = 40\sqrt{2}\) m

4. Final Answer:

• Distance traveled: \(60\pi\) m
• Displacement: \(40\sqrt{2}\) m

Therefore, the answer is \(60\pi, 40\sqrt{2}\).

Answer 2

The diameter of the circular track is 80 m. The radius \( r \) is half of the diameter: \[ r = \frac{80}{2} = 40 \, \text{m} \] The athlete runs along the circular track, covering \( \frac{3}{4} \) of the circle. The total circumference of the circle is: \[ C = 2\pi r = 2\pi \times 40 = 80\pi \, \text{m} \] Thus, the distance travelled by the athlete when he covers \( \frac{3}{4} \) of the circle is: \[ \text{Distance travelled} = \frac{3}{4} \times 80\pi = 60\pi \, \text{m} \] Next, we need to calculate the magnitude of the displacement. The displacement is the straight-line distance between the starting point and the final point. Since the athlete covers \( \frac{3}{4} \) of the circle, the angle subtended at the center by the path travelled is: \[ \theta = \frac{3}{4} \times 360^\circ = 270^\circ \] The displacement forms a right triangle, with the hypotenuse being the straight-line distance from the starting point to the final point. This is equivalent to the diagonal of a square formed by the two radii of the circle that meet at the angle of 270°. The magnitude of the displacement is given by: \[ \text{Displacement} = \sqrt{r^2 + r^2} = \sqrt{2r^2} = r\sqrt{2} = 40\sqrt{2} \, \text{m} \] Thus, the distance travelled is \( 60\pi \, \text{m} \) and the displacement is \( 40\sqrt{2} \, \text{m} \).