Question

A ball moves in a circle of radius $0.5\ \text{m}$ from A to B (quarter circle) in $\sqrt{2}\ \text{s}$. The average velocity is:

• A. 0.25
• B. 0.5
• C. 0.75
• D. 1.5
• E. 1.25

Hint:

Displacement is the straight-line distance between the start and end points, not the curved path length.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Average velocity is defined as the total displacement divided by the total time taken. It is a vector quantity, and its magnitude depends on the shortest distance between the initial and final points, not the path taken.
Step 2: Key Formula or Approach:
1. Average velocity \( \vec{v}_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \). 2. Displacement is the straight-line distance from the initial point (A) to the final point (B).
Step 3: Detailed Explanation:
From the diagram, the ball moves from point A to point B along a circular arc. The points A and B are on a diameter, meaning they are diametrically opposite. The path from A to B is a semicircle. - Radius of the circle, \( r = 0.5 \text{ m} \). - Total time taken, \( t = \sqrt{2} \text{ s} \). The displacement is the length of the straight line connecting A and B. Since A and B are at the ends of a diameter, this distance is equal to the diameter of the circle.
\[ \text{Displacement} = \text{Diameter} = 2 \times r = 2 \times 0.5 = 1.0 \text{ m} \] The question's diagram shows A and B separated by a quarter circle. Let's re-evaluate based on the visual. If A and B are at the ends of a quarter circle, the angle between OA and OB is 90 degrees. The displacement is the chord AB. Using Pythagoras' theorem on the triangle OAB: \[ \text{Displacement}^2 = OA^2 + OB^2 = r^2 + r^2 = 2r^2 \] \[ \text{Displacement} = \sqrt{2r^2} = r\sqrt{2} = 0.5\sqrt{2} \text{ m} \] Now, calculate the magnitude of the average velocity: \[ |\vec{v}_{avg}| = \frac{\text{Displacement}}{\text{Time}} = \frac{0.5\sqrt{2}}{\sqrt{2}} = 0.5 \text{ m/s} \] This result matches the correct answer. The diagram shows a quarter-circle path, not a semicircular one. Step 4: Final Answer:
The average velocity of the ball is 0.5 ms\(^{-1}\).