Question

As shown in the figure, a particle is moving with constant speed π m/s. Considering its motion from A to B, the magnitude of the average velocity is

• A. π m/s
• B. \(\sqrt3\,m/s\)
• C. \(1.5\sqrt3\,m/s\)
• D. \(2\sqrt3\,m/s\)

Answer 1

The Correct Option is C

The problem involves calculating the magnitude of the average velocity of a particle moving with constant speed along a circular path from point A to point B, subtending an angle of \(120^\circ\) at the center.

Given:

• Speed of the particle = \(\pi\) m/s
• Subtended angle at the center = \(120^\circ\)

Let's calculate the average velocity from A to B:

1. Average velocity is defined as the total displacement divided by the total time taken.
2. The displacement between points A and B is the chord length AB.
3. The formula for chord length \( AB \) in a circle is:
   AB = 2 \times R \times \sin\left(\frac{\theta}{2}\right)
   where \( R \) is the radius of the circle and \( \theta \) is the angle in radians.
4. As the angle \(\theta = 120^\circ\), convert this to radians:
   120^\circ = \frac{2\pi}{3} radians.
5. If the particle moves with a constant speed \(\pi\) m/s, the radius \( R \) of the circle can be calculated using the relation:
   v = \omega \cdot R \Rightarrow \pi = \frac{\pi}{R}\cdot R = 1\Rightarrow R = 1 m.
6. Substitute \( R = 1 \) m and \(\theta = \frac{2\pi}{3}\) radians into the chord length formula:
   AB = 2 \times 1 \times \sin\left(\frac{2\pi}{6}\right) = 2 \times \sin\left(\frac{\pi}{3}\right) = 2 \times \frac{\sqrt{3}}{2} = \sqrt{3} m.
7. The time taken to move from A to B with speed \(\pi\) m/s over the arc length:
   Arc length = R\theta = 1 \cdot \frac{2\pi}{3} = \frac{2\pi}{3} m.
   Time = \frac{\text{Arc length}}{v} = \frac{\frac{2\pi}{3}}{\pi} = \frac{2}{3} s.
8. The average velocity magnitude is:
   \text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{\sqrt{3}}{\frac{2}{3}} = \frac{3\sqrt{3}}{2} = 1.5\sqrt{3} m/s.

Therefore, the correct option is \(1.5\sqrt{3}\, \text{m/s}\).