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

Given, speed \(v = \pi \, \text{m/s}\), or \(R\omega = \pi\). Therefore, \(\omega = \frac{\pi}{R} \, \text{rad/s}\). Angular displacement \(\theta = 120^\circ = \frac{2\pi}{3}\). Using \(\theta = \omega t\), we get: \[ t = \frac{\theta}{\omega} = \frac{2\pi/3}{\pi/R} = \frac{2R}{3}. \] Linear displacement \(d = 2R \sin( \theta / 2 ) = 2R \sin(60^\circ) = 2R \times \frac{\sqrt{3}}{2} = R\sqrt{3}\).

Hence, average velocity \( = \frac{d}{t} = \frac{R \sqrt{3}}{2R/3} = 1.5 \sqrt{3} \, \text{m/s}\).