Question

A particle moves from A to B in uniform circular motion with a velocity $\omega$. Then which of the following is possible :

• A. The average acceleration during A to B = Avg acceleration during A to X.
• B. $\left|a_{avg}\left(AX\right)\right|=\left|a_{avg}\left(XB\right)\right|$
• C. $\Delta V_{AX}=\Delta V_{AB}=0$
• D. $\left|\Delta V_{AB}\right|=\sqrt{3}\left|\Delta V_{AX}\right|$

Answer 1

Answer: (D)

$\left|\Delta V_{AB}\right|=\sqrt{3}\left|\Delta V_{AX}\right|$

Let $v$ be its uniform speed

Magnitude of change in velocity at any two point $=2v\sin \left(\theta /2\right)$

$\left|\Delta v_{AB}\right|=2v\sin \left(120/2\right)=\sqrt{3}v$ and $\left|\Delta v_{AX}\right|=2v\sin \left(60/2\right)=v\Rightarrow \left|\Delta v_{AB}\right|=\sqrt{3}\left|\Delta v_{AX}\right|$

Let time $t_{AX}=t\therefore t_{AB}={2t}_{AX}=2t(\because \omega isuniform)$

Average acceleration $a_{avg}\left(AX\right)=\frac{\Delta v_{AX}}{t_{AX}}=\frac{v}{t}$ and $a_{avg}\left(AB\right)=\frac{\Delta v_{AB}}{t_{AB}}=\frac{\sqrt{3}v}{2t}$

Option $D$ is correct