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# Starting from rest, a particle rotates in a circle of radius $R=\sqrt 2 m$ with an angular acceleration $a=\large\frac{\pi}{4} $$rad/s^2. The magnitude of average velocity of particle over time when it rotates quarter circle is $(a)\;1.5 m/s \quad (b)\;2 m/s \quad (c)\;1.25m/s \quad (d)\;1 m/s$ Can you answer this question? ## 1 Answer 0 votes V_{av}=\large\frac{displacement}{time} \qquad=\large\frac{PQ}{t} \qquad=\large\frac{\sqrt 2 R}{t} \qquad=\large\frac{2}{t} (since R=\sqrt 2 m) now \theta=\large\frac{1}{2} \times a t^2 \qquad \large \frac{\pi}{2}=\large \frac{1}{2}$$ \times \large \frac{\pi}{4} $$t^2 \qquad t^2=4 \qquad t=2 v_{av}=\large \frac{2}{t}=\frac{2}{2}$$=1 m/s$
Hence d is the correct answer.

edited Mar 14, 2014

+1 vote