Question 293 of 308

An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin. When the object is at $x=+2m$, its velocity is $-4\hat{j}m{s}^{-1}$ . The object’s velocity $(v)$ and acceleration $(a)$ at $x=–2m$ will be

A$v=4\hat{i}m{s}^{-1},a=8\hat{j}m{s}^{-2}$B$v=4\hat{j}m{s}^{-1},a=8\hat{i}m{s}^{-2}$C$v=-4\hat{j}m{s}^{-1},a=8\hat{i}m{s}^{-2}$D$v=-4\hat{i}m{s}^{-1},a=-8\hat{j}m{s}^{-2}$

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Answer: (B) $v=4\hat{j}m{s}^{-1},a=8\hat{i}m{s}^{-2}$

As the speed is constant, only centripetal acceleration will act(towards the centre of the circular path), ${a}_{c}=\dfrac{{v}^{2}}{r}=\dfrac{{4}^{2}}{2}=8m{s}^{-2}$ towards the centre which at the given time will be towards right, hence ${\vec{a}}_{c}=8\hat{i}m{s}^{-2}$. Velocity of the particle at this instant will in upward direction and hence $\vec{v}=4\hat{j}m{s}^{-1}$.