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 = + 2   m , its velocity is - 4 j ^   m   s - 1 . The object’s velocity ( v ) and acceleration ( a ) at x = – 2   m will be

Question

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 = + 2 &#160; m , its velocity is - 4 j ^ &#160; m &#160; s - 1 . The object&rsquo;s velocity ( v ) and acceleration ( a ) at x = &#8211; 2 &#160; m will be

Aicharya · Accepted Answer

As the speed is constant, only centripetal acceleration will act(towards the centre of the circular path), $a_{c} = \frac{v^{2}}{r} = \frac{4^{2}}{2} = 8 m s^{- 2}$ 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 $\vec{v} = 4 \hat{j} m s^{- 1}$ .

An object moves at a constant speed along a circular path in a horizontal plane with centre at the origin.…

Solution