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 m , its velocity is - 4 j ^ m s - 1 . The object’s velocity ( v ) and acceleration ( a ) at x = – 2 m will be, v = 4 i ^ m s - 1 , a = 8 j ^ m s - 2, v = 4 j ^ m s - 1 , a = 8 i ^ m s - 2, v = - 4 j ^ m s - 1 , a = 8 i ^ m s - 2, v = - 4 i ^ m s - 1 , a = - 8 j ^ m s - 2

MARKS App · Accepted Answer

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

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