An object moves at a constant speed along a circular path in a horizontal `XY` plane, with the center at the origin. When the object is at `x=-2m`, its velocity is `-(4m//s)hatj`. What is the object's acceleration when it is `y=2m`

Options

A. `-(8m//s^(2))`
B. `-(8m//s^(2))`
C. `-(4 m//s^(2))hatj`
D. `(4 m//s^(2))hati`

Answer

A

Explanation

`r = 2m, v = 4 m//s`
`a = v^(2)/r = 16/2 = 8 m//s^(2)`
when y = 2m, `a = (-8 m//s^(2))hatj`
because acceleration is directed towards centre.