.A particle moves along a circle of radius (20)/(pi) m with constant tangential acceleration.If the velocity of the particle is "80m/s" at the end of the second revolution after motion has begun,the tangential acceleration is :-

Options

A. `640 pim//s^(2)`
B. `160 pi m//s^(2)`
C. `40 pi m//s^(2)`
D. `40 m//s^(2)`

Answer

D

Explanation

`omega=(v)/(r )=(80)/(20//pi)=4pi, omega_(0)=0, theta=2pixx2=4pi`
`omega^(2)=omega_(0)^(2)+2alphatheta`
implies `alpha=(omega^(2)-omega_(0)^(2))/(2theta)=(16pi^(2))/(2xx4pi)=2pi`
Tangential acceleration `(a_(t))=r alpha=(20)/(pi)xx2pi=40 m//s^(2)`