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

Options

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

Answer

D

Explanation

Here, applying the equation `omega^(2)-omega_(0)^(2)=2alphatheta`
`((v^(2))/(r ))^(2)=02((a)/(r )).theta`
`therefore`    `a=(v^(2))/(2theta.r)`    `(because (theta)=2xx2pirad)`
`a=(80xx80)/(2xx4pixx(20)/(pi))=40ms^(-2)`