A particle starts moving along a circle of radius `(20//pi)m` with constant tangential acceleration. If the velocity of the parthcle is `50 m//s` at the end of the second revolution after motion has began, the tangential acceleration in `m//s^(2)` is :

Options

A. `1.6`
B. 4
C. `15.6`
D. `31.2`

Answer

D

Explanation

As `v=omegar`
`rArr" "50=omega((20)/(pi))`
`rArr" "omega=(50pi)/(20)=(5pi)/(2)`
`rArr" "(2pi)/(T)=(5pi)/(2)rArrT=(4)/(5)s`
Given, `t=2T=(8)/(5)s`
`therefore" "omega=alphat`
`rArr" "(5pi)/(2)=alphaxx(8)/(5)rArr=(25pi)/(16)rads^(-2)`
`therefore" "a_(t)=r(alpha)=((20)/(pi))((25pi)/(16))=31.2rads^(-2)`