A particle is performing circular motion of radius 1 m. Its speed is `v=(2t^(2)) m//s`. What will be magnitude of its acceleration at `t=1s`.

Answer

`4sqrt(2) ms^(-2)`

Explanation

`v=2t^(2), a_(T)=(dv)/(dt)=4t rArr a_(T)(1)=4`
`a_(N)=v^(2)/R=((2xx1^(2))^(2))/1=4`
`a=sqrt(a_(T)^(2)+a_(N)^(2))=sqrt((4)^(2)+4^(2))=sqrt(32)`
`a=4sqrt(2)`