A point P moves in a counter-clockwise direction on a circular path as shown in the figure. The movement of P is such that it sweeps out a length `s= t^(3) +5` where s is in the metre and t is in seconds. The radius of the path is 27 m. The acceleration of P when t= 3 s is _________ `m//s^(2)`. (Take `sqrt(13) = 3.6`)

Answer

`32.40`

Explanation

As `s=t^(3)+5`
`(ds)/(dt)=3t^(2)=v`
`:.a_(g)=(dv)/(dt)=6t`
`|veca|=sqrt(a_(c)^(2)+a_(t)^(2))`
At t=3s
`|veca|=sqrt(((v^(2))/R)^(2)+a_(t)^(2))`
`=sqrt(((9t^(4))/R)^(2)+(6t)^(2))`
`=sqrt((27)^(2)+324)=sqrt(1053)=9sqrt(13)`
`:.|veca|=32.4ms^(-2)`