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_t=(dv)/(dt)=6t`
`absveca=sqrt((v^2/R)^2+a_t^2)`
`=sqrt((9t^4)/R^2+(6t^2))`
`=sqrt((27)^2+324)=sqrt(1053)=9`
`sqrt13`
`:. absveca=32.4m//s^2`