A point `p` moves in counter - clockwise direction on a circular path as shown in the figure . The movement of 'p' is such that it sweeps out in the figure . The movement of 'p' is such that it sweeps out a length `s = t^(3) + 5 ` , where `s` is in metres and ` t` is in seconds . The radius of the path is `20 m` . The acceleration of 'P' when ` t = 2 s` is nearly .

Options

A. `14 m//s^(2)`
B. `13 m//s^(2)`
C. `12 m//s^(2)`
D. `7.2 m//s^(2)`

Explanation

`s=t^(3)+5`
`v=(ds)/(dt)=3t^(2)`
At `t=2s, v=3(2)^(2)=12`
`a_(c)=(v^(2))/(R ) =(12^(2))/(20)=7.2m//s^(2)`
`v=3t^(2)`
`a_(t)=(dv)/(dt)=6t`
At `t=2s, a_(t)=12m//s^(2)`
Resultant acceleration
`a=sqrt(a_(c)^(2)+a_(t)^(2))=sqrt((7.2)^(2)+(12)^(2))`
`=14m//s^(2)`