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

Answer

a

Explanation

`s=t_(3)+5`
`v=(ds)/(dt)=3t^(2)`
`a_(T)=(dv)/(dt)=6t`
At `t=2,a_(T)=6xx2=12m//s^(2)`
The centripetal accelration is given by
`a_(C)=(v^(2))/(R)=([3(2)^(2)]^(2))/(20)=((12)^(2))/(20)=(144)/(20)=7.2m//s^(2)`
The net acceleration is given by
`a=sqrt(a_(C)^(2)+a_(T)^(@))=sqrt((7.2)^(2)+(12)^(@))=14m//s^(2)`