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

Answer

D

Explanation

`S=t^(3)+5`
Linear speed of the particle
`v=(dS)/(dt)=3t^(2)`
at t=2s `v=(3xx2^(2))m//s`
Linear acceleration
`a_(1)=(dv)/(dt)=6t`
at t=2s, `a_(1)=12 m//s^(2)`
The centripetal acceleration
`a_(2)=(v^(2))/R=(12^(2))/20 m//s^(2)`
The centripetal acceleration
`a_(2)=(v^(2))/R=(12^(2))/20 m//s^(2)`
`:. a_("net")=sqrt(a_(1)^(2)+a_(2)^(2))=sqrt(12^(2)+7.2^(2))=14 m//s^(2)`