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)`

Answer

A

Explanation

`S = t^(3) +5`
Linear speed of the particle
`v = (dS)/(dt) = 3 t^(2)`
at `t = 2s v = (3 xx^(2)) m//s`
`12 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)`
`= 7.2 m//s^(2)`
`:. a_(net) = sqrt(a_(1)^(2) +a_(2)^(2)) = sqrt(12^(2) +7.2^(2)) = 14m//s^(2)`