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 rArr velocity , v = (ds) /(dt) = 3 t^(2) ` ,brgt Tangential acceleration `a_(t) = (dv)/ (dt) = 6 t ` Radial acceleration ` a_(c) = (v^(2) /( R) = 9t^(4)/( R) `
At ` t= 2s , a _(t) = 6xx2 = 12 m//s^(2)`
`a _(c) = (9xx 16)/ (20) = 7.2 m//s^(2) `
`:.` Resultant acceleration
`= sqrt ( a_(1)^(2) + a_(c)^(2)) = sqrt(( 12) ^(2) + (7.2)^(2)) = sqrt( 144 + 51.84) `
` = sqrt(195.84 = 14 m//s^(2) `