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 a length ` s = t ^ 3 + 5` where s is in meters 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 `

Explanation

Distance time relation of the particle, `s=t^(3)+5`
speed of the particle `v=(ds)/(dt)=(d)/(dt) (t^(3)+5)=3t^(2)`
Tangential acceleration, `a_(t)=(dv)/(dt)=(d)/(dt) (3t^(2))=6t`
At time t=2s, Speed of the particle `v=3(2)^(2)=12m//s`
Tangential acceleration, `a_(x)=6(2)=12m//s^(2)`
Centripetal acceleration, `a_(c)=v^(2)/R=((12)^(2))/(20)=(144)/(20)=7.2 m//s^(2)`
Net acceleration `a=sqrt((a_(c))^(2)+(a_(t))^(2))=sqrt((7.2)^(2)+(12)^(2))=14m//s^(2)`