A point P moves in counter-clockwise direction on figure. The movement of P is such that it sweeps out a length `s=t^(3)+5`, where s is in metre and t is in second. The radius of the pathh is 20 m. the acceleration of P when t=2s is nearly

Options

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

Answer

A

Explanation

Given, `s=t^(3)+3`
`therefore`Tangential velocity of the particle is
`v=(ds)/(dt)=(d)/(dt)(t^(3)+3)=3t^(2)`
Tangential acceleration,
`a_(t)=(dv)/(dt)=(d)/(dt)(t^(3)+3)=3t^(2)`
tangential acceleration,
`a_(t)=(dv)/(dt)=(d)/(dt)(3t^(2))=6t`
at t=2s,
`v=3(2)^(2)=12`m/s.
`a_(t)=6(2)=12` m/s
`therefore` Centripetal acceleration.
`a_(c)=(v^(2))/(R)=((12)^(2))/(20)=(144)/(20)=7.2m//s^(2)`
Net acceleration, `a=sqrt((a_(c))^(2)+(a_(t))^(2))`
`=sqrt((7.2)^(2)+(12)^(2))~~14m//s^(2)`.