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^(2)+5`, wheres is in metres and t is in seconds. The radius of the path is 20m. The acceleration of 'P' when `t=5sqrt((3)/(10))` seconds is nearly:

Options

A. `13m//s^(2)`
B. `12m//s^(2)`
C. `7/2m//s^(2)`
D. `14m//s^(2)`

Answer

D

Explanation

Given `s=t^(3)+5`
`V=3t^(2)`
`a_(t)=6t`
and `a_(cp)=((3t^(2))^(2))/(R)`
at `t=2s" "a_(t)=12ms^(-2)`
`a_(cp)=(9(16))/(20)=(36)/(5)ms^(-2)`
`therefore a_("net")=sqrt(a_(t)^(2)+a_(cp)^(2))=sqrt((12)^(2)+((36)/(5))^(2))`
`=13.994~~14ms^(-2)`.