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A point P moves in counter-clockwise dir...

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

A

`14 m//s^(2)`

B

`113 m//s^(2)`

C

`12 m//s^(2)`

D

`7.2 m//s^(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
Given :` s= t^(3 ) + 5 `
` therefore v= (ds)/(dt) = 3 t^(2) ` when `t= 2s , v=12 m//s `
Tangential acceleration `a_(t) =(dv)/(dt)= 6t `
when `t= 2 s, a_(t) = 12 m//s ^(2)`
A particle moving aloong a circle , has centripetal and tangential accelerations.
C.P acceleration `(a_(C ))`
` =(v^(2))/(r ) =(12xx12)/(20) =(144)/(20) = 7.2 m//s`
and as `a_(T ) and a_(C )` are perpendicular to each other ,
`a=sqrt(a_(T )^(2)+a_(c)^(2))=sqrt(12^(2)+(7.2)^(2))`
`=sqrt(144+51.8)`
`=sqrt(195.8)=14 m//s^(2) `
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