Question

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

Answer 1

`S =t^(3) + 5 rArr v = (dS)/(dt) =3t^(2)`
For `t =2s, v =3xx 4 =12ms^(-1)`
Tangential acceleration, `a_(t) = (dv)/(dt) = 6t`
For `t =2s, a_(t) =12ms^(-2)`
Centripetal acceleraton
`a_(c) =(v^(2))/(R) =(144)/(20) = 7.2ms^(-2)`
net acceleration `=sqrt(a_(t)^(2) +a_(c)^(2))~~ 14ms^(-2)` .