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 metres and t is in seconds. The radius of the path is 20 ~m . The acceleration of ' P ' when t=2 ~s is nearly

Step-by-step solution

S=t^3+5 speed, v = ds dt =3 t ^2 and rate of change of speed = dv dt =6 t tangential acceleration at t =2 ~s , a _ t =6 2=12 ~m / s ^2 at t=2 ~s , v =3(2)^2=12 ~m / s centripetal acceleration, a_c= v^2 R = 144 20 ~m / s ^2 net acceleration = a _t^2+ a _ i ^2 14 ~m / s ^2