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=t3+5 , where s is in metres and t is in seconds. The radius of the path is 20m. The acceleration of ‘P’ when t = 2s is nearly.

s=t3+5 ∴  Speed v=dsdt=3t2         Tangential acceleration (at)=dvdt=6tAt t = 2s,  v=3×(2)2=12ms−1 and at=6×2=12ms−2Centripetal acceleration (ac)=v2R=(12)220=14420ms−2∴  Acceleration of P is                              a=at2+ac2Substituting the values of at  and ac , we get a=14ms−2 ,