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 meters and t is in seconds. The radius of the path is 20 m. The acceleration of 'P' when t = 2 s is nearly
Options:
A . 13 m/s2
B . 12 m/s2
C . 7.2 m/s2
D . 14 m/s2
Answer: Option D : D S=t3+5 Speed,v=dsdt=3t2and rate of change of speed =dvdt=6t ∴tangential acceleration at t=2S,at = 6 × 2=12m/s2 Att=2s,v=3(2)2=12m/s2 ∴centripetal acceleration, =v2R=14420m/s2 ∴ net acceleration=√a2t+a2i≈14m/s2
Submit Comment/FeedBack