S=t3+5 ∴ speed, v=dtds=3t2 and rate of change of speed =dtdv=6t ∴ tangential acceleration at t=2 s,at=6×2=12 m/s2 at t=2 s,v=3(2)2=12 m/s ∴ centripetal acceleration, ac=Rv2=20144 m/s2 ∴ net acceleration =at2+ai2 ≈14 m/s2
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 20 m. The acceleration of ' P ' when t=2 s is nearly 
Held on 30 Apr 2010 · Verified 6 Jul 2026.
13 m/s2
12 m/s2
7.2 m/s2
14 m/s2
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