Given: Distance, s = t3 + 5 At any instant t, velocity is v =
ds
dt
= 3t2 At time t = 2 s ⇒ v = 3×22 = 12 m/s Than , the centripetal acceleration that is produced radially and directed towards the centre is ac =
v2
r
=
122
20
= 7.2 m∕s2 Tangential acceleration is aT =
dv
dt
= 5t At time t = 2 s, magnitude of the tangential acceleration is given by aT = 6 × 2 = 12m∕s2 The magnitude of the resultant acceleration is a = √ac2+aT2 ⇒ a = √(7.2m∕s2)2+(12m∕s2)2 ⇒ a = 14m∕s2 Hence, option 'A' is correct.