A particle starts from rest with acceleration 2m/s2.  The acceleration of the particle decreases down to zero uniformly during the time interval of 4 s. The velocity of the particle after 2 s is

Since variation of acceleration is uniform so, a=mt+c.According to the problem, at t = 0, a=2m/s2. ∴   2=m×0+c  ⇒ c=2.When t=4s,  a=0   ∴  0=m×4+2  ⇒  m=−12.∴  a=−t2+2   ⇒  ∫0vdv=−12 ∫02t  dt  +  2 ∫02dt  ⇒  v=3m/s.