The acceleration a in `ms^-2` of a particle is given by `a=3t^2+2t+2`, where t is the time. If the particle starts out with a velocity `v=2ms^-1` at `t=0`, then find the velocity at the end of `2s`.

Explanation

` a= (dv)/(dt) =3 t^2 + 2 t =2`
or ` d v=( 3 t^2 + 2 t + 2) dt` ltBrgt Integarting it within the conditions of motion
we have ` int_2^v dv= int_0^2 ( 3 t^2 + 2 t+2) dt`
or ` (v)_2^v = (t^3+t^2 + 2 t)_0^2`
or ` v-2 =2^3 +2^2 + 2 xx 2` ltbRgt or `1 v= 18 m//s` .