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

Giveven, ` A= (dv)/ (dt) = 3 t^1 ^2 + 2 +2` ltbRgt or ` dv = ( 3 t^2 + 2 t + 2 t + 2) dt`
When time changes from ` o` to `3 s`, velocity changes from ` 2 ms^(-10` to ` v` (say)
Integrating the above relation withine the condittions of motion we ger
` int _2^v dv = int_)^3 ( 3t^2 + 2 t+ 2 ) dt`
` or ` (v-2) = ( t^3 + t^3 + 2)_0^3 = ( 27 + 9 + 2 xx 3)
`=42`
or ` v=2+ 42 = 44 ms^(-10` .