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`.

Options

A. `12 m//s`
B. `18 m//s`
C. `27 m//s`
D. `36 m//s`

Answer

B

Explanation

`a=(dv)/(dt)=3t^(2)+2t+2`
`int_(2)^(v)dv=int_(0)^(2)(3t^(2)+2t+2)dt`
`|v|_(2)^(v)=|t^(3)+t^(2)+2t|_(0)^(2)`
`v-2={(2)^(3)+(2)^(2)+2(2)}-{0}`
`v=18 m//s`