The acceleration a' in ms-2 of a particle is given by a = `3t^2` +2t+2, where is the time. If the particle starts with a velocity v = 2 m s-at i=0 then velocity at the end of 2 sec is:

Options

A. 12`m s^(-1)`
B. 18 `m s^(-1)`
C. 27 `m s^(-1)`
D. 36 `m s^(-1)`

Answer

b

Explanation

Here `v=u+int_(0)^(t)` adt
`=u+int_(0)^(t)(3t^(2)+2t+2)dt`
`=u+[(3t^(3))/(3)+(2t^(2))/(2)+2t]_(0)^(2)=2+8+4+4`
=18`ms^(-1)`