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 ms^(-1)`
B. `14 ms^(-1)`
C. `16 ms^(-1)`
D. `18 ms^(-1)`

Answer

C

Explanation

`d upsilon = a d t`
`therefore int_(2)^(upsilon)d upsilon=int_(0)^(2)(3t^(2)+2t+2)dt=[t^(3)+t^(2)+2t]_(0)^(2)`
or `upsilon=18 ms^(-1)`