The acceleration ‘a’ in ${\text{m/s}}^{\text{2}}$ of a particle is given by $\text{a}=3{\text{t}}^{\text{2}}+2\text{t}+2$ where t is the time. If the particle starts out with a velocity $\text{u}=2\text{m/s}$ at t = 0, then the velocity at the end of 2 second is              

$\text{a}=\frac{\text{dv}}{\text{dt}}$

$\text{dv}=\text{adt}$

$\text{dv}=\left(3{\text{t}}^{\text{2}}+2\text{t}+2\right)\text{dt}$

Integrating both sides

${\displaystyle \int \text{dv}}={\displaystyle \int \left(3{\text{t}}^{\text{2}}+2\text{t}+2\right)\text{dt}}$

${\displaystyle \int \text{dv}}={\displaystyle \int 3{\text{t}}^{\text{2}}\text{dt}+{\displaystyle \int 2\text{tdt}+{\displaystyle \int 2\text{dt}}}}$

$\text{v}=3\frac{{\text{t}}^{\text{3}}}{\text{3}}+2\frac{{\text{t}}^{\text{2}}}{\text{2}}+2+\text{C}$

$\text{at}\text{t}=\text{0,}\text{v}=\text{2}$  (According to problem)

$\text{v}=\text{3}\times \frac{0^3}{3}+2\frac{0^2}{2}+2+\text{C}$

$\text{2}=\text{0}+\text{C}$

$\text{v}=\text{2}=\text{C}$

$\text{C}=\text{2}\text{at}\text{t = 0}$

$\text{v}=\text{2}$

$\text{at}\text{t}=\text{2}$

$\text{v}=\frac{{\text{3t}}^{\text{3}}}{\text{3}}+2\frac{{\text{t}}^{\text{2}}}{\text{2}}+2\text{t}+\text{C}$

$\text{v}=\text{3×}\frac{2^3}{3}+2\times \frac{2^2}{2}+2\times 2+2$

$\text{v}=8+4+4+2$

$\text{v}=18\text{m/s}\text{at}\text{t}=2$