If the velocity of a particle is $\mathrm{\upsilon }=\mathrm{A}\mathrm{t}+\mathrm{B}{\mathrm{t}}^2$, where $\text{A}$ and $\text{B}$ are constants, then the distance travelled by it between $\text{1}\text{s}$ and $\text{2}\text{s}$ is:

• A

  $\frac{3}{2}\mathrm{A}+4\mathrm{B}$
• B

  $3\mathrm{A}+7\mathrm{B}$
• Correct answer
  C

  $\frac{3}{2}\mathrm{A}+\frac{7}{3}\mathrm{B}$
• D

  $\frac{\mathrm{A}}{2}+\frac{\mathrm{B}}{3}$

Hence, the correct answer is option: 'C'.