The velocity of a particle $V=U+mt+n{t}^2$ its position is $x=0$ at $t=0$. Then its displacement  after a unit time

$$\begin{gathered} \begin{array}{l}\text{ at }\mathrm{t}=0,\mathrm{x}=0\\ \mathrm{V}=\mathrm{U}+\mathrm{mt}+{\mathrm{nt}}^2\\ differential form of velocity=\frac{\mathrm{dx}}{\mathrm{dt}}, substitute in above equation\\ \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{U}+\mathrm{mt}+{\mathrm{nt}}^2\\ \Rightarrow \mathrm{dx}=\left(\mathrm{U}+\mathrm{mt}+{\mathrm{nt}}^2\right)\mathrm{dt}\\ integrating to get displacement,\\ \Rightarrow {\int }_0^{\mathrm{x}} \mathrm{dx}={\int }_0^{\mathrm{t}} \mathrm{Udt}+{\int }_0^{\mathrm{t}} \mathrm{mtdt}+{\int }_0^{\mathrm{t}} {\mathrm{nt}}^2\mathrm{dt}\end{array} \\ \begin{array}{l}\Rightarrow \mathrm{x}=\mathrm{Ut}+\frac{{\mathrm{mt}}^2}{2}+\frac{{\mathrm{nt}}^3}{3}\\ \text{ In unit time , substitute t=1sec}\\ \text{ The displacement }\mathrm{S}=\mathrm{U}+\frac{\mathrm{m}}{2}+\frac{\mathrm{n}}{3}\end{array} \end{gathered}$$