Question

the velocity of a particle is v₀+gt+ft² . if its position is x=0 at t=0, then its displacement after 1 sec is

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}} \\ \mathrm{dx}=\mathrm{v}\mathrm{dt} \\ \mathrm{integrating}\mathrm{both}\mathrm{side},\mathrm{we}\mathrm{get}, \\ {\int }_0^{\mathrm{x}}\mathrm{dx}={\int }_0^1\left({\mathrm{v}}_0+\mathrm{gt}+{\mathrm{ft}}^2\right)\mathrm{dt} \\ {\overline{)\mathrm{x}}}_0^{\mathrm{x}}={\overline{){\mathrm{v}}_0+\mathrm{g}\frac{{\mathrm{t}}^2}{2}+\mathrm{f}\frac{{\mathrm{t}}^3}{3}}}_0^1 \\ \mathrm{x}-0={\mathrm{v}}_0+\frac{\mathrm{g}{\left(1\right)}^2}{2}+\frac{\mathrm{f}{\left(1\right)}^3}{3}-0 \\ \mathrm{x}={\mathrm{v}}_0+\frac{\mathrm{g}}{2}+\frac{\mathrm{f}}{3} \end{gathered}$$