Question:
The velocity of a particle is $\mathrm{v}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$. Its position is $\mathrm{x}=0$ at $\mathrm{t}=0$; then its displacement after time $(\mathrm{t}=1)$ is :
Correct Option: , 2
Solution:
(2) $\mathrm{v}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$
$\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}$
$\int \mathrm{ds}=\int_{0}^{1}\left(\mathrm{v}_{0}+\mathrm{gt}+\mathrm{Ft}^{2}\right) \mathrm{dt}$
$\mathrm{s}=\left[\mathrm{v}_{0} \mathrm{t}+\frac{\mathrm{gt}^{2}}{2}+\frac{\mathrm{Ft}^{3}}{3}\right]_{0}^{\mathrm{l}}$
$\mathrm{s}=\mathrm{v}_{0}+\frac{\mathrm{g}}{2}+\frac{\mathrm{F}}{3}$