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