A force F is applied to the initially stationary cart. The variation of force with time is shown in the figure. The speed of cart at t = 5 sec is

$\text{F}={\text{kt}}^{\text{2}}$

$50=\text{k}{5}^2$

$\text{k}=\frac{50}{25}=2$

$\text{F}={\text{2t}}^{\text{2}}$

$\text{m}\text{.}\frac{\text{dv}}{\text{dt}}=2{\text{t}}^{\text{2}}$

${\displaystyle {\int }_0^{\text{v}}5\text{dv}}={\displaystyle {\int }_0^{5}2{\text{t}}^{\text{2}}\text{dt}}$

$5{\left(\text{v}\right)}_0^{\text{v}}=\frac{2}{3}{\left({\text{t}}^{\text{3}}\right)}_0^5$

$5\text{v}=\frac{2}{3}\times 125$

$\text{v}=\frac{2}{3}\times 25$

$\text{v}=\frac{50}{3}=16.6\text{m/s}$

$\text{v}=\text{16}\text{.6}\text{m/s}$