Let ${\mathrm{t}}_1$,${\mathrm{t}}_2$ and ${\mathrm{t}}_3$ be the time taken by the particles to cover the distance $2\mathrm{x}$, $4\mathrm{x}$ and $6\mathrm{x}$ respectively. Let $\mathrm{v}$ be the velocity of the particle at $\mathrm{B}$ $\mathrm{ie}$, maximum velocity. The particle moves with uniform acceleration from $\mathrm{A}$ to $\mathrm{B}$.

For motion from $\mathrm{A}$ to $\mathrm{B}$.

Average velocity $=\frac{0+\mathrm{v}}{2}=\frac{\mathrm{v}}{2}$

Time taken,  ${\mathrm{t}}_1=\frac{2\mathrm{x}}{{\displaystyle \raisebox{1ex}{$\mathrm{v}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Particle moves with uniform retardation from $\mathrm{C}$ TO $\mathrm{D}$, ${\mathrm{V}}_{\mathrm{avg}}=\frac{0+\mathrm{v}}{2}=\raisebox{1ex}{$\mathrm{v}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Time taken ,${\mathrm{t}}_3=\frac{6\mathrm{x}}{\left(\raisebox{1ex}{$\mathrm{v}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}$

Total time $={\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3$

$\mathrm{t}=\frac{2\mathrm{x}}{\raisebox{1ex}{$\mathrm{v}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{4\mathrm{x}}{\mathrm{v}}+\frac{6\mathrm{x}}{\raisebox{1ex}{$\mathrm{v}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{20\mathrm{x}}{\mathrm{v}}$

Average Velocity over $=\frac{6x+2x+4x}{\raisebox{1ex}{$20x$}\!\left/ \!\raisebox{-1ex}{$v$}\right.}=\frac{3}{5}v$

Required Ratio $=\frac{\mathrm{Average}Speed}{\mathrm{v}}=\frac{3}{5}$