A particle starts from rest and traverses a distance 2x with...

Let ${t}_{1}$,${t}_{2}$ and ${t}_{3}$ be the time taken by the particles to cover the distance $2x$, $4x$ and $6x$ respectively. Let $v$ be the velocity of the particle at $B$ i.e, maximum velocity. The particle moves with uniform acceleration from $A$ to $B$.

For motion from $A$ to $B$.

Average velocity $= \dfrac{0 + v}{2} = \dfrac{v}{2}$

Time taken, ${t}_{1} = \dfrac{2x}{\dfrac{v}{2}}$

Particle moves with uniform retardation from $C$ TO $D$, ${V}_{avg} = \dfrac{0 + v}{2} = \dfrac{v}{2}$

Time taken ,${t}_{3} = \dfrac{6x}{( \dfrac{v}{2} )}$

Total time = ${t}_{1} + {t}_{2} + {t}_{3}$

$t = \dfrac{2x}{v2} + \dfrac{4x}{v} + \dfrac{6x}{v2} = \dfrac{20x}{v}$

Average Velocity over= $\dfrac{6x + 2x + 4x}{20xv} = \dfrac{3}{5}v$
Required Ratio =& $\dfrac{Average Speed}{v} = \dfrac{3}{5}$