Explanation

(b) 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 $B$ $ie,$ maximum velocity. The particle moves with uniform acceleration from $A$ to $B$.
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For motion for $A$ to $B$.
Average velocity $= \frac{0 + v}{2} = \frac{v}{2}$
Time taken, $t_{1} = \frac{2x}{v/2} = \frac{4x}{v}$
Particle moves with uniform retardation from $C$ to $D$.
Time taken , $t_{3} = \frac{6x}{(0 + v)/2} = \frac{12x}{v}$
Total time $= t_{1} + t_{2} + t_{3}$
$= \frac{4x}{v} + \frac{4x}{v} + \frac{12x}{v} = \frac{20x}{v}$
$v_{av} = \frac{2x + 4x + 6x}{20x/v} = \frac{12v}{20}\ $
or $\frac{v}{v} = \frac{12}{20} = \frac{3}{5}$.