A body has a speed V, 2V and 3V in the first $\frac{1}{3}$,second $\frac{1}{3}$and the remaining $\frac{1}{3}$of the total distance respectively. Its average speed is

Let a body covered total distance AB = s
(i) It covered $\mathrm{AC}=\mathrm{s}/3$
Time t₁= ?
Velocity = v
We know $\mathrm{t}=\frac{\mathrm{s}}{\mathrm{v}}$
$\therefore {\mathrm{t}}_1=\frac{\mathrm{s}}{3\mathrm{v}}$
ii) It covered $\mathrm{CD}=\mathrm{s}/3$
Velocity = 2v
Time taken ${\mathrm{t}}_2=?$
We can have ${\mathrm{t}}_2=\frac{\mathrm{s}}{6\mathrm{v}}$
iii) It covered D to B  $\raisebox{1ex}{$\mathrm{s}$}\!\left/ \!\raisebox{-1ex}{$3$}\right.$
Velocity = 3v
Time taken ${\mathrm{t}}_3=?$
${\mathrm{t}}_3=\frac{\mathrm{s}}{9\mathrm{v}}$
iv) We know
$\begin{array}{l}\mathrm{v}=\frac{\mathrm{s}}{\mathrm{t}}=\frac{{\mathrm{s}}_1+{\mathrm{s}}_2+{\mathrm{s}}_3}{{\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3}=\frac{\mathrm{s}}{{\mathrm{t}}_1+{\mathrm{t}}_2+{\mathrm{t}}_3}\\ {\mathrm{v}}_{\text{Avg }}=\frac{\mathrm{s}}{\frac{\mathrm{s}}{3\mathrm{v}}+\frac{\mathrm{s}}{6\mathrm{v}}+\frac{\mathrm{s}}{9\mathrm{v}}}\\ {\mathrm{v}}_{\text{Avg }}=\frac{\mathrm{s}}{\mathrm{s}\left[\frac{6+3+2}{18\mathrm{v}}\right]}\\ {\mathrm{v}}_{\text{Avg }}=\frac{18\mathrm{v}}{11}\end{array}$