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

Let distance AB = BC = CD = X
Speed at AB = V
Time at AB  ${\mathrm{t}}_1=\frac{\mathrm{X}}{\mathrm{V}}$
Speed at BC = 2V
Time at BC  ${\mathrm{t}}_2=\frac{\mathrm{X}}{2\mathrm{V}}$
Speed at CD = 3V
Time at CD  ${\mathrm{t}}_3=\frac{\mathrm{X}}{3\mathrm{V}}$
Total Distance  $=\mathrm{X}+\mathrm{X}+\mathrm{X}=3\mathrm{X}$
Total time take  $=\frac{\mathrm{X}}{\mathrm{V}}+\frac{\mathrm{X}}{2\mathrm{V}}+\frac{\mathrm{X}}{3\mathrm{V}}=\frac{6\mathrm{X}+3\mathrm{X}+2\mathrm{X}}{6\mathrm{V}}=\frac{11\mathrm{X}}{6\mathrm{V}}$
Average speed  $=\frac{3\mathrm{X}}{\frac{11\mathrm{X}}{6\mathrm{V}}}=\frac{3\times 6\mathrm{V}}{11}=\frac{18}{11}\mathrm{V}$