A particle covers half of its journey with a constant speed of ‘v’, half of the remaining part of journey with a constant speed of 2v and the rest of the journey with a constant speed of 4v. Its average speed during the entire journey is            

Given that

A particle travelled from A to D

AD = s

It covered A to B

i.e. $\text{AB}=\frac{\text{s}}{\text{2}}$

With speed $={\text{v}}_{\text{1}}=\text{v}$

It covered B to C

i.e. BC $\text{=}\frac{\text{s}}{\text{4}}$

With speed $={\text{v}}_{\text{2}}=2\text{v}$

It covered C to D

i.e. CD $=\frac{\text{s}}{\text{4}}$

With speed ${\text{v}}_{\text{3}}=4\text{v}$

We can have

${\text{V}}_{\text{BD}}=\frac{{\text{2V}}_1{\text{V}}_2}{{\text{V}}_1+{\text{V}}_2}$

${\text{V}}_{\text{BD}}=\frac{2\left(\text{2V}\right)\left(\text{4V}\right)}{\text{2V}+4\text{V}}=\frac{{\text{16V}}^2}{\text{6V}}=\frac{\text{8V}}{\text{3}}$

            ${\text{V}}_{\text{AD}}=\frac{{\text{2V}}_1{\text{V}}_2}{{\text{V}}_1+{\text{V}}_2}=\frac{2\left(\text{V}\right)\left(\frac{8}{3}\text{V}\right)}{\left(\text{V}+\frac{8}{3}\text{V}\right)}$

            $\frac{\frac{16}{3}{\text{V}}^2}{\frac{\text{3V+8V}}{3}}=\frac{16{\text{V}}^2}{11\text{V}}=\frac{16\text{V}}{11}$

            ${\text{V}}_{\text{AD}}=\frac{16}{11}\text{V}$