If a car covers ${\text{2/5}}^{\text{th}}$ of the total distance with ${\text{v}}_{\text{1}}$ speed and ${\text{3/5}}^{\text{th}}$ distance with${\text{v}}_{\text{2}}$  then average speed is

Let total distance = s

${\text{v}}_{\text{1}}={\text{v}}_{\text{1}},{\text{t}}_{\text{1}}={\text{t}}_{\text{1}},{\text{s}}_1=\frac{\text{2s}}{5}$

$\Rightarrow {\text{v}}_{\text{1}}=\frac{{\text{s}}_1}{{\text{t}}_{\text{1}}}=\frac{\text{2s}}{{\text{5t}}_1}\Rightarrow {\text{t}}_1=\frac{\text{2s}}{{\text{5v}}_1}\to \left(1\right)$

${\text{s}}_2=\frac{\text{3s}}{5},{\text{v}}_{\text{2}}={\text{v}}_{\text{2}},{\text{t}}_{\text{2}}={\text{t}}_{\text{2}}$

$\Rightarrow {\text{v}}_{\text{2}}=\frac{{\text{s}}_2}{{\text{t}}_{\text{2}}}=\frac{\text{3s}}{{\text{5t}}_2}\Rightarrow {\text{t}}_2=\frac{\text{3s}}{{\text{5v}}_2}\to \left(2\right)$

Total time$\text{t}={\text{t}}_1+{\text{t}}_2=\frac{\text{2s}}{{\text{5v}}_1}+\frac{\text{3s}}{{\text{5v}}_2}=\frac{\text{s}}{\text{5}}\left[\frac{2}{{\text{v}}_{\text{1}}}+\frac{3}{{\text{v}}_{\text{2}}}\right]$

$=\frac{\text{s}}{\text{5}}\left[\frac{2{\text{v}}_{\text{2}}+3{\text{v}}_{\text{1}}}{{\text{v}}_{\text{1}}{\text{v}}_{\text{2}}}\right]$

$=\frac{\text{s}\left(3{\text{v}}_{\text{1}}+2{\text{v}}_{\text{2}}\right)}{{\text{5v}}_{\text{1}}{\text{v}}_{\text{2}}}$

$\left|{\overrightarrow{v}}_{avg}\right|=\frac{\left|\overrightarrow{s}\right|}{t}=\frac{s}{\frac{\text{s}\left(3{\text{v}}_{\text{1}}+2{\text{v}}_{\text{2}}\right)}{{\text{5v}}_{\text{1}}{\text{v}}_{\text{2}}}}$

$v_{avg}=\frac{{\text{5v}}_{\text{1}}{\text{v}}_{\text{2}}}{3{\text{v}}_{\text{1}}+2{\text{v}}_{\text{2}}}$