A particle moving in a straight line travelled half of the distance with a speed ${\text{V}}_{\text{0}}$ . The remaining part of the distance is covered with a speed ${\text{V}}_1$  for half of the time and with a speed ${\text{V}}_2$  for the other half of the time. The average speed of the particle over the whole time of the motion is

Given that

A particle covered the journey from A to D = 2s

It covered A to B i.e. = s

With speed $={\text{v}}_{\text{0}}$

Time taken ${\text{t}}_{\text{0}}=\frac{\text{s}}{{\text{v}}_{\text{0}}}\left(\because \text{t}=\frac{\text{s}}{\text{v}}\right)$

It covered B to D i.e. = 2s

In time = t

Such that $\text{BC}={\text{v}}_{\text{1}}\frac{\text{t}}{\text{2}},\text{CD}={\text{v}}_{\text{2}}\frac{\text{t}}{\text{2}}$

We get

$\text{BD}=\text{BC}+\text{CD}$

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

$\text{t}=\frac{\text{2s}}{{\text{v}}_{\text{1}}+{\text{v}}_{\text{2}}}$

Average speed $=\frac{\text{total}\text{distance}}{\text{total}\text{time}}$

Average speed $=\frac{\text{AD}}{{\text{t}}_{\text{0}}+\text{t}}$

$=\frac{2\text{s}}{\frac{\text{s}}{{\text{v}}_{\text{0}}}+\frac{2\text{s}}{\left({\text{v}}_{\text{1}}+{\text{v}}_{\text{2}}\right)}}$

$=\frac{{\text{2v}}_{\text{0}}\left({\text{v}}_{\text{1}}+{\text{v}}_{\text{2}}\right)}{{\text{2v}}_{\text{0}}+{\text{v}}_{\text{1}}+{\text{v}}_2}$