A particle travels at a constant velocity of $3{\text{\hspace{0.17em}ms}}^{-1}$ for  20 s. Then it stops for 10 s before completing its journey in another 30 s at a constant velocity of $2{\text{\hspace{0.17em}ms}}^{-1}$.Its average velocity for the whole journey is

Let the particle covered the journey from A to B

It has reached C in 20s with constant velocity ${\text{v}}_1=3{\text{ms}}^{-1}$

So $\text{AC}={\text{v}}_{\text{1}}{\text{t}}_{\text{1}}=20\times 3{\text{ms}}^{-1}$

Stopped at C for ${\text{t}}_{\text{2}}=10\text{s}$

It has reached from C to B with constant velocity ${\text{v}}_{\text{2}}=2{\text{ms}}^{-1}$

${\text{t}}_{\text{2}}=30\text{s}$

So CB =$30\times 2{\text{ms}}^{-1}$

${\text{V}}_{\text{Avg}}=\frac{\text{s}}{\text{t}}=\frac{3\times 20+0\times 10+2\times 30}{20+10+30}$$=\frac{60+60}{60}=\frac{120}{60}=2\text{m/s}$

${\text{V}}_{\text{Avg}}=2\text{m/s}$