A car covers AB distance with first one-third at velocity v _ 1 ms...

Let the total distance be $d$.

Time taken to cover ${\left(\dfrac{1}{3}\right)}^{rd} of total distance is${t}_{1} = \dfrac{d}{3{v}_{1}}$.</p><p>Time taken to cover${\left(\dfrac{1}{3}\right)}^{rd} of total distance is ${t}_{2} = \dfrac{d}{3{v}_{2}}$.

Time taken to cover ${\left(\dfrac{1}{3}\right)}^{rd} of total distance is${t}_{3} = \dfrac{d}{3{v}_{3}}$.</p><p>Average speed$= \dfrac{Totaldistance}{Totaltime}$</p><p>${v}_{avg} = \dfrac{d}{\dfrac{d}{3{v}_{1}} + \dfrac{d}{3{v}_{2}} + \dfrac{d}{3{v}_{3}}}$</p><p>$⇒ {v}_{avg} = \dfrac{3\left({v}_{1}{v}_{2}{v}_{3}\right)}{{v}_{1}{v}_{2} + {v}_{2}{v}_{3} + {v}_{3}{v}_{1}}$</p><p>Here,${v}_{1} = 11m{s}^{- 1}$, then,${v}_{3} = 33m{s}^{- 1}, {v}_{2} = 22m{s}^{- 1}$</p><p>${V}_{avg} = \dfrac{3 \times 11 \times 22 \times 33}{\left(11 \times 22\right) + \left(22 \times 33\right) + \left(33 \times 11\right)} = 18m{s}^{- 1}$