Physics - Kinematics Question with Solution | TestHub

Let the total distance be $d$.

Time taken to cover ${\left(\frac{1}{3}\right)}^{\mathrm{rd}}$ of total distance is $t_1=\frac{d}{3v_1}$.

Time taken to cover ${\left(\frac{1}{3}\right)}^{\mathrm{rd}}$ of total distance is $t_2=\frac{d}{3v_2}$.

Time taken to cover ${\left(\frac{1}{3}\right)}^{\mathrm{rd}}$ of total distance is $t_3=\frac{d}{3v_3}$.

Average speed$=\frac{\mathrm{Total} \mathrm{distance}}{\mathrm{Total} \mathrm{time}}$

$v_{avg}=\frac{d}{{\displaystyle \frac{d}{3v_1}}+{\displaystyle \frac{d}{3v_2}}+{\displaystyle \frac{d}{3v_3}}}$

$\Rightarrow {\mathrm{v}}_{avg}=\frac{3\left(v_1{v}_2{v}_3\right)}{v_1{v}_2+v_2{v}_3+v_3{v}_1}$

Here, $v_1=11 \mathrm{m} {\mathrm{s}}^{-1}$, then, $v_3=33 \mathrm{m} {\mathrm{s}}^{-1}, v_2=22 \mathrm{m} {\mathrm{s}}^{-1}$

$V_{\mathrm{avg} }=\frac{3\times 11\times 22\times 33}{\left(11\times 22\right)+\left(22\times 33\right)+\left(33\times 11\right)}=18 \mathrm{m} {\mathrm{s}}^{-1}$