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A point traversed half of the distance with a velocity $v_0$. The remaining part of the distance was covered with velocity $v_1$ for half the time and with velocity $v_2$ for the other half of the time. The mean velocity of the point averaged over the whole time of motion is
$\frac{v_0+v_1+v_2}{3}$
$\frac{2 v_0+v_1+v_2}{3}$
$\frac{v_0+2 v_1+2 v_2}{3}$
$\frac{2 v_0\left(v_1+v_2\right)}{\left(2 v_0+v_1+v_2\right)}$
Solution
(a)
$v_1 t+v_2 t=s$
$t=\frac{s}{v_1+v_2}$
$\text { Total time }=t_1+2 t \text { and total displacement }=2\,s$
$\text { Mean velocity }=\frac{\text { Total displacement }}{\text { Total time }}$
$=\frac{2 s}{\left(s / v_0\right)+\left(2 s / v_1+v_2\right)}$
$=\frac{2}{\left(1 / v_0\right)+\left(2 / v_1+v_2\right)}$
$=\frac{2 v_0\left(v_1+v_2\right)}{\left(2 v_0+v_1+v_2\right)}$
