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A point traversed half of the distance with a velocity $v_{0}$. The half of remaining part of the distance was covered with velocity $v_{1}$ & second half of remaining part by $v_{2}$ velocity. The mean velocity of the point, averaged over the whole time of motion is
Explanation
Let the total distance be $d$. Then for first half distance, time = $d/2v_{0}$, next distance = $v_{1}t$ and last half distance = $v_{2}t$ $\therefore v^{1}t + v^{2}t = \frac{d}{2};$ $t = d/2(v_{1} + v_{2})$ Now average speed $t = \frac{d}{\frac{d}{2v^{0}} + \frac{d}{2\left( v_{1} + v_{2} \right)} + \frac{d}{2\left( v_{1} + v_{2} \right)}}$ $= \frac{2v_{0}(v_{1} + v_{2})}{(v_{1} + v_{2}) + 2v_{0}}$