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Negative of a Vector

A point trave...

Question

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 one-third of the remaining time and with velocity $v_{2}$ for the remaining time. The mean velocity of the point averaged over the whole time of motion

\frac{v_{0} + v_{1} + v_{2}}{3}

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\frac{2 v_{0} + v_{1} + v_{2}}{3}

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\frac{3 v_{0} (v_{1} + 2 v_{2})}{(2 v_{0} + v_{1} + 2 v_{2})}

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\frac{2 v_{0} (v_{1} + 2 v_{2})}{(3 v_{0} + v_{1} + 2 v_{2})}

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Solution

The correct option is D $\frac{2 v_{0} (v_{1} + 2 v_{2})}{(3 v_{0} + v_{1} + 2 v_{2})}$
Let $2 S$ be the total distance travelled

For first half of the journey
$S = v_{0} \times t_{1}$ (where $t_{1}$ is the time taken to complete half of the distance)

Let $t$ be the time taken to complete other half of the journey

For the remaining part of the journey we can say,
$v_{1} \times \frac{t}{3} + v_{2} \times \frac{2 t}{3} = S$
$∴ t = \frac{3 S}{v_{1} + 2 v_{2}}$
Total time $= t_{1} + t$

$Mean velocity = \frac{Total displacement}{Total time}$
$= \frac{2 S}{(\frac{S}{v_{0}}) + (\frac{3 S}{v_{1} + 2 v_{2}})}$
$= \frac{2}{(\frac{1}{v_{0}}) + (\frac{3}{v_{1} + 2 v_{2}})}$
$= \frac{2 v_{0} (v_{1} + 2 v_{2})}{(3 v_{0} + v_{1} + 2 v_{2})}$

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