A point traversed half of the distance with velocity v0. The remaining part of the distance was covered with…

A point traversed half of the distance with velocity v0. The remaining part of the distance was covered with velocity $v_1$ & then velocity $v_2$ for equal time. The mean velocity of the point, averaged over the whole time of motion is
  1. $\frac{\mathrm{v}_{0}+\mathrm{v}_{1}+\mathrm{v}_{2}}{3}$
  2. $\frac{2 \mathrm{v}_{0}+\mathrm{v}_{1}+\mathrm{v}_{2}}{3}$
  3. $\frac{v_{0}+.2 v_{1}+2 v_{2}}{3}$
  4. $\frac{2 \mathrm{v}_{0}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}{\left(2 \mathrm{v}_{0}+\mathrm{v}_{1}+\mathrm{v}_{2}\right)}$

Solution

Let the total distance be $\mathrm{d}$. Then for first half distance, time $=\frac{\mathrm{d}}{2 \mathrm{v}_{0}}$, next distance. $=\mathrm{v}_{1} \mathrm{t}$ and last half distance $=\mathrm{v}_{2} \mathrm{t}$
$\therefore \quad \mathrm{v}_{1} \mathrm{t}+\mathrm{v}_{2} \mathrm{t}=\frac{\mathrm{d}}{2} ; \mathrm{t}=\frac{\mathrm{d}}{2\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}$
Now average speed $\mathrm{t}=\frac{\mathrm{d}}{\frac{\mathrm{d}}{2 \mathrm{v}_{0}}+\frac{\mathrm{d}}{2\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}+\frac{\mathrm{d}}{2\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}}$
$=\frac{2 \mathrm{v}_{0}\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)}{\left(\mathrm{v}_{1}+\mathrm{v}_{2}\right)+2 \mathrm{v}_{0}}$ ~

Asked in: JEE Mains - Motion In One Dimension - Test 1

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