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# A particle traversed along a straight line for first half time with velocity $$V_0$$. For the remaining part, half of the distance is traversed with velocity $$V_1$$ and other half distance with velocity $$V_2$$. Find the mean velocity of the particle for the total journey.

A
$$\dfrac{2V_0(V_1+V_2)}{V_1+V_2+2V_0}$$
B
$$\dfrac{V_0(V_1+V_2)+2V_1V_2}{2(V_1+V_2)}$$
C
$$\dfrac{V_1+V_2}{2}$$
D
$$\dfrac{V_0+2V_1V_2}{2(V_1+V_2)}$$
Solution
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#### Correct option is B. $$\dfrac{V_0(V_1+V_2)+2V_1V_2}{2(V_1+V_2)}$$Lets total time $$=t$$$$s_1 =V_0 x\left(\dfrac {t}{2}\right)(A \ to \ B)$$$$t_2 =\left(\dfrac {3-S}{2}\right)/V_1$$$$t_2 =\left(\dfrac {S-V_0 (t/2)}{2}\right)/V_1$$$$t+2 = \dfrac {S}{2V_1}-\dfrac {V_0}{V_1}\times \dfrac {t}{4}$$$$\left(\dfrac {t}{2}-t_2 \right)=\dfrac {\left(\dfrac {S-S_1}{2}\right)}{V_2}$$$$\dfrac {t}{2}-\left(\dfrac {3}{2V_1}-\dfrac {V_0t}{4V_1}\right)=\dfrac {S-S_1}{2V_2}$$$$\dfrac {t}{2}-\dfrac {S}{2V_1}+\dfrac {V_0t}{4V_1}=\dfrac {3-V_0t/2}{2V_2}$$$$\dfrac {t}{2}-\dfrac {3}{2V_1}+\dfrac {V_0t}{4V_1}=\dfrac {S}{2V_2}-\dfrac {V_0}{4V_2}t$$$$t\left(\dfrac {1}{2}+\dfrac {V_0}{4V_1}+\dfrac {V_0}{4V_2}\right)=S\left(\dfrac {1}{2V_2}+\dfrac {1}{2V_1}\right)$$$$V_{arg}=\dfrac {S}{t}=\dfrac {2V_1 V_2+V_0(V_1 +V_2)}{4V_1V_2}/ \dfrac {14+V_2}{2V_1V_2}$$$$\dfrac {S}{t}=\dfrac {\dfrac {2V_1V_2 +V_0 (V_1+V_2)}{2V_1 V_2}}{\dfrac {(V_1+V_2)}{2V_1V_2}}$$$$V_{avg} = \dfrac {V_0 (V_1+V_2)+2V_1V_2}{2(V_1+V_2)}$$

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