A particle moves in a straight line .1/3rd of its journey with velocity V1,next 1/3rd with velocity V2,next 1/3rd with velocity V3.How to prove that its average velocity is ;3v1v2v3 ÷ v1v2 + v2v3 + v1v3?

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A particle moves in a straight line .1/3rd of its journey with velocity V1,next 1/3rd with velocity V2,next 1/3rd with velocity V3.How to prove that its average velocity is ;3v1v2v3 ÷ v1v2 + v2v3 + v1v3?

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Answer 1 · 2018-03-31T10:19:34

Total distance ÷ total time

Explanation:

Total distance=L/3 +L/3 +L/3 = L

Time = distance / speed

So for first 1/3, time = $(\frac{L}{3}) \div v_{1}$ $(L/3)÷ v_1$

For second 1/3, time $(\frac{L}{3}) \div v_{2}$ $(L/3)÷ v_2$

For third 1/3, time = $(\frac{L}{3}) \div v_{3}$ $(L/3)÷v_3$

Add all these times to get total time
$\frac{\frac{L}{3}}{v_{1}} + \frac{\frac{L}{3}}{v_{2}} + \frac{\frac{L}{3}}{v_{3}}$ $(L/3)/v_1 + (L/3)/v_2 + (L/3)/v_3$

Make it to one fraction

Note: $\frac{1/3}{X} = \frac{1}{3 X}$ $("1/3")/X = 1/(3X)$
SO
$\frac{L}{3 v_{1}} + \frac{L}{3 v_{2}} + \frac{L}{3 v_{3}} = \frac{L \cdot (v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2})}{3 v_{1} v_{2} v_{3}}$ $L/(3v_1) + L/(3v_2) + L/(3v_3) = (L*(v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3)$

This is total time

Substitute into average speed formula: $\frac{Total distance}{total time}$ $"Total distance" / "total time"$

$v_{ave} = \frac{L}{\frac{L \cdot (v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2})}{3 v_{1} v_{2} v_{3}}}$ $v_"ave"= L/((L*(v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3))$

We can cancel the 2 "L"s giving us

$v_{ave} = \frac{1}{\frac{(v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2})}{3 v_{1} v_{2} v_{3}}}$ $v_"ave"= 1/(((v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3))$

Since its one divided by $\frac{(v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2})}{3 v_{1} v_{2} v_{3}}$ $((v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3)$

The ans is reciprocal of that fraction
Which is

$\frac{3 v_{1} v_{2} v_{3}}{v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2}}$ $(3v_1v_2v_3) / (v_2v_3+v_1v_3+v_1v_2)$

which is equivalent to the expression used in the question:

$3 v_{1} v_{2} v_{3} \div v_{2} v_{3} + v_{1} v_{3} + v_{1} v_{2}$ $3v_1v_2v_3 ÷ v_2v_3+v_1v_3+v_1v_2$

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A particle moves in a straight line .1/3rd of its journey with velocity V1,next 1/3rd with velocity V2,next 1/3rd with velocity V3.How to prove that its average velocity is ;3v1v2v3 ÷ v1v2 + v2v3 + v1v3?

1 Answer

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