# 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?

Mar 31, 2018

Total distance ÷ total time

#### Explanation:

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

Time = distance / speed

So for first 1/3, time = (L/3)÷ v_1

For second 1/3, time (L/3)÷ v_2

For third 1/3, time = (L/3)÷v_3

Add all these times to get total time
$\frac{\frac{L}{3}}{v} _ 1 + \frac{\frac{L}{3}}{v} _ 2 + \frac{\frac{L}{3}}{v} _ 3$

Make it to one fraction

Note: $\frac{\text{1/3}}{X} = \frac{1}{3 X}$
SO
$\frac{L}{3 {v}_{1}} + \frac{L}{3 {v}_{2}} + \frac{L}{3 {v}_{3}} = \frac{L \cdot \left({v}_{2} {v}_{3} + {v}_{1} {v}_{3} + {v}_{1} {v}_{2}\right)}{3 {v}_{1} {v}_{2} {v}_{3}}$

This is total time

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

${v}_{\text{ave}} = \frac{L}{\frac{L \cdot \left({v}_{2} {v}_{3} + {v}_{1} {v}_{3} + {v}_{1} {v}_{2}\right)}{3 {v}_{1} {v}_{2} {v}_{3}}}$

We can cancel the 2 "L"s giving us

${v}_{\text{ave}} = \frac{1}{\frac{\left({v}_{2} {v}_{3} + {v}_{1} {v}_{3} + {v}_{1} {v}_{2}\right)}{3 {v}_{1} {v}_{2} {v}_{3}}}$

Since its one divided by $\frac{\left({v}_{2} {v}_{3} + {v}_{1} {v}_{3} + {v}_{1} {v}_{2}\right)}{3 {v}_{1} {v}_{2} {v}_{3}}$

The ans is reciprocal of that fraction
Which is

$\frac{3 {v}_{1} {v}_{2} {v}_{3}}{{v}_{2} {v}_{3} + {v}_{1} {v}_{3} + {v}_{1} {v}_{2}}$

which is equivalent to the expression used in the question:

3v_1v_2v_3 ÷ v_2v_3+v_1v_3+v_1v_2