Question

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?

Answer 1

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
`(L/3)/v_1 + (L/3)/v_2 + (L/3)/v_3`

Make it to one fraction

Note: `("1/3")/X = 1/(3X)`
SO
`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: `"Total distance" / "total time"`

`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"= 1/(((v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3))`

Since its one divided by `((v_2v_3 + v_1v_3+v_1v_2))/ (3v_1v_2v_3)`

The ans is reciprocal of that fraction
Which is

`(3v_1v_2v_3) / (v_2v_3+v_1v_3+v_1v_2)`

which is equivalent to the expression used in the question:

`3v_1v_2v_3 ÷ v_2v_3+v_1v_3+v_1v_2`