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

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#