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Formula for average of two velocities on same distance 
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#1
Apr2208, 01:56 PM

P: 205

1. The problem statement, all variables and given/known data
(translated from Portuguese) during the performance test of a new automobile model, the pilot goes through the first half of a track with an average velocity of 60 km/h and the second half with 90 km/h. what is the average velocity during the complete test, in km/h? 2. Relevant equations [tex]v = \frac{2v_1v_2}{v_1 + v_2}[/tex] 3. Attempt at solution initially, we weren't sure about how we would solve it; we thought it was through arithmetic average (average v between v1 and v2 = v = (v1+v2)/2). but it was solved in class, and the teacher said we were supposed to use that special formula, 2*v1*v2/(v1 + v2) in all situations similar to this, that is, when the problem asks us to find the average of velocities when the distance traveled is the same, but with different velocities. look: since the pilot goes through the same distance (half of the path) with different velocities, v1 = 60 km/h and v2 = 90 km/h, the general formula for this situation is: [tex]v = \frac{2v_1v_2}{v_1 + v_2} \Rightarrow v = \frac{2(60)(90)}{60 + 90} = \frac{10800}{150} = 72 km/h.[/tex] our question is: how do we obtain this formula, 2*v1*v2/(v1+v2)? why do we need to use this formula specifically, that is, why can't we just calculate the arithmetic average between 60 and 90, which would be (60 + 90) / 2 = 150 / 2 = 75? thank you in advance. 


#2
Apr2208, 02:02 PM

P: 277

anytime you see something of the form a*b / a+b > its from adding things inversely (in my experience).
lets think of it in terms of time. total time t = t1 + t2; by v=d/t we can transform this to d/v = d1/v1 + d2/v2 (where v is average, and v1 and v2 are the 2 different velocities) you also know that d1 = d/2 = d2 so you can solve for v. 


#3
Apr2208, 02:22 PM

P: 205

so, from your reply, we can think of 2v1v2/(v1+v2) as:
2 * inverse of [tex]\frac{1}{v_1} + \frac{1}{v_2} = 2\frac{1}{\frac{1}{v_1} + \frac{1}{v_2}}=[/tex] [tex]= \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}[/tex] but wouldn't this be the harmonic mean? 


#4
Apr2208, 02:58 PM

P: 50

Formula for average of two velocities on same distance



#5
Apr2208, 03:04 PM

P: 277

I'm not super familiar with harmonic means, but i think that in this case > the harmonic mean of the distances will be the mean of the velocities? ... something to that extent.
1/2v1 + 1/2v2 = 1/v = (v2/2*v1*v2)+(v1/2*v1*v2) = ... combine fractions and invert both sides. 


#6
Apr2208, 03:06 PM

P: 50




#7
Apr2208, 03:26 PM

P: 277

The reason you can't take a direct average is because we want to know the average velocity over all time, while 60+90/2 would be the average over distance. I.e. if the car spent half of the TIME going 60 and half of the TIME going 90, then the average would be 75. Because the car spends more time going 60 (to cover the same distance as going 90), the true average will be lower than 75. Does that make sense? 


#8
Apr2208, 03:29 PM

P: 50

Yes, you're right. Thanks.



#9
Apr2208, 03:50 PM

P: 205

salutations,
lzkelley: working on your suggestion, we found: [tex]\frac{d}{v} = \frac{d_1}{v_1} + \frac{d_2}{v_2}[/tex] then, as [tex]d_1 = d_2 = \frac{d}{2}[/tex], [tex]\frac{d}{v} = \frac{d}{2v_1} + \frac{d}{2v_2}[/tex] then, [tex]\frac{1}{v} = \frac{1}{2v_1} + \frac{1}{2v_2}[/tex] summing the fractions: [tex]\frac{1}{v} = \frac{v_2 + v_1}{2v_1v_2}[/tex] and inverting: [tex]v = \frac{2v_1v_2}{v_1 + v_2}[/tex] < solution. a question: does it make sense to say this: [tex]\frac{d}{v} = \frac{d}{2v_1} + \frac{d}{2v_2} = \frac{v_2d_1}{v_1v_2} = \frac{d_2v_1}{v_1v_2} = \frac{v_2d}{2v_1v_2} = \frac{v_1d}{2v_1v_2}[/tex]? SheldonG: we tried to work on your suggestion: [tex]\frac{d}{2} = \frac{v_1t_1}{1} = \frac{v_2t_2}{1}[/tex] then: [tex]d = 2v_1t_1 = 2v_2t_2; v_1t_1 = v_2t_2[/tex] so: [tex]v_1t_1 + v_2t_2 = \frac{d}{t_1 + t_2}[/tex] is it right? we got stuck here. thank you in advance. 


#10
Apr2208, 03:57 PM

P: 50

And for each half: [tex] D/2 = v_1t_1\quad D/2 = v_2t_2 [/tex] Solve these last equations for time, and then substitute them in the first equation. 


#11
Apr2208, 04:14 PM

P: 205

SheldonG: from your reply,
[tex]t_1 = \frac{d}{2v_1}[/tex] and [tex]t_2 = \frac{d}{2v_2}[/tex] then: [tex]v = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{d}{\frac{v_2d + v_1d}{2v_1v_2}} = \frac{d \cdot 2v_1v_2}{v_2d + v_1d} = \frac{d \cdot 2v_1v_2}{d(v_2 + v_1)} = [/tex] [tex] = \frac{2v_1v_2}{v_1 + v_2}[/tex] < solution. thank you, all answers were very helpful. 


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