# Formula for average of two velocities on same distance

1. Apr 22, 2008

### pc2-brazil

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
$$v = \frac{2v_1v_2}{v_1 + v_2}$$
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:
$$v = \frac{2v_1v_2}{v_1 + v_2} \Rightarrow v = \frac{2(60)(90)}{60 + 90} = \frac{10800}{150} = 72 km/h.$$
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?

Last edited: Apr 22, 2008
2. Apr 22, 2008

### lzkelley

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. Apr 22, 2008

### pc2-brazil

2 * inverse of $$\frac{1}{v_1} + \frac{1}{v_2} = 2\frac{1}{\frac{1}{v_1} + \frac{1}{v_2}}=$$
$$= \frac{2}{\frac{1}{v_1} + \frac{1}{v_2}}$$
but wouldn't this be the harmonic mean?

Last edited: Apr 22, 2008
4. Apr 22, 2008

### SheldonG

Average velocity is the displacement divided by the time it takes to travel that displacement. Your problem doesn't give the distance or the time, but it does say that 1/2 of the displacement is traveled at each velocity. Since velocity is constant, D/2 = v1*t1 and D/2 = v2*t2. You want the average velocity, that is v_avg = D/total_time. So v_avg = D/(t1 + t2). If you work out the algebra from here, you will see where your teacher got the formula.

5. Apr 22, 2008

### lzkelley

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. Apr 22, 2008

### SheldonG

Sorry, forgot to answer this bit. There is a period where the car is traveling at 60 and a period where it is traveling at 90. But since the change is not instantaneous, there will be a 3rd period where the car is traveling between 60 and 90. So the average of the end points will be too high in this case, since it ignores the middle part.

7. Apr 22, 2008

### lzkelley

Thats incorrect. The problem specifies that half the distance is average 60, and half is average 90. We can treat this as no acceleration; and velocities are constant in each domain and then change instantaneously inbetween.
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. Apr 22, 2008

### SheldonG

Yes, you're right. Thanks.

9. Apr 22, 2008

### pc2-brazil

salutations,

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

10. Apr 22, 2008

### SheldonG

You want to eliminate t. $$v_{avg} = D/(t_1 + t_2)$$

And for each half: $$D/2 = v_1t_1\quad D/2 = v_2t_2$$

Solve these last equations for time, and then substitute them in the first equation.

11. Apr 22, 2008

### pc2-brazil

$$t_1 = \frac{d}{2v_1}$$ and $$t_2 = \frac{d}{2v_2}$$
$$v = \frac{d}{\frac{d}{2v_1} + \frac{d}{2v_2}} = \frac{d}{\frac{v_2d + v_1d}{2v_1v_2}} = \frac{d \cdot 2v_1v_2}{v_2d + v_1d} = \frac{d \cdot 2v_1v_2}{d(v_2 + v_1)} =$$
$$= \frac{2v_1v_2}{v_1 + v_2}$$ <-- solution.