Harmonic mean: car trip mean velocity

But the correct answer is 40 mi/hr. This is because the average of two speeds is not the same as the speed of the entire distance covered at those two speeds.So, for the problem at hand, the average velocity is the total distance (1/2 TD + 1/2 TD) divided by the total time (1/2 TD/v1 + 1/2 TD/v2).Here, the total time is (1/2 TD)/v1 + (1/2 TD)/v2, or 1/2(v1+v2)TD, so the average velocity is (1/2 TD + 1/2 TD)/(1/2(v1+v2)TD), or (v1
  • #1
CynicusRex
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Homework Statement


For the first half of a trip a car has velocity v1; for the second half of a trip it has velocity v2. What is the mean velocity of the car?

(The book does not mention a direction.)

Homework Equations


Arithmetic mean: $$\frac{v_{1}+v_{2}}{2}$$
Harmonic mean:
$$\frac{1}{(\frac{1}{v_{1}}+\frac{1}{v_{2}})\frac{1}{2}}$$

The Attempt at a Solution


[/B]
The velocity is v1 for 1/2 of a trip and v2 for the other 1/2.
How far do they get if they if they were driving at 1 km/h?
$$\frac{v_{1}}{v_{1}}km/h
= 1 km/h \rightarrow
\frac{1}{2v_{1}}trip$$
$$\frac{v_{2}}{v_{2}}km/h
= 1 km/h \rightarrow
\frac{1}{2v_{2}}trip$$

Now, what is the arithmetic mean between those two trip distances they've traveled at 1 km/h? Or, the mean trip distance at 1 km/h is:
$$\frac{\frac{1}{2v_{1}}+\frac{1}{2v_{2}}}{2}=\frac{1}{v_{1}}+\frac{1}{v_{2}} trip$$

But, we need the mean velocity of 1/2 of a trip.
$$\frac{\frac{\frac{1}{v_{1}}+\frac{1}{v_{2}}}{\frac{1}{v_{1}}+\frac{1}{v_{2}}}}{2}=\frac{1}{2}trip\rightarrow v\text{̅}=\frac{1}{(\frac{1}{v_{1}}+\frac{1}{v_{2}})\frac{1}{2}}km/h$$

I know the harmonic mean is the correct solution, but do I get there correctly?
 
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  • #2
If both velocities are the same, both means will work. You have to look at different speeds to see the difference.

"Half of the trip" is ambiguous: Half of the time or half of the distance?
You can derive the correct formula in both cases.
 
  • #3
"Half of the trip" is ambiguous indeed. If "half of distance" is what is meant then the answer is the harmonic mean. If "half of time" is what is meant then the answer is the arithmetic mean.

If we suppose that it means "half of (total) distance" then another way to arrive at the result is the following:
Suppose ##s## is half the distance so total trip is ##2s##. The mean velocity definition is the constant velocity at which the car would have travel the same total distance in the same total time. So by definition it would be
##v=\frac{total distance}{total time}=\frac{s_1+s_2}{t_1+t_2}=\frac{s+s}{\frac{s}{v_1}+\frac{s}{v_2}}=\frac{1}{\frac{1}{2}(\frac{1}{v_1}+\frac{1}{v_2})}##.

If "half of time" is instead what is meant then
##v=\frac{s_1+s_2}{t_1+t_2}=\frac{v_1\frac{t}{2}+v_2\frac{t}{2}}{\frac{t}{2}+\frac{t}{2}}=\frac{v_1+v_2}{2}##.
 
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  • #4
There's a huge error in my first post. It makes no sense in finding out the mean distance per 1km/h because the velocity is constant.
$$
\frac{\frac{1}{(\frac{1}{v_{1}}+\frac{1}{v_{2}})}}{2}
\neq\frac{1}{(\frac{1}{v_{1}}+\frac{1}{v_{2}})\frac{1}{2}}
but = \frac{1}{(\frac{1}{v_{1}}+\frac{1}{v_{2}})2}$$

The latter is not the harmonic mean.

Instead.
Given: v1 km/h during 1/2 TD (Total Distance), v2 km/h during 1/2 TD.

Which means it takes (1/2 TD)/v1 = 1/(2v1) hours to drive 1/2 TD at a speed of v1 km/h and 1/(2v2) hours to drive 1/2 TD at a speed of v2 km/h.

So, to calculate the mean velocity: Δd/Δt
$$
\frac{\frac{1}{2}TD+\frac{1}{2}TD}{\frac{1}{2v_{1}}+\frac{1}{2v_{2}}}=\frac{1TD}{\frac{1}{2}(\frac{1}{v_{1}}+\frac{1}{v_{2}})}=\frac{1}{\frac{1}{2}(\frac{1}{v_{1}}+\frac{1}{v_{2}})}km/h
$$

to traverse 1 Total Distance. I know it's the same what Delta2 said, but wanted to do it myself again.
 
  • #5
One thing to beware of is "averaging" average speeds/velocities. The way you calculate the average velocity is the total distance divided by the total time.

There's an old problem stating that a car goes up a one-mile-long hill at an average speed of 30 mi/hr. The problem is to find the average speed on the downhill leg so that the average speed for both legs is 60 mi/hr.

The intuitive (and completely wrong) answer is 90 mi/hr.
 

Related to Harmonic mean: car trip mean velocity

1. What is harmonic mean?

The harmonic mean is a type of average that is used to calculate the central tendency of a set of numbers. It is calculated by dividing the number of values in the set by the sum of the reciprocals of the values.

2. How is harmonic mean different from other types of averages?

Harmonic mean is different from other types of averages such as arithmetic mean and geometric mean because it takes into account the effects of extreme values in the dataset. It is more appropriate for averaging rates or ratios.

3. How is harmonic mean used in calculating car trip mean velocity?

In the context of car trip mean velocity, harmonic mean is used to calculate the average speed of a car during a trip. This is useful when the car travels at varying speeds, as it takes into account the time spent at each speed.

4. What are the limitations of using harmonic mean in calculating car trip mean velocity?

One limitation of using harmonic mean is that it is strongly influenced by extreme values in the dataset. This means that a few very high or low speeds can significantly affect the calculated average. Additionally, it may not be suitable for trips with a large number of stops or interruptions.

5. When is it appropriate to use harmonic mean in calculating car trip mean velocity?

Harmonic mean is most appropriate for calculating car trip mean velocity when the trip involves varying speeds and the purpose is to determine an overall average speed. It is also useful when the trip has a small number of stops or interruptions.

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