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Harmonic mean: car trip mean velocity

  1. Jan 3, 2017 #1

    TheBlackAdder

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    Gold Member

    1. The problem statement, all variables and given/known data
    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.)

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

    3. The attempt at a solution

    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?
     
  2. jcsd
  3. Jan 3, 2017 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    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.
     
  4. Jan 3, 2017 #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}##.
     
    Last edited: Jan 3, 2017
  5. Jan 9, 2017 #4

    TheBlackAdder

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    Gold Member

    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.
     
  6. Jan 9, 2017 #5

    Mark44

    Staff: Mentor

    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.
     
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