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Acceleration does not matter when driving

  1. Oct 1, 2014 #1
    In my physics lecture, my professor "proved" that acceleration does not affect the amount of energy used in driving a car. Assume car is driven along flat road. In the short,

    (assume air resistance is neglected, and the W done by the car's engine, [tex] W_{engine} [/tex] is non conservative, [tex] \Delta PE \Delta KE [/tex] are change in potential and change in kinetic energy, respectively )

    \Delta PE = 0 \\
    \Delta KE &= W_{engine}\\
    &= 1/2 m {v_f}^2 - 1/2 m {v_o}^2

    m is constant mass, vf and vi are final and initial velocity, respectively.
    Proof done: change in kinetic energy does NOT depend on acceleration; only the mass and final velocity of the car.

    Okay now lets add air resistance:
    W_{res} = \frac{ \rho D A v^2}{2}\\
    \Delta KE &= W_{engine} - W_{Res}\\
    &= 1/2 m {v_f}^2 + \frac{\rho D A v^4}{8a}

    where ## \rho ## is the air density (constant), D is coefficient of drag, and A is cross section.

    Doesn't this mean that the energy used (in this case the gas in the car) depends on the acceleration of car, if there is air resistance?

    Is friction done by the ground on the car affected by acceleration?
    Last edited: Oct 1, 2014
  2. jcsd
  3. Oct 1, 2014 #2
    Is the second part (with air resistance) part of the professor's proof or it is your contribution?
  4. Oct 1, 2014 #3
    My professor's proof.

    Let me fix the equation that my professor gave. The ##W_{engine}=## is the correct formula my prof gave.

    W_{res} = \frac{ \rho D A v^2}{2}\\

    \Delta KE = W_{engine} - W_{Res}\\
    From this we get

    W_{engine} = 1/2 m {v_f}^2 + \frac{\rho D A v^4}{8a}

    I didnt want to show the proof for it but I can if need be. Also, I have the "ability" to post the slides here but I am concerned about posting my teacher's lectures because its his property; he has rights to his work which he has not "released".
  5. Oct 1, 2014 #4
    What you call Wres looks like the drag force and not the work.
    How do you get that last formula? What kind of motion is assumed for the car?
    I mean, if the engine force is fixed, the drag force will increase until the two will be equal, for example. But it's not the only possibility.
  6. Oct 4, 2014 #5

    D H

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    Staff Emeritus
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    It's hard to tell if your professor was pulling your legs or truly believes this nonsense.

    Of course how you accelerate matters, particularly if you have a vehicle with an internal combustion engine. Things are a bit different with an electronic vehicle, but nobody who has the $70,000 to buy a Tesla drives a Tesla at 10 miles per hour (the optimal speed for a Tesla with regard to mileage).

    Internal combustion engines are rather lousy at producing torque when the engine is running slowly and also when the engine is operating close to its maximum. A plot of torque versus engine speed reveals a curve that reaches a maximum at some point. This is the vehicle's torque curve. There's a related curve, the horsepower curve, that shows how much power the car can produce at a given engine rotation rate. These curves don't quite capture the dynamics of fuel consumption versus acceleration. What you need to look at is the quantity of fuel needed to produce a given acceleration. This too varies with engine speed. Acceleration wouldn't matter if this curve was flat. This curve is anything but flat. Acceleration matters.

    Suppose you are entering a freeway. If you accelerate too cautiously / too slowly, you are doing two bad things: wasting gas and risking lives. On the other hand, if you accelerate too quickly, you are also wasting gas and risking lives. There's a happy medium that optimizes gas consumption, and a slightly different happy medium that brings you up to speed in a reasonable amount of time and thereby minimizes the chance of an accident due to incompatible velocities. Personally, I go for the second happy medium.
  7. Oct 5, 2014 #6
    As D H pointed out, acceleration DOES matter in real life for the same reason you use energy when holding a book: both you and a car engine are inefficient machines. However, as professor pointed out, it doesn't matter when dealing purely with work done on the car. The car's initial and final kinetic energies are only functions of its speed and mass. How fast you got to that point is irrelevant.
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