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Every time we have to stop the car, it is costing us extra money, because we have to then accelerate to full speed again. I would like to know how much.

In order to simplify the situation, we can make these assumptions.

  • we disregard cost of car maintenance (brake pads wearing out etc.). This is focused entirely on fuel economy.
  • A car is always traveling at 60km/h when it needs to stop
  • The car's fuel economy is 8L/100kms when traveling at 60km
  • The cost of fuel is $1.25 per litre.
  • The car weighs 1000kg

I will be happy with either a definitive answer, an explanation of how to arrive at the answer, or an explanation of why there is not enough information in the question to come up with the answer.

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Off topic. Some vehicle brake system may connected to a engine to generate electricity and it is used later for acceleration, so the cost is small in those vehicle :) –  hwlau Dec 17 '10 at 12:32
2  
The crucial parameter is how your fuel consumption depends on the acceleration (and rpm) because the major part of your cost is due to a much lower efficiency of accelerating car compared to a car traveling at a constant speed. –  gigacyan Dec 17 '10 at 12:44
    
Where do you live that they both charge in dollars and measure in liters? –  Mark Eichenlaub Dec 17 '10 at 21:03
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@Mark : australia. –  RoboShop Dec 18 '10 at 11:41
    
It depends on whether or not you ran the light first and got a ticket:) –  Gordon Feb 3 '11 at 5:19

2 Answers 2

up vote 8 down vote accepted

Here is a very basic estimation:

The kinetic energy of a 1000 kg car moving at 60 km/h is $$E=\frac{mv^{2}}{2}=\frac{1000kg(16.7m/s)^{2}}{2}=138.9 kJ$$

The heat of gasoline combustion is 47 MJ/kg = 35000 kJ/litre. Assuming 10% efficiency of the car's engine, you would need to burn $$\frac{138.9 kJ}{0.1\cdot35000kJ/l}=0.04 litre$$ of gasoline to accelerate the car to a given speed which would cost you 5 cents.

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Given the amount of information provided, this is the best we can get. –  Malabarba Dec 18 '10 at 20:33

I believe more information is needed to formulate an answer. Either you'd need the potential energy per liter of the gas and the necessary information to determine how much energy is lost in breaking (in which case, assuming no other loss of energy for simplicity's sake, you can work it out based on pure energy considerations) or you need more detail regarding the acceleration and deceleration phases of motion (in which case you can use forces and work).

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What about the new info I've added about the weight. If you know the weight, and you know how many litres it is required to travel 100kms, shouldn't you somehow be able to derive the potential energy per litre? –  RoboShop Dec 17 '10 at 12:40
    
can you clarify what you mean about more details about acceleration and deceleration? –  RoboShop Dec 17 '10 at 12:41

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