Maximum travelling distance of car from 1 L of fuel

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SUMMARY

The discussion centers on calculating the maximum distance a car can travel using 1 liter of petrol, given specific parameters. A car traveling at 60.0 km/h faces a retarding force of 1500 N due to air resistance, and 1.00 L of petrol releases 40,000,000 J of energy with an engine efficiency of 5%. The effective energy available for motion is 2,000,000 J, which must overcome the retarding force to determine the distance traveled.

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  • Understanding of basic physics concepts such as force, energy, and efficiency.
  • Familiarity with kinematic equations, specifically those related to motion.
  • Knowledge of energy conversion and work-energy principles.
  • Ability to perform calculations involving units of energy (Joules) and distance (kilometers).
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  • Calculate the effective distance using the formula: distance = energy / force.
  • Research the concept of engine efficiency and its impact on fuel consumption.
  • Explore the relationship between speed, force, and energy in automotive physics.
  • Learn about the work-energy theorem and its application in real-world scenarios.
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Students studying physics, automotive engineers, and anyone interested in understanding the relationship between fuel efficiency and distance traveled by vehicles.

shannon.leigh
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Sorry 1 and 2 aren't filled in very well but I have absolutely no idea what to do for this question. My brain wasn't designed for physics :(

Homework Statement



Question: A car traveling at 60.0 km/h has a total retarding force of 1500 N due to air resistance. If 1.00 L of petrol can release 40 000 000 J of energy, how many kilometres can the car travel at 60.0 km/h on 1.00 L of fuel (assume that the engine is 5% efficient)?

Homework Equations



I really have no idea, but making a rough guess. . .

s=ut+0.5at^2

or

v^2=u^2 +2as

The Attempt at a Solution



?

Really, I have no idea where to start.

Any help would be greatly appreciated!
 
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All the energy that the engine gets from the petrol goes into opposing the work done by the retarding force.
 

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