Friction Cars Racing - 200m Race Results

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SUMMARY

The forum discussion centers on a physics problem involving two cars racing 200 meters, where one car has its tires slicked with oil, reducing its coefficient of friction to 0.07. The first car, with a static coefficient of friction of 0.8, reaches a final velocity of 39.6 m/s, while the second car, with a kinetic coefficient of friction of 0.07, accelerates at 0.686 m/s² and achieves a final velocity of only 1.2 m/s. The analysis concludes that the first car will win the race by a significant distance, as the second car only travels 14 meters in the same time frame.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of friction coefficients
  • Knowledge of kinematic equations
  • Basic algebra for solving equations
NEXT STEPS
  • Study the effects of different friction coefficients on vehicle acceleration
  • Learn about kinematic equations in depth, particularly Vf^2 = Vi^2 + 2a * x
  • Explore real-world applications of friction in automotive engineering
  • Investigate the impact of tire materials on performance and safety
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in the dynamics of vehicle performance and frictional forces.

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Homework Statement


Two cars are going to begin at rest and race 200 meters to a stoplight. One driver sneaks over and puts oil on the other car's tires, which reduces it coefficient of friction to .07. Assuming the first car does not slip then by how much distance will it win?


Homework Equations


Ff = μ(Fn)
a = μ(9.8)
Vf^2 = Vi^2 + 2a * x
Static coefficient of friction of rubber on concrete = 1.0
Kinetic coeff. of friction of rubber on concrete = 0.8

The Attempt at a Solution


1st car:
Vf^2 = 2(.8*9.8)200
vf = 39.6 m/s

2nd car:
a = .07(9.8)
a = .686 m/s^2

vf^2 = 2(.686)
vf = 1.2 m/s

Not sure what to do now.
 
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All things being equal for the first car then I'd say it's acceleration would be .07 of the second.

So for the unslicked car 200 = 1/2a*T2

The slicked car would only have gone only .07 (200) = 14 m in the same time.
 

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