Friction Cars Racing - 200m Race Results

AI Thread Summary
In a 200-meter race between two cars, one car has its tires slicked with oil, reducing its coefficient of friction to 0.07, while the other maintains a standard friction coefficient of 0.8. The first car, with a higher friction, reaches a final speed of 39.6 m/s, while the slicked car only achieves a final speed of 1.2 m/s due to significantly lower acceleration. Calculations indicate that the slicked car would cover only 14 meters in the same time it takes the first car to complete the race. Thus, the first car wins the race by a substantial margin. The analysis highlights the critical impact of friction on vehicle acceleration and race outcomes.
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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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