Friction Cars Racing - 200m Race Results

[Intrusionv2]

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.

 

[LowlyPion]

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.

 

[nlightner]

I would like to clarify some points before providing a response to this content. First, it is important to specify the units used in the equations, as they are crucial for accurate calculations. Additionally, it is necessary to mention the mass of the cars and the type of tires used, as these factors can also affect the coefficient of friction and the cars' overall performance.

Now, to answer the question, assuming both cars have the same mass and are using the same type of tires, the first car will win the race by a distance of approximately 198.8 meters. This can be calculated by using the equations provided in the statement.

The first car's final velocity (vf) can be calculated using the equation Vf^2 = Vi^2 + 2a * x, where Vi is the initial velocity (0 m/s) and a is the acceleration (0.8 m/s^2). Plugging in these values, we get:

Vf^2 = 2(0.8)(200)
Vf^2 = 320
Vf = √320
Vf = 17.9 m/s

This means that the first car will reach a velocity of 17.9 m/s before reaching the stoplight. On the other hand, the second car's final velocity can be calculated using the same equation, but with a different acceleration value (0.686 m/s^2) due to the reduced coefficient of friction. Therefore, we get:

Vf^2 = 2(0.686)(x)
Vf^2 = 1.372x
Vf = √(1.372x)

Since both cars will reach the same final velocity (0 m/s) at the stoplight, we can set the two equations equal to each other:

17.9 = √(1.372x)
17.9^2 = 1.372x
320.41 = 1.372x
x = 233.7 m

This means that the second car will only travel a distance of 233.7 meters before reaching the stoplight. Therefore, the first car will win the race by a distance of approximately 198.8 meters (200 - 1.2 = 198.8 m).

In conclusion, the first car will win the race by a considerable margin due to the reduced coefficient of friction on the second car's tires. However, it