Friction cars racing

  • #1

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.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
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.
 

Related Threads on Friction cars racing

  • Last Post
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
6K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
6
Views
882
  • Last Post
Replies
3
Views
1K
Top