Relating friction with acceleration

Click For Summary
SUMMARY

The discussion focuses on calculating the shortest stopping distance of a car with a coefficient of kinetic friction of 0.80 while traveling at 28.7 m/s. The key equation used is derived from the kinematic equation v² = 0 + 2a(x - 0), leading to the conclusion that the maximum deceleration is determined by the formula a = μg, where μ is the coefficient of friction and g is the acceleration due to gravity. The mass of the car is irrelevant in this calculation, simplifying the process significantly. The correct stopping distance can be calculated using these established principles.

PREREQUISITES
  • Understanding of kinematic equations
  • Knowledge of friction coefficients
  • Basic physics concepts of acceleration and deceleration
  • Familiarity with gravitational acceleration (g = 9.81 m/s²)
NEXT STEPS
  • Calculate stopping distances for different speeds using the same friction coefficient
  • Explore the effects of varying coefficients of friction on stopping distances
  • Learn about the role of mass in dynamics and its impact on acceleration
  • Investigate real-world applications of friction in vehicle safety systems
USEFUL FOR

This discussion is beneficial for physics students, automotive engineers, and anyone interested in understanding vehicle dynamics and safety measures related to braking and friction.

elimenohpee
Messages
64
Reaction score
0

Homework Statement


If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop a car by locking the brakes when traveling at 28.7 m/s?


Homework Equations


The problem doesn't state the mass of the car, so I'm not sure how to exactly relate kinetic friction with the acceleration.


The Attempt at a Solution


v^2 = 0 + 2a(x - 0)
v^2/2a= x

a?
 
Physics news on Phys.org
Mass will drop out.

What is the maximum deceleration the car will tolerate?

Won't that be when m*a = μ*m*g ?
 
OHHHH so it doesn't even depend on mass, so I can plug a = ug, which just gave me the correct answer :)

Thanks!
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
Replies
18
Views
3K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
6K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
3
Views
2K