Distance to bring car to a full stop

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Homework Help Overview

The discussion revolves around calculating the distance required for a car traveling at 64 mi/h to come to a full stop under different conditions: when skidding and when the wheels are not locked. The problem involves concepts of friction, acceleration, and the physics of motion.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of the equation vf^2=v0^2+2ax and question the values used for initial velocity and acceleration. There are inquiries about the relevance of static and kinetic friction coefficients in the calculations.

Discussion Status

Some participants have attempted calculations using different values for acceleration based on friction, while others are questioning the methods and assumptions made in the original attempts. There is an ongoing exploration of how to correctly apply the principles of physics to this problem.

Contextual Notes

Participants note that the acceleration should not be assumed to be 9.81 m/s² without justification, as the car is not in free fall. There is also mention of previous posts being deleted due to providing too much assistance, indicating a focus on guiding rather than solving.

DOA_Kasumi22
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Homework Statement


A car travels at 64 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect to the road; otherwise, when the tires roll, normally the point where the tire contacts the road is stationary. Assume the coefficients of friction between the tires and the road are μK = 0.80 and μS = 0.90.

(a) Calculate the distance required to bring the car to a full stop when the car is skidding.

(b) Calculate the distance required to bring the car to a full stop when the wheels are not locked up.

(d) Can antilock brakes make a big difference in emergency stops? Explain.


Homework Equations



Possibly

vf^2=v0^2+2ax

The Attempt at a Solution



(0m/s)^2 =(47)^2 +2(9.81m/s/s)

x^2=2228.62
x=47 m
 
Last edited:
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DOA_Kasumi22 said:

Homework Statement


A car travels at 64 mi/h when the brakes are suddenly applied. Consider how the tires of a moving car come in contact with the road. When the car goes into a skid (with wheels locked up), the rubber of the tire is moving with respect to the road; otherwise, when the tires roll, normally the point where the tire contacts the road is stationary. Assume the coefficients of friction between the tires and the road are μK = 0.80 and μS = 0.90.

(a) Calculate the distance required to bring the car to a full stop when the car is skidding.

(b) Calculate the distance required to bring the car to a full stop when the wheels are not locked up.

(d) Can antilock brakes make a big difference in emergency stops? Explain.


Homework Equations



Possibly

vf^2=v0^2+2ax

The Attempt at a Solution



(0m/s)^2 =(47)^2 +2(9.81m/s/s)

x^2=2228.62
x=47 m
Where did you get v0=47 from?
What units are you using?
Where did you use the info about static friction and kinetic friction?
Why would the acceleration be 9.81 m/s2?
Where is x in your equation?
Where did x2=2228.62 come from?
 
The acceleration is not 9.81 m/s2. (The car is not in free fall.)

Try using Newton's second law to find the acceleration for each case.
 
My earlier reply was deleted by the mods as it provide too much help. I'll try again..

DOA_Kasumi22 - Explain how you got the acceleration term in your equation. I can see why "g" is in there but do you?
 
CWatters said:
My earlier reply was deleted by the mods as it provide too much help. I'll try again..

DOA_Kasumi22 - Explain how you got the acceleration term in your equation. I can see why "g" is in there but do you?

Well I reworked the problem again and this time to get the acceleration for staic friction I multipled -0.90 x 9.81= -8.83 and to get the accelration for the kinetic friction multpiled -0.80 x 9.81 = -7.85. Finally I used vf^2=v0^2+2a(d) for each of the two acclrations but my answer was still wrong.
 
Hope this will help you on μs(rolling) and μk(skidding)

http://imageshack.us/a/img651/93/50797244.jpg
 
Last edited by a moderator:
DOA_Kasumi22 said:
Well I reworked the problem again and this time to get the acceleration for staic friction I multipled -0.90 x 9.81= -8.83 and to get the accelration for the kinetic friction multpiled -0.80 x 9.81 = -7.85. Finally I used vf^2=v0^2+2a(d) for each of the two acclrations but my answer was still wrong.
It would help if, instead of merely describing what you did, you showed us what you actually did. You made mistakes other than using the wrong acceleration in your original attempt, and you probably did again. We can't see where these mistakes are unless we can see your actual work.
 

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