Distance to bring car to a full stop

  • Thread starter Thread starter DOA_Kasumi22
  • Start date Start date
  • Tags Tags
    Car
AI Thread Summary
The discussion focuses on calculating the stopping distance of a car traveling at 64 mi/h under two conditions: skidding with locked wheels and rolling with normal friction. The coefficients of kinetic and static friction are provided as μK = 0.80 and μS = 0.90. Participants emphasize the need to correctly apply Newton's second law to determine acceleration, with specific calculations for each friction type yielding -8.83 m/s² for static friction and -7.85 m/s² for kinetic friction. The effectiveness of antilock brakes in reducing stopping distance during emergencies is also questioned, highlighting the importance of understanding friction dynamics in braking scenarios.
DOA_Kasumi22
Messages
7
Reaction score
0

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:
Physics news on Phys.org
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.
 
Back
Top