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Distance to bring car to a full stop

  • #1

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:

Answers and Replies

  • #2
vela
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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
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?
 
  • #3
TSny
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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.
 
  • #4
CWatters
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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?
 
  • #5
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.
 
  • #6
1,065
10
Hope this will help you on μs(rolling) and μk(skidding)

http://imageshack.us/a/img651/93/50797244.jpg [Broken]
 
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  • #7
vela
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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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