Distance to bring car to a full stop

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In summary, the conversation discussed a problem involving a car traveling at 64 mi/h and the distance required to bring the car to a full stop when the brakes are suddenly applied. It was assumed that the coefficients of friction between the tires and the road were μK = 0.80 and μS = 0.90. The problem was solved using Newton's second law and the equation vf^2=v0^2+2ax for both the cases of skidding and when the wheels are not locked up. The use of antilock brakes in emergency stops was also discussed.
  • #1
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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  • #2
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?
 
  • #3
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
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
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.
 
  • #6
Hope this will help you on μs(rolling) and μk(skidding)

http://imageshack.us/a/img651/93/50797244.jpg
 
Last edited by a moderator:
  • #7
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.
 

FAQ: Distance to bring car to a full stop

What is the formula for calculating the distance to bring a car to a full stop?

The formula for calculating the distance to bring a car to a full stop is: distance = (initial velocity^2) / (2 * deceleration).

What factors affect the distance required to bring a car to a full stop?

There are several factors that can affect the distance required to bring a car to a full stop, including the initial velocity, road conditions (such as wet or icy roads), tire traction, and the type and condition of the brakes on the car.

How does the weight of the car impact the distance needed to bring it to a full stop?

The weight of the car does impact the distance needed to bring it to a full stop. A heavier car will require a longer distance to stop because it has more momentum to overcome. This is why it is important to consider weight when purchasing a car and to regularly maintain the brakes on heavier vehicles.

Is there a recommended safe following distance to allow enough space to bring a car to a full stop?

Yes, the recommended safe following distance is three seconds. This means that when the car in front of you passes a fixed object, you should be three seconds behind it. This allows enough time and distance for you to react and bring your car to a full stop if necessary.

How does the speed limit affect the distance needed to bring a car to a full stop?

The speed limit does have an impact on the distance needed to bring a car to a full stop. The higher the speed, the longer the distance needed to stop. This is why it is important to always follow the posted speed limit and adjust your driving accordingly in different weather and road conditions.

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