Kinetic Friction to find velocity

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Homework Help Overview

The problem involves determining the speed of a car just before it braked, based on the length of skid marks and the coefficient of kinetic friction. The context is set within a scenario related to an automobile accident.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using kinematic equations to relate distance, initial velocity, and acceleration, with some uncertainty about how to calculate acceleration. There is mention of the formula a = (mu x g) and its derivation, prompting questions about its validity and origin.

Discussion Status

Some participants have expressed confusion regarding the application of certain formulas and concepts, while others have provided clarifications and insights into the relationships between force, friction, and acceleration. There is an ongoing exploration of the underlying principles without a definitive consensus on the approach.

Contextual Notes

Participants note that their educational resources have not adequately covered certain concepts, leading to questions about foundational physics principles such as the coefficient of friction and its application in this context.

SuPaFly43218
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Homework Statement



You have been called to testify as an expert witness in a trial involving an automobile accident. The speed limit on the highway where the accident occurred was 40 mph. The driver of the car slammed on his brakes, locking his wheels, and left skid marks as the car skidded to a halt. You measure the length of these skid marks to be 219 ft, 9 in., and determine that the coefficient of kinetic friction between the wheels and the pavement at the time of the accident was 0.400.

How fast was this car traveling (to the nearest number of mph) just before the driver hit his brakes?

Homework Equations



I would guess using the old formula of x1-x0 = (Vf^2-Vi^2)/2a would work. I am not sure how to find a though.

The Attempt at a Solution



0-66.9798 (that is the 219.75 ft converted to meters) = (0 - Vi^2)/2a

I am not sure where to go from here; please help! I know the answer is 51.2 mph, but I am lost on where to go.
 
Last edited:
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Bump! This is very important!
 
SuPaFly43218 said:

Homework Statement



You have been called to testify as an expert witness in a trial involving an automobile accident. The speed limit on the highway where the accident occurred was 40 mph. The driver of the car slammed on his brakes, locking his wheels, and left skid marks as the car skidded to a halt. You measure the length of these skid marks to be 219 ft, 9 in., and determine that the coefficient of kinetic friction between the wheels and the pavement at the time of the accident was 0.400.

How fast was this car traveling (to the nearest number of mph) just before the driver hit his brakes?

Homework Equations



I would guess using the old formula of x1-x0 = (Vf^2-Vi^2)/2a would work. I am not sure how to find a though.

The Attempt at a Solution



0-66.9798 (that is the 219.75 ft converted to meters) = (0 - Vi^2)/2a

I am not sure where to go from here; please help! I know the answer is 51.2 mph, but I am lost on where to go.

32.2 f/s2 is the acceleration from gravity. And they tell you the coefficient of friction is .4 of that.

So that yields your velocity in f/s from V2 = 2*a*x

Then to get mph divide by 1.46666.

Easy peasy. Nice and easy.
 
Right, i got the answer... but what is the formula a= (mu x g). I have never seen this formula before. Is this derived from something else??
 
SuPaFly43218 said:
Right, i got the answer... but what is the formula a= (mu x g). I have never seen this formula before. Is this derived from something else??

F = u*m*g = m*a

a = u*g
 
Thank you, our physics book and teacher never explained that to us. Just to get this right, you said that (Force) F = mu x m (mass) x g (gravity)?
 
Nvm, i figured it out. Thank you for your help LowlyPion, your help is much appreciated.
 
SuPaFly43218 said:
Thank you, our physics book and teacher never explained that to us. Just to get this right, you said that (Force) F = mu x m (mass) x g (gravity)?

The retarding force of friction divided by the normal force to the surface is the coefficient of friction.
That's what mu is.
 

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