Solving a Problem: Car Speed When Driver Slammed Brakes

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SUMMARY

The discussion focuses on calculating the speed of a car at the moment the driver applied the brakes, using the length of skid marks (88 m) and the coefficient of friction (0.42). The key formula derived is the deceleration, where acceleration (a) equals the coefficient of friction (μ) multiplied by the acceleration due to gravity (g). The final velocity is then calculated using the equation v² = u² + 2as, where 's' is the distance of the skid marks. The confusion arose from misinterpreting the forces acting on the vehicle during deceleration.

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  • Understanding of Newton's Second Law (F=ma)
  • Knowledge of friction coefficients and their application in physics
  • Familiarity with kinematic equations, specifically v² = u² + 2as
  • Basic concepts of forces acting on a vehicle during motion
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  • Learn more about kinematic equations and their derivations
  • Explore the effects of different road conditions on vehicle braking distances
  • Investigate advanced topics in physics related to motion and forces, such as energy conservation
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kbrowne29
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I'm a bit confused as to how to finish this problem:
At an accident scene on a level road, inestigators measure a car's skid marks to be 88 m long. It was a rainy day and the coefficient of friction was estimated to be .42. Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes.

OK, here's what I can get. Using F=ma, I know the net force is equal to the force of friction minus the applied force. And so we get F(friction)-F(applied)=ma. We also know that F(friction)=
mu(kinetic)F(normal). And the normal force is equal to mg. And so the force of friction is mu(kinetic)mg. So, mu(kinetic)mg-force(applied)=ma. Now, the applied force can be rewritten as ma(applied). Now the mass can be canceled out of each term, and the result is: mu(kinetic)g-a(applied)=a. This is where I get stuck. I can't seem to figure out how to determine the applied acceleration. Once I get that, then using v(final)^2=v(initial)^2+2a(delta x), I can find the initial velocity. I would appreciate any help on this problem. Thanks.
 
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Well, here's how I solved the problem:

The force of friction = μR

R = mg

So,

Friction = μmg

When the car's brakes are applied, the force causing the deceleration is equal to the force of friction, is it not?

Therefore

ma = μmg

And

a = μg

Then just use v^2 = u^2 + 2as, like you said.
 
Originally posted by kbrowne29

OK, here's what I can get. Using F=ma, I know the net force is equal to the force of friction minus the applied force.
What "applied" force? The only horizontal force on the car is the friction of the road!
 


Originally posted by Doc Al
What "applied" force? The only horizontal force on the car is the friction of the road!

That's exactly where I was going wrong. For some reason, I kept solving the problem as if the car were accelerating, not decelerating.
Thank you both for your help.
 

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