Solving a Problem: Car Speed When Driver Slammed Brakes

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The discussion focuses on calculating the speed of a car at the moment the driver slammed on the brakes, using the length of skid marks and the coefficient of friction. Participants clarify that the force causing deceleration is solely the force of friction, which equals the product of the coefficient of friction and the normal force. The correct approach involves recognizing that the applied force should not be included, as the car is decelerating due to friction alone. The final formula to determine the initial velocity incorporates the acceleration derived from the frictional force. The conversation highlights the importance of understanding the distinction between acceleration and deceleration in solving physics problems.
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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