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Tire Marks

  • Thread starter kbrowne29
  • Start date
  • #1
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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 cancelled 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.
 

Answers and Replies

  • #2
AD
72
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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.
 
  • #3
Doc Al
Mentor
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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!
 
  • #4
12
0


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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