Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Tire Marks

  1. Nov 23, 2003 #1
    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.
  2. jcsd
  3. Nov 23, 2003 #2


    User Avatar

    Well, here's how I solved the problem:

    The force of friction = μR

    R = mg


    Friction = μmg

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


    ma = μmg


    a = μg

    Then just use v^2 = u^2 + 2as, like you said.
  4. Nov 23, 2003 #3

    Doc Al

    User Avatar

    Staff: Mentor

    What "applied" force? The only horizontal force on the car is the friction of the road!
  5. Nov 23, 2003 #4
    Re: Re: Tire Marks

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook