University of Wisconsin Green Bay

You are driving your car along a level residential street at 35 mph (16 m/s) when you see a ball roll out into the street in front of you. You slam on your brakes in case anyone follows the ball out into the street. If the coefficient of friction between your car and the road is 0.67, how far do you travel before you come to a stop?

  • In this problem, you are asked to relate motion (the car comes to a stop) to force (friction). Force and motion of a single object are always related through Newton’s Second Law, so this is a force or 2nd Law problem even though you are asked about distance! You should always approach a problem first by thinking about the key interactions described, regardless of what quantity you are asked to find. Going from acceleration to distance will be a second step in this problem.

    If you have already studied energy, you will recognize that this problem can be solved more efficiently using energy. Click here for the energy solution.

  • n car moving right car moving right Brakes mg n fr +x mg
  • Stopping Distance

    The key equation for any problem that relates forces and motion is Newton’s Second Law. Regardless of what quantity you are asked to find, begin with the Second Law. If additional information is needed, it will become apparent as you proceed even when that additional information is the quantity you are asked to find.

  • Step 1

    0 mg 0 0 0 0 -fr a a a un

    In this problem, you are asked to find the stopping distance of the car. Because you were given information about the strength of the braking force, you were able to find the acceleration of the car. Acceleration can now be used to look at variables as distance, which describe the motion. Scroll down to continue to the second part of this problem.


    Step 2

    At this stage in the problem, you know acceleration and velocities and are asked to find distance. In other words, you are asked to describe motion, or do a kinematics problem.

    Stopping Distance

    To select the appropriate kinematic equation, recognize that you are asked for the location when the car has stopped. In other words, you are asked to relate position (x) and velocity (v).

    v2 = v02 + 2a(Δx)
    0 = (16 m/s)2 + 2(-6.6 m/s2) (Δx)
    -256 m2/s2 = -13.2 m/s2(Δx)
    19 m = Δx

  • ΣF=ma
    -fr = ma
    -6.6 m/s2 = a

    In this problem, the car comes to a stop (accelerates) because of the frictional force. Therefore, Newton’s 2nd Law was used to find the acceleration on the car. The negative sign on 'a' tells you that acceleration is opposite to the (+ direction) motion of the car—friction is acting to slow the car down. The relatively high value (about 2/3 the acceleration due to gravity) indicates that, indeed, you slammed on the brakes.

    Once we know the acceleration that results from the forces present, we are able to describe the motion. Because we are asked to relate velocity and location, the most productive equation is v2 = v02 + 2a(Δx), which gives a stopping distance of 19 m. City blocks are typically 100-200 m, so this is a stopping distance of about 1/5 of a block or less.