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Physics car breaking problem

  1. Jul 9, 2005 #1
    Travelling at a speed of 58.0 km/h, the driver of an automobile suddenly locks the wheels by a slamming on the brakes. The coefficient of kinetic friction between the tires and the road is 0.720. How far does the car skid before coming to a halt? Ignore effects of air resistence.

    How to solve this problem? I have no idea at all how to start since seem like lack of some others value. We have only the value of v and coefficient of kinetic friction. How to solve this??
     
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  3. Jul 9, 2005 #2

    Pyrrhus

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    Start by drawing a FreeBody Diagram (find the acceleration, using Newton's 2nd Law), and then work the time with the uniform acceleration equations from kinematics.
     
  4. Jul 9, 2005 #3
    How to find out acceleration as we don`t know the mass of the car?
     
  5. Jul 9, 2005 #4

    Pyrrhus

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    You don't need the mass of the car, why don't you set up the equations algebraicly and see where it takes you?
     
  6. Jul 10, 2005 #5

    Päällikkö

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    Or you can use the conservation of energy principle.
     
  7. Jul 10, 2005 #6
    Be v=58 km/h the velocity and f=0.720 the friction coefficient.

    The only force acting on the car is friction. Newton tells us that the sum of all forces acting on a point of mass equals m*a. So you know the friction equals m*a. That is (on a horizontal plane):

    [tex]m\cdot a=f\cdot m \cdot g[/tex]
    Divide by m on both sides and you get:
    [tex]a=f\cdot g[/tex]

    Now you know the acceleration.

    Be the time at the moment, where the car starts breaking (the wheels stand still), [tex]t_{i}=0[/tex] and the distance at that point of time [tex]s_{i}=0[/tex].
    Notice: we have reduced the problem thru a dynamic consideration (fricition is the only force) to a purely kinematic one. We have now an accelerated motion (no initial distance and no movement with a constant speed). That's it. So we use the kinematic formulas you're surely familiar with:

    [tex]s=\frac{1}{2}\cdot a t^2[/tex]
    We also know that acceleration a=v/t. Therefore, t=v/a. We can thus compute s:
    [tex]s=\frac{1}{2}\cdot a \left(\frac{v}{a}\right)^2=\frac{1}{2}\cdot \frac{v^2}{a}[/tex]

    That was it.
     
  8. Jul 10, 2005 #7
    Thanks..but I am still not really understand why ma = f (mg). Even the car is moving with a deceleration, isn`t there any force which cause the car moving forward?
    Some more, why consider the initial speed = 0 when finding the distance?
     
  9. Jul 10, 2005 #8

    Doc Al

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    The only horizontal force on the car is that of friction. And the value of the kinetic friction force is [itex]F_f = \mu N[/itex], where N is the normal force between the surfaces (the tires and the road). In this case, the normal force is just the weight of the car (mg). Note that the mass cancels, so you don't need to know what it is.
    A force is only needed to change the motion. If there were no forces acting to slow the car down, the car would keep moving forward.
    It may be less confusing if you took the initial speed to be 58.0 km/h, the final speed to be 0, and the acceleration to be negative. (Don't forget to change everything to standard units. For example, speed should be in m/s.)
     
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