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Kinematics - Constant acceleration and friction

  1. Sep 21, 2006 #1
    Unsolvable? Kinematics - Constant deceleration and friction

    Here's the way the problem is laid out:

    A car travels at 52.4 km/hr on a flat highway. If the coefficient of kinetic friction is 0.151, what is the minimum distance needed for the car to stop?

    Vi = 52.4 km/hr (calculated as 14.555(...) m/s)
    Vf = 0 m/s
    muk= 0.151
    g = 9.81 m/s2

    /\ x, /\ t, a, m

    It seems to me that I have too many missing variables to solve the problem.

    I know that Vavg = (Vi + Vf)/2, so the average velocity would be around 7. 277 m/s, but without /\t, I can't carry out any of the normal kinematic equations to find /\x, and without m, I can't figure out Fk, Fg, or FN.

    Is there something I'm missing, or is the problem actually unsolvable?
    Last edited: Sep 21, 2006
  2. jcsd
  3. Sep 21, 2006 #2
    Why don't you think you can find t? You know the average speed the car will have over the distance. If you have the average rate of change of distance with respect to time, and you know the total distance then you can determine time.
  4. Sep 21, 2006 #3
    But that's just the problem-- I'm trying to find the distance (/\x), and to do that, I would need a, which requires /\t, none of which I have...
  5. Sep 22, 2006 #4


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    Homework Helper

    You can calculate the decceleration of the car from Newton's equation of motion, where the decceleration must be proportional to the resultant force acting on the car, which is the force of friction. Generally, this force equals F = k N, where k is the coefficient of friction (dynamic friction, in your case), and N is the reaction from the ground acting on the car, which equals the weight of the car. Now, since you know the decceleration, you can use the equation [tex]v_{final}=v_{0}-at[/tex] to calculate the time of decceleration from 52.4 km/hr to 0 km/hr. In the end, you can use the equation for displacement of the car, [tex]x(t)=v_{0}t-\frac{1}{2}at^2[/tex], to find the distance which the car passes until it stops.
  6. Sep 22, 2006 #5


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    Science Advisor

    If you don't know the mass of the car, the "coefficient of friction" is irrelevant.
  7. Sep 22, 2006 #6


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    Homework Helper

    Absolutely right, I oversaw that the mass is not given.
  8. Sep 22, 2006 #7

    After asking my teacher, I solved the problem-- I was assuming that there was a force in the opposite direction of friction.

    Basically, the acceleration was

    -(mu(k)Fn)/m, which equals

    -(mu(k)mg)/m, canceling to

    a = -(mu(k)g).

    Then I could use the vf2 = vi2 + 2a/\x equation to solve for distance.

    Thank you all for your help anyway!
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