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Help with explanation

  1. Oct 29, 2004 #1
    ok, i have a problem and its answer, but im not sure how to get the answer:
    Police lieutenants, examining the scene of an accident involving two cars, measure the skid marks of one of the cars, which nearly came to a stop before colliding, to be 81 m long. The coefficient of kinetic friction between rubber and the pavement is about 0.70. Estimate the initial speed of that car assuming a level road.
    the answer is 33.34.
    I can't figure out how to get that answer.
  2. jcsd
  3. Oct 29, 2004 #2
    Essentially, this is a problem with requires the use of kinematics and Newton's second law.
    The only force acting on the car (neglecting things like drag) is the force of the friction, [tex]F_k[/tex]
    Now, we know that [tex] F_k = \mu _k N [/tex]
    Here, the normal force [tex] N = mg [/tex]
    Now, let us substitute N into [tex]F_k[/tex]:
    [tex]F_k = \mu _k mg[/tex]
    From Newton's second law we know that: [tex] F_{net} = ma [/tex]
    But since we know that the only force acting on the car is [tex]F_k[/tex], we can substitute [tex]F_k[/tex] for [tex]F_{net}[/tex]. We get:
    [tex]F_k = ma [/tex]
    [tex]\mu _k mg = ma [/tex]
    Canceling the m-s out,
    [tex]\mu _k g = a[/tex]
    Now, let us use kinematics equations:
    [tex]v_f^2 = v_0^2 + 2a(\Delta x)[/tex]
    From which we get,
    [tex]v_0 = \sqrt{2a(\Delta x) - v_f^2}[/tex]
    Substituting for a
    [tex]v_0 = \sqrt{2(mu _k g)(\Delta x) - v_f^2}[/tex]
    Substituting our givens,
    [tex]v_0 = \sqrt{2(0.70)(9.81)(81) - 0}[/tex]
    [tex]v_0 = 33.34[/tex]

    Hope this helped.
    Last edited: Oct 29, 2004
  4. Oct 30, 2004 #3
    yes thank you.
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