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Work-Energy Theorem: Effect of increasing speed vs stopping distance.

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data
    If the speed of a car is increased by 50%, by what factor will the minimum braking distance be increased assuming all else is the same? Ignore the driver's reaction time.


    2. Relevant equations
    [tex] W = Fd \cdot cos\theta [/tex]
    [tex] W = \Delta{E_k} [/tex]


    3. The attempt at a solution
    First i declared:
    [tex] Fd = \frac{1}{2}mv^2 [/tex]

    Where d would be the stopping distance, there is no final velocity because its stopping. I then stated:

    [tex] Fd = \frac{1}{2}m(1.5v)^2 [/tex]
    [tex]Fd = \frac{1}{2}m\cdot2.25v^2
    [/tex]

    so due that F, the applied braking force is constant, the mass remains constant of course, the only thing that changes is the kinetic energy is multiplied by 2.25, so for Fd to correspond, d would have to be 2.25 times larger correct?

    This question has no answer in the key..

    Personal Reference: Q24 p. 145

    Im always so horrible with these questions that don't give any data for the variables .. :S

    Thanks,
    Senjai
     
  2. jcsd
  3. Oct 11, 2009 #2
    bump again :( can someone please.. pretty please, confirm :)
     
  4. Oct 12, 2009 #3
    You're first declaration: F*d = (1/2)*m*v^2 seems logical, you're assuming all kinetic energy must be reduced to zero by the work done through braking.

    The assumption that the braking force is constant in both situations seems adequate for the problem.

    If speed is increased to 1.5 that of the original, you look good in saying that d should be 2.25 times as long as originally determined. Your work makes sense to me.
     
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