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Driver going 40km/hr slams on the brakes & goes 3m before stopping - how far for 80?

  1. Jun 21, 2006 #1
    When the driver applies the brakes of a light truck traveling at [itex]40\,\frac{km}{hr}[/itex], it skids 3 m before stopping. How far will the truck skid if it is traveling [itex]80\,\frac{km}{hr}[/itex] when the brakes are applied?

    Here is what I have done, we are supposed to use the work energy formula, but I can't figure out how to relate the small amount of information. There is no mass specified, no kinetic friction coeeficient specified? Please help!



    I figured [itex]t_1[/itex] using kinematics:







    Then I use these numbers in another kinematic equation for the 80 km/hr instance:




    The answer is actually 12m though.
    Last edited: Jun 21, 2006
  2. jcsd
  3. Jun 21, 2006 #2
    If the distance it travels before stopping is 3 meters, taking s0 as 0 like you've done then then s should be 3 not 6 since the total displacement is only 3 meters. Also it may have been easier to use a different equation from the start, one that doesn't take into account time since you aren't given a stopping time, such as

    vf2 - v02 = 2as
  4. Jun 21, 2006 #3




    That is right if you round up! Thanks for the help.

    NOTE: I think we were supposed to solve it using a work-energy formula (i.e. - [itex]\frac{1}{2}\,m\,v_0^2\,+\,\sum\,U\,=\,\frac{1}{2}\,m\,v_f^2[/itex]) somehow.
    Last edited: Jun 21, 2006
  5. Jun 21, 2006 #4
    Your welcome, I'm glad I could help.
  6. Jun 21, 2006 #5
    Note that you don't need to actually find the deceleration to get to the answer. This can be done as follows,
    [tex]v^2 = 2aS=>v^2\varpropto S[/tex]
    since acceleration is constant in both cases , we get
    Now you can solve by inserting given values.
    Do you follow ?
    Last edited: Jun 21, 2006
  7. Jun 21, 2006 #6


    But it's really 12!
    Last edited: Jun 21, 2006
  8. Jun 21, 2006 #7


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    You forgot to square the velocities. Re calculate your value and you should obtain 12m.
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