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Homework Help: Stopping distance

  1. Aug 24, 2010 #1
    1. The problem statement, all variables and given/known data
    The maximum straight-line deceleration of a racing car under braking is 5 m s-2. What is the minimum stopping distance of the car from a velocity of 108 km h-1?



    2. Relevant equations

    ½ V x t

    3. The attempt at a solution

    V = 108km/h=30m/s

    t = 30/ 5 = 6 m s-2

    ½ V x t = ½ x 30 x 6 = 90

    Answer: Distance = 90 m

    ___________

    I found this formula ½ V x t (to determine stopping distance) on the internet, but im not sure if its correct.

    Are there any mistakes?

    Thank You
     
  2. jcsd
  3. Aug 24, 2010 #2
    It seems like your formula for distance is the average distance travelled in a given time for an object moving at constant speed. I would Instead try the formula:
    vf 2 = vo 2 + 2 a [tex]\Delta[/tex] x

    Note that you are decelerating so the plus in front of the 2 is actually a minus here.
     
    Last edited: Aug 24, 2010
  4. Aug 24, 2010 #3
    so: vf = 30 m/s -2 x 5 m/s...
    what is this delta x? sorry, haven't slept this night at all, feeling a bit dizzy now:/
     
  5. Aug 24, 2010 #4
    Sorry, delta x here stands for the change in position of the car.

    Basically, how far it goes in stopping.

    Edit: Sorry I'm without a calculator so I hadn't noticed but this gives the same answer you already had.
     
  6. Aug 24, 2010 #5
    90 meters?
     
  7. Aug 24, 2010 #6
    Yes, 90 meters. Aside from the units in your 't', everything seems to be in order.
     
  8. Aug 24, 2010 #7
    Thank you.

    By the way, whats wrong with t?
     
  9. Aug 24, 2010 #8
    Units: I'm a little mixed up by your scheme but it looks like you have t (time) in meters per second per second, instead of just seconds.
     
  10. Aug 24, 2010 #9
    Oh seriously... I fixed that. Thanks again:)
     
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