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Stopping distance problem

  1. Oct 12, 2009 #1
    1. Need help with daughters physics....been 35 yrs since I was in class.

    100kg car moves at speed of 29.3m/s on horizontal road
    Static CoF = 0.145
    Kinetic CoF = 0.1015
    Acceleration of gravity = 9.8m/s squared

    Road is wet
    Neglect Reaction time of driver

    What is shortest possible stopping distance for car under these conditions? answer in "m"

    2. I think you might star with 2nd law: Fnet=m*a but I can't figure out how to work with the CoF's
     
    Last edited: Oct 12, 2009
  2. jcsd
  3. Oct 12, 2009 #2

    lanedance

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    Homework Helper

    in the static case - the maximum sustainable friction froce is, (before the wheels lock up & slide)
    Ff = cs.N

    where cs is the static coefficient of friction, N is the reaction force form teh road on the tires, which in this case will be exactly equal (as it is balancing) to the graviataional force

    in the sliding case - the friction force is
    Ff = ck.N

    where ck is the kinetic coefficient of friction

    use these to find the acceration to use in the 2nd law
     
  4. Oct 13, 2009 #3
    Many Thanks!
     
  5. Oct 13, 2009 #4
    OK...synapse lapse again....

    I have Velocity initial (Vi) = 29.3m/s and Velocity final (Vf)= 0ms/

    Calculated Ff Static = 1.421 (m/sec sq?)
    Calculated Ff Kinetic = 0.9947(m/sec sq?)

    Where do I go from here to find the distance?
     
  6. Oct 13, 2009 #5
    The units of force are Newton - [tex]\frac{kg\cdot m}{s^2}[/tex]

    You posted the correct formula, you just didn't apply it properly. Look at [tex]F=ma[/tex] again.
     
  7. Oct 13, 2009 #6
    Since the car has an initial velocity (Vi) and is being accelerated by the force of friction which is slowing the car to a final speed of (Vf) of zero, you need to find the acceleration of the car by the formula F=ma where F equals the force of friction static (since the road is wet and the car is sliding), m the mass of the car, and then solve for a. Once you have Vi, Vf, a, you should be able to find the distance traveled through the equation

    Vf2 = Vi2 + 2ad where d is the distance the car travels
     
  8. Oct 13, 2009 #7
    Thanks to all! I was almost there....just circling around the final. Appreciate the help!
     
  9. Oct 13, 2009 #8
    Man...have I spent too much time in financial spread sheets today

    OK...
    Static CoF (Ff-s) = M x a

    Ff-s = 0.145 x 9.8m/sec squared = 1.421 m/sec sqrd

    1.421 m/sec sqrd = 1000kg x a
    1.421 m/sec sqrd / 1000kg = a
    0.001421 m kg/sec sqrd = a

    Vf sqrd = Vi sqrd + 2ad
    (0 m/s) sqrd = (29.3 m/s) sqrd = 2(0.001421 m kg/sec sqrd)(d)
    (0 m sqrd/sec sqrd) - (858.49 m sqrd/sec sqrd) = .002842 m kg/sec sqrd(2d)
    (-858.49 m sqrd/sec sqrd) / .002842 m kg/sec sqrd = 2d
    -302072.48 Label =???? = 2d
    -151036.24 label =???? = d

    I'm guessing that utlimatly the stopping distance will be 151.03624m, but I can't seem to rationalize how I would get there....

    where am I missing it??
     
  10. Oct 13, 2009 #9
    You're second part finding the distance is correctly done but with the wrong numbers. Once you correct the numbers plug in the new correct numbers and you'll have correct distance...

    For the first part, you are a little off...

    First of all we want to find the Ff of the car. We know that Ff=(cof static friction)*N
    N=mg = 100*9.8
    Ff=.145(N) and for a note of reference, force is measured in newtons or N (don't be confused with the N represented by normal force)

    Once you have Ff, set that equal to F=ma where the force is the Ff we just calculated.
    Ff=m*a
    Solve for a which is measured in m/s2

    Now that you have correct a, refind the distance as you did correctly before.

    That should give you the correct answer
     
  11. Oct 13, 2009 #10
    OK...think I have muddled through

    Force = CoF Static )Ff-s x N
    Force = 0.145 x (100 x 9.8m/sec sq)
    Force = 142.1

    F=ma
    142.1 = 1000kg x a
    0.1421m/sec sq = a

    Vf sq = Vi sq =2ad
    (0m/sec) sq = (29.3m/sec)sq + 2ad
    -858.49 = 2(.1421)2d
    -858.49 / .2842 = 2d
    3020.73/2 = d

    1510.36 m = d

    How'd I do??
     
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