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Am I on the right track?

  1. Sep 14, 2007 #1
    Question:
    A. A car that weighs 12000.0 N is initially moving at a speed of 57.0 km/hr when the brakes are applied and the car is brought to a stop in 2.6 s. Find the magnitude of the force that stops the car, assuming it is constant.

    B. What distance does the car move during this time?

    Okay so I drew the diagram and I am planning on using the f=ma equation but I think I need to solve for a first using Vf=Vo + at. Am I on the right track or am I completely off??
     
  2. jcsd
  3. Sep 14, 2007 #2

    learningphysics

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    Yes, you're on the right track.
     
  4. Sep 14, 2007 #3
    Okay so I did...
    (57 km/hr)^2 = a x (2.6s)
    and I got 1249.6 m/s^2...which way wrong...where am I going wrong?
     
    Last edited: Sep 14, 2007
  5. Sep 14, 2007 #4
    Wait I wrote that backwards sorry...

    I got...

    0 = (57 km/hr)^2 +a(2.6s)
     
  6. Sep 14, 2007 #5

    learningphysics

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    Why are you squaring 57?
     
  7. Sep 14, 2007 #6

    PhanthomJay

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    Why are you squaring the velocity? And you've got to be consistent with your units...you can't have seconds on one side of the equation and hours on the other without introducing a conversion factor.
     
  8. Sep 14, 2007 #7
    oh my gosh I don't know why I am squaring it...I am looking at the wrong equation. Okay I changed 57 km/hr to 0.01583 km/s. Now I have...
    0 = 0.01583 km/s + a (2.6s)...
    (-0.01583 km/s) / (2.6s) = a

    so now I can use f=ma
    f= (12000.0 N) x (0.0060897 km/s^2)
     
  9. Sep 14, 2007 #8

    learningphysics

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    careful... you should use m/s for velocity and m/s^2 for acceleration (because your force equation requires kg for mass and m/s^2 for acceleration... to get the result in N)...

    and also 12000.0N isn't the mass, it's the weight... get the mass of the object.
     
  10. Sep 14, 2007 #9
    okay so this is what I have now...

    0=15.83 m/s + a x (2.6s)
    (-15.83m/s) / (2.6s) = a = (-6.08846)

    w=mg
    12000/9.8=1224.49

    f=ma
    1224.49 * -6.08846 = -7455.26 N
    That is really big!!
     
  11. Sep 14, 2007 #10

    learningphysics

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    Looks good. That's the right answer.
     
  12. Sep 14, 2007 #11
    It says its incorrect....I don't know what is wrong.
     
  13. Sep 14, 2007 #12

    learningphysics

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    I don't think you carried enough decimal places... I get -7456.8N
     
  14. Sep 14, 2007 #13
    should I be entering it as positive since its the magnitude?
     
  15. Sep 14, 2007 #14

    learningphysics

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    Yes! I didn't notice that it asked for magnitude!
     
  16. Sep 15, 2007 #15
    Yay it's right!! Hehe...Thank you so much for your time!! I really do appreciate it!!
     
  17. Sep 15, 2007 #16

    learningphysics

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    no prob.
     
  18. Sep 16, 2007 #17
    How do I start the second part of this question?
     
  19. Sep 16, 2007 #18
    Hey,

    If I remember right I believe you should use the displacement function,

    [tex]
    {x}(t) = {x}_{0} + {v}_{x_{0}}{t} + \frac{1}{2}{a}_{x}{t}^{2}
    [/tex]

    I think that should work.

    Thanks,

    -PFStudent
     
  20. Sep 16, 2007 #19
    Thats what I used but it says incorrect?
    I put...

    Xt = 0 + (57 km/hr) x (2.6s) + (1/2)(21.9km/s^2)(2.6s^2)
     
  21. Sep 16, 2007 #20

    learningphysics

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    don't mix up your units... do everything in m, s, m/s, m/s^2
     
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