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Please look at my method forces

  1. Nov 8, 2009 #1
    1. The problem statement, all variables and given/known data
    a car moving initially at a speed of 80km/h and weighing 3000 lb is brought to a stop in a distance of 61 m.

    a) find the braking force
    b) the time required to stop
    c) the distance and time required to stop if the car was going 40 km/h



    2. Relevant equations

    V^2 = Vo^2 + 2a(p-po)


    3. The attempt at a solution

    this is how I see it, the guy presses his breaks so a force is exerted and so there is an acceleration/deceleration... 0 = 80^2 + 2a(0.061)

    a = 80^2 / 2(0.061)

    well, I would what to do after getting the acceleration, but this is a huge number.. doesn't seem right.. what have I done wrong? thank you
    but also, I'm assuming that 80 km/h is the speed the instant he STARTS BREAKING.. and not his initial velocity for his overall car ride.. this might be wrong.. also, this is assuming that the breaks doesn't spontaneously stop the car?
     
    Last edited: Nov 8, 2009
  2. jcsd
  3. Nov 8, 2009 #2

    jgens

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    Your work is fine.
     
  4. Nov 8, 2009 #3
    thank you for the reply.. but the acceleration I get is -52459 :S did I mess up something with the units?
     
  5. Nov 8, 2009 #4

    jgens

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    Well, your acceleration is in km/h2! :rolleyes:
     
  6. Nov 8, 2009 #5
    thank you for replying.

    huh? I thought I needed it in km/h^2? if I make it km/h it isn't acceleration anymore is it :S
     
  7. Nov 8, 2009 #6

    jgens

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    My point: You have units of acceleration, but your units of choice make the acceleration exceptionally large. As I said earlier, everything you've done so far is fine.
     
  8. Nov 8, 2009 #7
    ohh I see, thank you :) hmm I thought km/h was a standard kind of thing? it isn't like I used m/s or something.. what should I use? Also if -52459 km/h^2 is the right answer, that means that the car slows that super super super fast right? that might make sense I think, since 61 m isn't a long way and it is a brake anyway
     
  9. Nov 8, 2009 #8

    jgens

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    Moving on, since you want to find the breaking force, you probably want to convert km/h2 into m/s2 so that way you can apply Newton's Second Law.
     
  10. Nov 8, 2009 #9
    okay, I just saw a SI chart and I see what you mean .. then for the rest of the questions, I just plug in the acceleration I found into a equation with time.. and then do the same thing but start with v = v0 + at solve for time and then solve for x from x = x0 + vx0t + 1/2at^2 right?
    thanks
     
  11. Nov 8, 2009 #10
    this seriously doesn't look right.. in m/s^2 I have that a = -52 459 000 meter/square second :S
     
  12. Nov 8, 2009 #11

    jgens

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    Not quite right. You didn't convert hours into seconds!
     
  13. Nov 8, 2009 #12
    I looked that up in a calculator, that is what it gave me (i trust the calculator more for conversion haha) . My own work shows this though:
    -188852 m/s^2



    by [tex]\frac{80^2}{2(0.061)}[/tex] * [tex]\frac{3600}{1}[/tex] and then times [tex]\frac{1}{1000}[/tex]

    but still, it looks really big.. does it physically make sense? because the braking effect happens so quickly it is almost instantaneous? but then the question says that it takes 61 m for him to stop.. he must have been going pretty fast then :S thank you
     
  14. Nov 8, 2009 #13

    jgens

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    Well, neither is correct! Write out all of your units and then do the conversions!
     
  15. Nov 8, 2009 #14

    isn't that the way I'm supposed to do it? I had 80^2 km / 2(0.061) hours.. so I did that conversion strategy with the 1 / relative equivalency thing
     
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