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Homework, Brake Question

  1. Nov 30, 2006 #1
    1. The problem statement, all variables and given/known data
    Driving home from school one day you spot a ball rolling out into the street (Figure 5-18). You brake for 1.15 s, slowing your 1020 kg car from 16.0 m/s to 9.50 m/s.

    a) What was the average force exerted on your car during braking?
    Magnitude
    _____ kN


    (b) How far did you travel while braking?

    2. Relevant equations
    xf=xi+.5(vi+vf)t
    vf^2=vi^2+2a(xf-xi)

    3. The attempt at a solution

    I have been unable to figure out this problem!

    From the information gathered, I know the following
    vi=16.0
    vf=9.5
    t=1.15

    now I do not know where to go from here, ive tried using the above formulas to find the acceleration so I can plug it into the Fnet=MA equation, but still im kind of lost. How the heck do I find the average force exerted on the car while brakeing
     
  2. jcsd
  3. Nov 30, 2006 #2

    PhanthomJay

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    Use your relevant equations and solve for Xf and a, setting xi = 0. Then yes use f=ma to solve for the average force during braking.
     
  4. Nov 30, 2006 #3
    im still kind of confused, I think the equations I need to use for acceleration are vf=vi+axt, but I still got the question wrong, even for part b where it asks me how far it went! Somebody please halp!

    vf=vi+axt then
    9.5=16+1.15a

    -6.5=1.15a

    a=-5.65 am I not correct?
     
  5. Dec 1, 2006 #4

    OlderDan

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    Your acceleration looks OK. How do you find the force? What distance did you get for b? Your equations are OK.
     
  6. Dec 1, 2006 #5
    well b should be -5.65(1.15) which = -6.4975 but that is not correct for b.

    and if I was to do the Fnet I would assume that Fnet=MA

    Fnet= 1020(-5.65)

    right?
     
  7. Dec 1, 2006 #6

    OlderDan

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    The force is correct once you get the units included. Multiply it out. The distance is not correct. Keeping track of units will help you know when things are not right. What you calculated is acceleration times time, which has units of velocity and is in fact the change in velocity. Go back to either of your two oringinal equations and compute xf - xi. That is the distance.
     
  8. Dec 1, 2006 #7
    but how do I find xf or xi if I know xi = 0 and xf = ?
     
  9. Dec 1, 2006 #8
    I plugged it into the position versus time graph and got xf=14.663

    xf=xi+vixt+.5at^2

    xf=0+16(1.15)+.5(-5.65)(1.15)^2
     
  10. Dec 1, 2006 #9
    Woohoo I got b. correct! But now for a. Fnet=MA then Fnet= 1020(-5.65)

    Fnet = -5763

    Then the average force exerted on the car while breaking is -5763 kN?

    kN im guessing is kilonewtons?
     
  11. Dec 1, 2006 #10

    PhanthomJay

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    You are not keeping track of your units. 1Kg*m/s^2 = 1 newton
     
  12. Dec 1, 2006 #11
    so how do I convert -5763 to kilonewtons?

    Infact what is -5763? If the mass is kilograms and the acceleration was in.. meters?

    Then.. -5763 would be kilograms right? 1 kilogram = 9.8 newtons? and 1000 newtons = 1 kilonewton? so -5763 would be -57630?
     
  13. Dec 1, 2006 #12

    PhanthomJay

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    You have to study up on the units. The acceleration is in meters/second^2. Velocity is in meters/second. Distance is in meters. And 1 newton is shorthand for 1Kg*m/second^2. and 1000N = 1kN .

    F=ma
    F=
    1050Kg(-5.65m/s^2) = -5763Kg*m/s^2 = -5763N
    -5763N(1kN/1000N) = - 5.763kN

    As an aside, you should note that the fprmula for weight, which is unrelated to this problem, is W=mg, so 1Kg weighs (1)9.8 Kg(m)/s^2 = 9.8 Newtons
     
  14. Dec 1, 2006 #13
    -5.763 kN was not correct, I still dont understand!
     
  15. Dec 1, 2006 #14

    PhanthomJay

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    Maybe the car was moving to the left. Try using +5.763kN.
    EDIT: Oh, now i see the problem asked for the magnitude of the force, not the direction. The minus sign means that the force is opposite the direction of the car's motion. The magnitude of the force is 5.763kN.
     
    Last edited: Dec 1, 2006
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