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Angular motion problem involving a car coming to a stop

  1. Oct 26, 2008 #1
    The driver of a car traveling at 12.0 m/s applies the brakes and undergoes a constnant decelertion of 1.90m/s^2.
    How many revolutions does each tire make before the car comes to a stop assuming that the car does not skid and that the tires of radii of 0.40 m? answer in units of rev.

    I used the equation Vf^2=Vi^2+2ad
    0=144+2(-1.90)(d)
    d=37.89meters
    d=r(θ)
    θ=37.89/0.40=94.725 radians
    revolutions=94.725/(2pi)=115.076 revolutions.
    However, I did not get a correct answer. Can anyone help?
     
  2. jcsd
  3. Oct 26, 2008 #2

    hage567

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    Is this a typo? Check your math here.
     
  4. Oct 26, 2008 #3
    I meant to write 15.075 and it's still not right though.
     
  5. Oct 27, 2008 #4

    hage567

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    I don't see what's wrong with that answer. Are you using correct sig figs when you enter your answer?
     
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