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Angular motion problem involving a car coming to a stop

  • Thread starter xregina12
  • Start date
  • #1
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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?
 

Answers and Replies

  • #2
hage567
Homework Helper
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revolutions=94.725/(2pi)=115.076 revolutions.
Is this a typo? Check your math here.
 
  • #3
27
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I meant to write 15.075 and it's still not right though.
 
  • #4
hage567
Homework Helper
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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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