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?