The driver of a car traveling at 25.0 m/s applies the brakes and undergoes a constant negative acceleration of magnitude 2.50 m/s2. 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 have radii of 0.27 m?

Answer 1

V^2 = Vo^2 + 2a*d = 0
d = -Vo^2/2a = -(25^2)/-5 = 125 m.

Circumference = pi * 2r = 3.14 * 0.54 =
1.70 m.

125m * 1rev/1.70m = 73.5 Revolutions.