An automobile traveling 70.0 km/h has tires of 65.0 cm diameter.?

(a) What is the angular speed of the tires about their axles?
rad/s

(b) If the car is brought to a stop uniformly in 25.0 turns of the tires (without skidding), what is the magnitude of the angular acceleration of the wheels?
rad/s^2

(c) How far does the car move during the braking?
m

♦ v=70km/h = 70e3/3600 =19.444 m/s;
w=v/R = 19.444/(0.65/2) =59.83 rad/s;
♦ brought to a stop means w=b*t, where t is time of deceleration, b is magnitude of deceleration, hence t=w/b;
angle done within time t is u=25 turns = 25*2pi =50pi rad;
and u= w*t –0.5*b*t^2; u=w*(w/b) –0.5*b*(w/b)^2;
b=w^2*(1-0.5)/u = 59.83^2*0.5/(50pi) =11.39 rad/s^2;
♦ distance =u*R =50pi*(0.65/2) =51.0 m;