1. Apr 4, 2006

### chazgurl4life

The tires of a car make 85 revolutions as the car reduces its speed uniformly from 90 km/h to 30 km/h. The tires have a diameter of 0.80 m.
(a) What was the angular acceleration of the tires?
angular acceleration = Omega final- Omega intial / Delta Time

does that mean that in this case angular acceleration= 30 km/hr-90km/hr /Delta T ...but we dont have Delta T so is there another equation i could use?

(b) If the car continues to decelerate at this rate, how much more time is required for it to stop?
s

2. Apr 4, 2006

### Gale

[tex] \alpha = a_T /r [/ tex] you can find the linear acceleration (you know the initial and final velocities and you can find the distance travelled) and you have the radius. 30hm/hr and 90km/hr are NOT angular speeds.

just use angular kinetmatics for the second part.

Last edited: Apr 4, 2006
3. Apr 4, 2006

for further help:

there are three essential kinematics equations. for angular kinematics:

1) theta = theta_initial + omega_initial*t + 1/2 alpha*t
2) omega = omega_initial + alpha*t
3) omega^2 = omega_initial^2 + 2*alpha*theta.

notice that 1 and 2 involve time, will 3 does not. (i have used the typical greek symbols for angle, angular velocity, and angular accelaration.)

you are given enough information to find alpha using just one of these three equations...

(remember to convert linear quantities to angular quantities!)