The tires of a car make 85 revolutions as the car reduces its speed uniformly from 110 km/hr to 40 km/hr. The tires have a diameter of 0.80 m. (a) What was the angular acceleration? rad/sec^2 (b) If the car continues to decelerate at this rate, how much more time is required for it to stop?

To solve this problem, we need to use the equations of motion.

(a) First, we need to convert the speed from km/hr to m/s.

110 km/hr = 30.56 m/s 40 km/hr = 11.11 m/s

The total distance covered during deceleration can be calculated using the formula for the circumference of a circle (C = πd) multiplied by the number of revolutions.

C = π * 0.80 m = 2.51 m Total distance = 2.51 m/rev * 85 rev = 213.4 m

We can use the equation of motion v^2 = u^2 + 2as to find the acceleration (a). Here, v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance.

(11.11 m/s)^2 = (30.56 m/s)^2 + 2a213.4 m a = - [(30.56 m/s)^2 - (11.11 m/s)^2] / (2*213.4 m) a = -7.14 m/s^2

The angular acceleration (α) can be found by dividing the linear acceleration (a) by the radius of the tires (r = d/2 = 0.40 m).

α = a / r α = -7.14 m/s^2 / 0.40 m α = -17.85 rad/s^2

(b) To find the time required for the car to stop, we can use the equation of motion v = u + at. Here, v is the final velocity (which is 0 m/s when the car stops), u is the initial velocity, a is the acceleration, and t is the time.

0 m/s = 11.11 m/s + (-7.14 m/s^2)*t t = - 11.11 m/s / -7.14 m/s^2 t = 1.56 s

So, the car would require an additional 1.56 seconds to stop if it continues to decelerate at this rate.