Calculate the linear acceleration of a car, the 0.260-m radius tires of which have an angular acceleration of 14.0 rad/s2. Assume no slippage and give your answer in m/s2. 3.64 Correct: Your answer is correct. m/s2 (b) How many revolutions do the tires make in 2.50 s if they start from rest

a. 3.64 m/s b. 7 revolutions

Explanation:

a. Linear acceleration

The linear acceleration,a = rα where r = radius of tyre = 0.260 m and α = angular acceleration = 14.0 rad/s².

a = rα = 0.260 m × 14.0 rad/s² = 3.64 m/s

b. Number of revolutions in 2.50 s

Using θ = ωt + 1/2αt² where θ = angular distance,ω = initial angular speed = 0(since it starts from rest) and t = 2.50 s

θ = 0 × 2.50s + 1/2(14.0 rad/s²)(2.50)² = 0 + 43.75 rad = 43.75 rad. So the number of revolutions is 43.75rad/2π = 6.96 revolutions ≅ 7.0 revolutions (since 1 revolution = 2π)

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees. 

answer; equivalent to a one(1) horsepower motor;