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)

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prolly a beacuse the momentuem the car has increases while going down hill