a) Calculate the linear acceleration of a car, the 0.310-m radius tires of which have an angular acceleration of 11.5 rad/s2. Assume no slippage and give your answer in m/s2
b) How many revolutions do the tires make in 2.50 s if they start from rest?
c) What is their final angular velocity in rad/s?
d) What is the final velocity of the car in m/s?
a) Linear acceleration can be calculated using the formula:
a = r * α
where a is the linear acceleration, r is the radius of the tires, and α is the angular acceleration.
Substitute the given values:
a = 0.310 m * 11.5 rad/s^2
a = 3.565 m/s^2
Therefore, the linear acceleration of the car is 3.565 m/s^2.
b) The number of revolutions can be calculated using the formula:
θ = ωi*t + 0.5*α*t^2
where θ is the angular displacement, ωi is the initial angular velocity (0 as the tires start from rest), α is the angular acceleration, and t is the time.
Substitute the given values:
θ = 0.5 * 11.5 rad/s^2 * (2.50 s)^2
θ = 28.75 rad
Since one revolution is equivalent to 2π rad, the number of revolutions is:
θ = 28.75 rad / (2π rad/revolution) ≈ 4.57 revolutions
Therefore, the tires make approximately 4.57 revolutions in 2.50 s.
c) The final angular velocity can be calculated using the formula:
ωf = ωi + α*t
Substitute the given values:
ωf = 0 rad/s + 11.5 rad/s^2 * 2.50 s
ωf = 28.75 rad/s
Therefore, the final angular velocity of the tires is 28.75 rad/s.
d) The final velocity of the car can be calculated using the formula:
v = r * ωf
Substitute the given values:
v = 0.310 m * 28.75 rad/s
v = 8.8625 m/s
Therefore, the final velocity of the car is 8.8625 m/s.