To solve this problem, you can use the equations of motion for rotational motion under constant acceleration:
ωf = ωi + αt --(1)
θ = ωit + 0.5αt^2 --(2)
where ωi is the initial angular velocity, ωf is the final angular velocity, α is the angular acceleration, t is the time elapsed, and θ is the angular displacement.
Using equation (1), we can find the angular acceleration of the wheel:
α = (ωf - ωi)/t
= (30.0 rad/s - 10.0 rad/s)/20.0 s
= 1.0 rad/s^2
Using equation (2), we can find the total angular displacement of the wheel during the 20.0 seconds:
θ = ωit + 0.5αt^2
= 10.0 rad/s × 20.0 s + 0.5 × 1.0 rad/s^2 × (20.0 s)^2
= 400.0 rad
To find the time at which the wheel has undergone half of this angular displacement, we can use equation (2) again:
θ/2 = ωit + 0.5αt^2
Rearranging and solving for t, we get:
t = [(-ωi) ± sqrt(ωi^2 + 2αθ)]/α
Since we are looking for a positive time, we take the positive root:
t = [(-10.0 rad/s) ± sqrt((10.0 rad/s)^2 + 2 × 1.0 rad/s^2 × 400.0 rad)]/1.0 rad/s^2
≈ 12.4 s
Therefore, the answer is B, 12.4 s.