Starting from rest at t = 0 s, a wheel undergoes a constant angular acceleration. When t = 1.6 s, the angular velocity of the wheel is 5.9 rad/s. The acceleration continues until t = 19 s, when the acceleration abruptly changes to 0 rad/s2. Through what angle does the wheel rotate in the interval t = 0 s to t = 47 s?

v = Vi + a t

theta = theta zero + Vi t + (1/2) a t^2

when t = 0, v = 0
v = a t
5.9 = a(1.6)
so a = 3.69 radians/s^2

now at 19 seconds:
v = 0 + 3.69 (19) = 70.1 radians/s
theta = 0 + 0 t + (.5)(3.69)(19)^2
= 666 radians

NOW continues at 70.1 rad/s for
47 - 19 = 28 seconds
so goes 28*70.1 = 1963 radians more

1963 + 666 = 2629 radians

To calculate the angle through which the wheel rotates in the interval t = 0 s to t = 47 s, we need to break the problem into two parts:

1. From t = 0 s to t = 19 s (constant angular acceleration)
2. From t = 19 s to t = 47 s (constant angular velocity)

Let's calculate the angle for each part separately.

1. From t = 0 s to t = 19 s:
We know that angular acceleration (α) is constant during this time interval.

The formula to calculate angular velocity (ω) at any given time (t) is:
ω = ω0 + αt, where ω0 is the initial angular velocity and t is the time.

Given:
Initial angular velocity, ω0 = 0 rad/s (starting from rest)
Angular acceleration, α = ?
Time, t = 19 s
Final angular velocity, ω = ?

We are given the final angular velocity at t = 1.6 s: ω = 5.9 rad/s.
Using this information, we can calculate α:

ω = ω0 + αt
5.9 = 0 + α(1.6)
5.9 = 1.6α
α = 5.9 / 1.6
α ≈ 3.69 rad/s^2

Now we can calculate the angle (θ1) for this time interval using the formula:
θ1 = ω0t + 0.5αt^2

θ1 = 0.5αt^2
θ1 = 0.5(3.69)(19)^2
θ1 ≈ 662.92 rad

2. From t = 19 s to t = 47 s:
We know that angular acceleration (α) is 0 during this time interval, so the angular velocity remains constant.

We can use the formula: θ2 = ωt

Given:
Initial angular velocity, ω = 5.9 rad/s
Time, t = 47 s
θ2 = ωt
θ2 = 5.9 * 47
θ2 ≈ 276.7 rad

The total angle (θ) through which the wheel rotates in the interval t = 0 s to t = 47 s is the sum of θ1 and θ2:

θ = θ1 + θ2
θ ≈ 662.92 + 276.7
θ ≈ 939.62 rad

Therefore, the wheel rotates approximately 939.62 radians in the interval t = 0 s to t = 47 s.

To find the angle through which the wheel rotates, we need to calculate the total change in angular displacement over the given time interval.

First, let's find the angular displacement between t = 0 s and t = 1.6 s. We can use the formula:

θ = ω_initial * t + (1/2) * α * t^2

where:
θ is the angular displacement,
ω_initial is the initial angular velocity at t = 0 s,
α is the angular acceleration, and
t is the time interval.

Given: ω_initial = 0 rad/s (starting from rest), α is constant but unknown, and t = 1.6 s.

1. Calculate the angular displacement from t = 0 s to t = 1.6 s:
θ_1 = 0 * 1.6 + (1/2) * α * (1.6)^2
θ_1 = 0 + 0.8α

Next, let's find the angular displacement between t = 1.6 s and t = 19 s. The wheel has a constant angular velocity during this time interval, so the angular acceleration is 0 rad/s^2. Therefore, the formula becomes:

θ = ω_initial * t

Given: ω_initial = 5.9 rad/s and t = 19 s.

2. Calculate the angular displacement from t = 1.6 s to t = 19 s:
θ_2 = 5.9 * (19 - 1.6)
θ_2 = 5.9 * 17.4

Finally, let's find the angular displacement between t = 19 s and t = 47 s. The angular acceleration is 0 rad/s^2, so the angular velocity remains constant.

3. Calculate the angular displacement from t = 19 s to t = 47 s:
θ_3 = 5.9 * (47 - 19)

To find the total angular displacement, add up all the individual angular displacements:

Total angular displacement = θ_1 + θ_2 + θ_3

Calculate the value based on the given angular acceleration and substitute it into the formula.