Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 4.0 s, it has rotated 30 rad.

Assuming that the acceleration does not change, through what additional angle will the disk turn during the next 7.0 s?

I calculated the acceleration to be 3.75 rad/s^2

Displacement=1/2 alpha*t^2

alpha=2*30/16

2)
d=wi*t+1/2 alpha(7^2)
where wi=alpha*4
d=alpha*4+1/2 alpha*49
and solve for additional displacemetn d.

so, (3.75)(4) + (.5)(3.75)(49) or 106.875?

or just the second part to find the additional angle?

(.5)(3.75)(49) = 91.875 but it is incorrect, what am I missing?

To find the additional angle through which the disk will rotate during the next 7.0 s, you can use the equation of motion for rotational motion:

θ = ω_i * t + (1/2) * α * t^2

where θ is the total angle rotated, ω_i is the initial angular velocity, α is the angular acceleration, and t is the time.

Given that the disk starts from rest (ω_i = 0) and rotates 30 rad in 4.0 s, we can find the angular acceleration using the equation:

θ = (1/2) * α * t^2

Rearranging the equation, we have:

α = (2 * θ) / t^2

Substituting the given values, we get:

α = (2 * 30 rad) / (4.0 s)^2
α = 3.75 rad/s^2

Now, let's find the additional angle rotated during the next 7.0 s. Plug the values into the equation of motion:

θ = ω_i * t + (1/2) * α * t^2

Since the disk starts from rest, ω_i = 0:

θ = (1/2) * α * t^2
θ = (1/2) * 3.75 rad/s^2 * (7.0 s)^2

Calculating the result:

θ = 0.5 * 3.75 rad/s^2 * 49 s^2
θ ≈ 91.88 rad

Therefore, the disk will rotate approximately 91.88 rad during the next 7.0 s.