Starting from rest, a DVD steadily acceleration to 580 rpm in 1.0 s, rotates at this angular speed for 3.0 s, then steadily decelerates to a halt in 2.0 s. How many revolutions does it make?

it makes N = 43.493≈ 43 revolutions

Explanation

there are 3 stages

1) steady acceleration from rest to 580 rpm in 1.0 seg

2) constant angular speed for 3.0 s

3) steady acceleration to halt in 2.0 s

knowing that

θ final = θ initial + ω initial * t + 1/2 α * t²

where θ angle respect with the horizontal axis, ω = angular velocity α = angular acceleration

since ω initial = 0, θ initial=0, α = (ω final - ω initial) / t = ω final / t

θ final = 1/2 ω final * t

since f = ω / (2π) and N = (θ final - θ initial) / (2π), where f=frequency and N = number of revolutions

N1 = 1/2 f final * t

N1 = 1/2 * 580 revolutions/minute * 1 s * 1 min/60 sec = 4. 833 rev

for 2) θ final - θ initial = ω initial + ω final * t

θ final = ω final * t

N2 = f final * t = 580 revolutions/minute * 3 s * 1 min/60 sec = 29 rev

for 3)

θ final = θ initial + ω initial * t + 1/2 α * t²

since ω final = 0, α = (ω final - ω initial) / t = - ω initial / t

θ final - θ initial = ω initial * t + 1/2 α * t² = ω initial * t - 1/2 ω initial * t = 1/2 ω initial * t

θ final - θ initial = 1/2 ω initial * t

therefore

N3 = 1/2 f initial * t = 1/2 * 580 revolutions/minute * 2 s * 1 min/60 sec = 9.66 rev

therefore

N total = N1 + N2 + N3 = 4. 833 rev + 29 rev + 9.66 rev = 43.493 revolutions