The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 7.0 rev/s in 9.0 s. at this point, the person doing the laundry opens the lid, and a safety switch turns off the washer. the tub slows to rest in 11.0 s. through how many revolutions does the tub turn during this 20 s interval

69.970 rev

Explanation:

Case 1: until the washer reaches its top spin

Initial angular speed ωi = 0 rev /s

Final angular speed ωf = 7 rev /s

Time t = 9 s

The angular acceleration is

  ωf - ωi = α t

  α = 7 - 0 / 9

The angular displacement

= 31.185 rev

Case II: the washer coming to rest from top spin

Initial angular speed ωi = 7 rev /s

Final angular speed ωf = 0 rev /s

Time t = 11 s

The angular acceleration is

ωf - ωi = α t

α = 0 - 7 / 11

= - 0.63 rev/s^2

The angular displacement

=38.885 rev

Total number of revolutions

θ1 + θ2 = 31.185 rev + 38.885 rev

  = 69.970 rev

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this is a good question! i'll do some research and come back when i have the answer!