The basketball's angular velocity ω at time t is
ω = 27 rad/s + (-0.15 rad/s²) t
It comes to a stop when ω = 0, which happens for
0 = 27 rad/s - (0.15 rad/s²) t
t = (27 rad/s) / (0.15 rad/s²)
t = 180 s
In this time, the ball would undergoes an angular displacement θ of
θ = (27 rad/s) t + 1/2 (-0.15 rad/s²) t ²
Plug in t = 180 s and solve for θ :
θ = (27 rad/s) (180 s) - 1/2 (0.15 rad/s²) (180 s)² = 2430 rad
One complete revolution corresponds to a turn of 2π rad, so the ball makes
(2430 rad) / (2π rad/rev) ≈ 386.747 rev
or about 387 revolutions as it slows down.
