Answer:
1)  The magnitude of the angular acceleration = 67.92 rad/
2) Magnitude of the linear acceleration = 2.744 m/
3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s
4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m 
5) the final velocity is 6.21 m/s
Explanation:
the given information :
Bowling mass m = 3.6 kg
Radius = 0.101 m
Initial speed  = 8.7 m/s
 = 8.7 m/s
Coefficient of kinetic friction μ = 0.28
1) he magnitude of the angular acceleration of the bowling ball is
F = m a 
 = μ N  ,   N = m g
  = μ N  ,   N = m g
 = μ m g
  = μ m g
1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:
momen inersia of Bowling ball I = (2/5) m 
torque τ = I α
τ = F R
I α = F R
 (2/5) m  α = μ m g R
  α = μ m g R
α = (5 μ g / 2R) μ g R
   = (5 x 0.28 x 9.8/ 2 x 0.101)
   = 67.92 rad/
2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane
F = -  ,
 ,  is the force of kinetic friction
 is the force of kinetic friction
m a = - μ m g, remove m
the magnitude of linear accelaration is
a = μ g
   = (0.28) (9.8)
   = 2.744 m/
3) The bowling ball takes time to begin rolling without slipping:
The linear speed,  =
 =  - a t
 - a t
                              =
  =   - μ g t
 - μ g t
the angular speed, ω = ω0 + α t
                                 ω = ω0 + (5  μ g/2R ) t
 = ω R
 = ω R
 - μ g t = ω0 R + (5  μ g/2R ) t R
 - μ g t = ω0 R + (5  μ g/2R ) t R
 7 μ g t/2 =  + ω0 R
 + ω0 R
hence,
t = (2  + ω0 R)/  7 μ g
 + ω0 R)/  7 μ g
ω0 = 0 (no initial spin), therefore
t = 2  / 7 μ g
 / 7 μ g
  = 2 x 8.7 / 7 (0.28) (9.8)
  = 0.906 s
4) How long it takes for the bowling ball to begin rolling without slipping, S
S =  t - (1/2) a
  t - (1/2) a 
   = (8.7) (0.906) - (1/2) (2.744) 
   = 6.75 m
5) The final velocity
 =
 =  - a t
 - a t
 = 8.7 - (2.744) (0.906)
 = 8.7 - (2.744) (0.906)
 = 6.21 m/s
 = 6.21 m/s