We have the next information
ω_i= initial angular speed= 0.96 rev/s
ω_f= final angular speed= 0 rev/s
t=2.4 min= 144 s
We will use the next formula to calculate the angular acceleration
[tex]\alpha=\frac{\omega_f-\omega_i}{t}=\frac{0.96-0}{144}=-0.006667rev/s^2[/tex]Then we will use
[tex]\theta=\omega_it+\frac{1}{2}\alpha t^2[/tex][tex]\theta=0.96(144)-\frac{1}{2}(0.0066)(144)^2=69.81\text{rev}[/tex]Then we transform to degrees
1 rev =360°
[tex]69.81\times360=25132.03\text{\degree}[/tex]ANSWER
angular acceleration=-0.006667 rev/s^2
angle= 69.81 rev= 25132.03°