(a) 88.0 rad/s
The angular speed of the wheel is
![\omega = 14 rev/s](https://img.qammunity.org/2020/formulas/physics/high-school/6ba8rnrgkdoojev94bbp89tjlbn74gdkj7.png)
Keeping in mind that
1 revolution =
radians
The angular speed in radians/second can be found by solving the proportion:
![14 rev/s : x = 1 rev : 2 \pi rad](https://img.qammunity.org/2020/formulas/physics/high-school/ocn1udsnspjj2xc4rw2p2m0y9vi6s3to7y.png)
From which we find
![x=(14\cdot 2 \pi)/(1)=88.0 rad/s](https://img.qammunity.org/2020/formulas/physics/high-school/ozwvv858dtwymt5da65l6lyg453j8s2ndb.png)
(b) 440 radians
Assuming the wheel is rotating at constant angular speed, the angular displacement of the wheel at time t is given by
![\theta= \omega t](https://img.qammunity.org/2020/formulas/physics/high-school/aiyw2hla34tst5dz0se83hlu7gemf1mvm8.png)
where
is the angular speed
t is the time
Substituting
t = 5 s
we find the angle through which the wheel has rotated after 5 seconds:
![\theta=(88.0)(5)=440 rad](https://img.qammunity.org/2020/formulas/physics/high-school/4hx47vake3t63u5cuxmkaaptoybnq7v72q.png)
(c) 94.5 rad/s
The angular speed after a time t is given by
![\omega(t) = \omega_o + \alpha t](https://img.qammunity.org/2020/formulas/physics/high-school/cmyejxmlqzyso36bp3p9rq0w17w9p51rrs.png)
where
is the angular speed at t=10 s, when the acceleration starts
is the angular acceleration
The duration of the acceleration is
t = 15 s - 10 s = 5 s
So substituting this value into the equation, we find the new angular speed:
![\omega(15) = 88.0+(1.3)(5)=94.5 rad/s](https://img.qammunity.org/2020/formulas/physics/high-school/j41ety1iuwpy776nwu7drneetj7wnp3wjg.png)