Answer
a) $\alpha=-1.38rad/s^2$
b) $\theta=+33.5rad$
Work Step by Step
a) The motorcycle's initial linear speed $v_0=6.6m/s$ decelerates to $v=2.1m/s$ in $t=5s$. Its linear deceleration is $$a=\frac{v-v_0}{t}=-0.9m/s^2$$
In rolling motion, we have $a=a_T=r\alpha$
The wheel has a radius of $r=0.65m$. Since we know $a$, we can calculate the wheel's angular deceleration: $$\alpha=\frac{a}{r}=-1.38rad/s^2$$
b) To find the wheel's displacement $\theta$, we still need to know $\omega_0$, which can be calculated by $$\omega_0=\frac{v_0}{r}=+10.15rad/s$$
The wheel's angular displacement in $t=5s$ is $$\theta=\omega_0t+\frac{1}{2}\alpha t^2=+33.5rad$$
