Answer
A complete revolution is an angular displacement of $\Delta \theta=2 \pi$ rad, so the angular velocity in rad/s is given by $\omega=\Delta \theta / T=2 \pi / T$ . The angular acceleration is given by
$$
\alpha=\frac{d \omega}{d t}=-\frac{2 \pi}{T^{2}} \frac{d T}{d t}
$$
For the pulsar described in the problem, we have
$$
\frac{d T}{d t}=\frac{1.26 \times 10^{-5} \mathrm{s} / \mathrm{y}}{3.16 \times 10^{7} \mathrm{s} / \mathrm{y}}=4.00 \times 10^{-13} \text { . }
$$
$$
\alpha=-\left[\frac{2 \pi}{(0.033 \mathrm{s})^{2}}\left(4.00 \times 10^{-13}\right)=-2.3 \times 10^{-9} \mathrm{rad} / \mathrm{s}^{2} \text { . }\right.
$$
the negative sign shows that angular acceleration is reverse (angular-velocity) and pulsar is slowing down
Work Step by Step
A complete revolution is an angular displacement of $\Delta \theta=2 \pi$ rad, so the angular velocity in rad/s is given by $\omega=\Delta \theta / T=2 \pi / T$ . The angular acceleration is given by
$$
\alpha=\frac{d \omega}{d t}=-\frac{2 \pi}{T^{2}} \frac{d T}{d t}
$$
For the pulsar described in the problem, we have
$$
\frac{d T}{d t}=\frac{1.26 \times 10^{-5} \mathrm{s} / \mathrm{y}}{3.16 \times 10^{7} \mathrm{s} / \mathrm{y}}=4.00 \times 10^{-13} \text { . }
$$
$$
\alpha=-\left[\frac{2 \pi}{(0.033 \mathrm{s})^{2}}\left(4.00 \times 10^{-13}\right)=-2.3 \times 10^{-9} \mathrm{rad} / \mathrm{s}^{2} \text { . }\right.
$$
the negative sign shows that angular acceleration is reverse (angular-velocity) and pulsar is slowing down
