Answer
(a) $-2.22\times 10^{-9}rad/s^2$
(b) $2718.2years$
(c) $20ms$
Work Step by Step
(a) We can find the required angular acceleration as follows:
The initial angular speed is given as:
$ \omega=\frac{\Delta \theta}{\Delta t}$
$\omega=\frac{2\pi rad}{33.0\times 10^{-3}s}$
$\omega=190.399rad/s$
and $\Delta t_{100}=33.0\times 10^{-3}s+1.26\times 10^{-3}s=0.03426s$
The final angular speed is given as:
$\omega_{100}=\frac{\Delta \theta}{\Delta t_{100}}$
$\omega_{100}=\frac{2\pi rad}{0.03426s}$
$\omega_{100}=183.397rad/s$
We can convert 100 years into seconds as
$\Delta t=(100 years)(365days/yr)(24h/yr)(60min/h)(60s/min)=3.1536\times 10^9s$
Now $\alpha_{avg}=\frac{\Delta \omega}{\Delta t}$
We plug in the known values to obtain:
$\alpha=\frac{183.397rad/s-190.399rad/s}{3.1536\times 10^9s}$
$\alpha=-2.22\times 10^{-9}rad/s^2$
(b) We can find the required time taken as follows:
$\Delta t=\frac{\Delta \omega}{\alpha_{avg}}$
We plug in the known values to obtain:
$\Delta t=\frac{-190.3rad/s}{-2.22\times 10^{-9}rad/s}$
$\Delta t=8.572\times 10^{10}s$
Now we convert this time into years as
$\Delta t=(8.572\times 10^{10}s)(\frac{1min}{60s})(\frac{1h}{60min})(\frac{1day}{24h})(\frac{1year}{365days})$
$\Delta t=2718.2years$
(c) As given that $1054A.D=1054$
$\implies 2008-1054=954years$
Now the time period of the pulsar when it was created is
$(1.25\times 10^{-5}s/y)(954y)=0.01202s$
and $t=0.033s-0.01202s$
$t=0.02098s$
$t=20ms$
