using equation for a straight line $(y=m x+c)$ where, $m=\tan 60^{\circ}=\sqrt{3}(\mathrm{slop}) C=0$ from graph
relation b/w $a_{T}$ and $t$ is obtained as
$a_{T}=\tan 60^{\circ} t+0$
this tangential acceleration increases velocity $(v)$ of the particless as.
$a_{T}=\frac{d v}{d t} \Rightarrow \frac{d v}{d t}=\sqrt{3} t$
$\Rightarrow \int d v=\int \sqrt{3}+d t \Rightarrow v=\frac{\sqrt{3} t^{2}}{2}$
so, $a_{T}$ a particular time, centripetal accin $\left(a_{c}\right)$ is given by $a_{c}=\frac{v^{2}}{r}=\frac{3}{4} t^{4}(r=1) \vec{a}_{T}$
let a time $\vec{a}$ masses angle $30^{\circ}$ with $\vec{a}_{c} .$ so it makes angles $60^{\circ}$ with $\overrightarrow{a_{T}}(\overrightarrow{a_{c}} K \overrightarrow{a_{T}} are\,pendicular)$
$\Rightarrow \overrightarrow{a_{T}}=\sqrt{3 t^{2}+\frac{9}{16}+8 \frac{1}{2}}$ that given
$\Rightarrow 3 t^{2}=\frac{1}{11}\left(3 t^{2}+\frac{9}{16}+8\right)$
$\Rightarrow t=0$ or $t=2^{2 / 3}$