The velocity-time graph of a body is shown in figure. 

From $t=0$ to $t=A$ velocity increases from 0 to v where v is the velocity at $t=A$
From $t=A$ to $t=B$ velocity increases from v to 0 where v is the velocity at $t=A$
Hence, $v_{avg}$ during interval  0 to A is same as $v_{avg}$ during interval  A to B

$\frac{v_c-0}{A-0}=\tan 60^{\circ }=\sqrt{3}$;
$\therefore v_c=\sqrt{3}\times OA$
$\frac{v_c-0}{A-0}=\tan 150^{\circ }=-\tan 30^{\circ }=-\frac{1}{\sqrt{3}}$
$\therefore v_c=\frac{1}{\sqrt{3}}\times AB$
Equating both $v_c$, $\sqrt{3}\times OA=\frac{1}{\sqrt{3}}\times AB$
$\Rightarrow \frac{AB}{OA}=3$
$\Rightarrow \frac{OA}{OA+AB}=\frac{1}{1+3}$
$\Rightarrow \frac{OA}{OB}=\frac{1}{4}$

Distance covered on in time OA: $d_{OA}=\frac{1}{2}\times OA\times v_A$

Distance covered on in time AB: $d_{AB}=\frac{1}{2}\times AB\times v_A$

$|a_{OA}|=\tan 60^{\circ }=\sqrt{3}$
$|a_{AB}|=\tan 30^{\circ }\frac{1}{\sqrt{3}}$
$\frac{|a_{OA}|}{|a_{AB}|}=3$

$\therefore \frac{d_{OA}}{d_{AB}}=\frac{\frac{1}{2}\times OA\times v_A}{\frac{1}{2}\times AB\times v_A}$
$\therefore \frac{d_{OA}}{d_{AB}}=\frac{\frac{AC}{\cot 60^{\circ }}}{\frac{AC}{\cot 30^{\circ }}}=\tan 60^{\circ }\times \cot 30^{\circ }=3$