Position-Time Graphs

Velocity = slope of the displacement-time graph: $v = \frac{dx}{dt} = \tan\theta$ $\frac{v_1}{v_2} = \frac{\tan 30^\circ}{\tan 45^\circ} = \frac{1/\sqrt{3}}{1} = \frac{1}{\sqrt{3}}$ So $v_1 : v_2 = 1 : \sqrt{3}$

Positive acceleration means $\frac{dv}{dt}>0$, i.e. velocity (slope of the x-t graph) keeps increasing. So the x-t graph is a curve that bends upward (concave up).

In a distance-time graph, the instantaneous speed at any instant is given by the slope of the curve at that instant: $v = \dfrac{ds}{dt}$.
So the instantaneous velocity is maximum where the curve is steepest.
Near A and B the curve is almost flat, so the slope (speed) is small.
Near D the curve flattens again, so the slope is small.
The curve rises most steeply around the point C, so the slope is maximum there.
Hence the maximum instantaneous velocity of the particle is around the point C.