The velocity-displacement graph describing the motion of a bicycle is shown in the figure.

The acceleration-displacement graph of the bicycle’s motion is best described by

Given velocity – displacement graph equation of line AB
$\begin{array}{l}\mathrm{v}=\frac{(50-10)}{(200-0)}\times (\mathrm{x}-0)+10(0\le \mathrm{x}\le 200)\\ \Rightarrow \mathrm{v}=\frac{40}{200}\times \mathrm{x}+10\Rightarrow \mathrm{v}=\frac{\mathrm{x}}{5}+10\dots .\left(\mathrm{i}\right)\end{array}$
Differentiate equation (i) w.r.t.x $\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{5}\dots \left(\mathrm{ii}\right)$
Acceleration, $\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{dv}}{\mathrm{dx}}\times \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{v}\frac{\mathrm{dv}}{\mathrm{dx}}$
$\mathrm{a}=\left(\frac{\mathrm{x}}{5}+10\right)\times \frac{1}{5}$ (Using (i) and (ii)
$\mathrm{a}=\left(\frac{\mathrm{x}}{25}+2\right){\mathrm{ms}}^{-2},\text{ At }\mathrm{x}=0,\mathrm{a}=2{\mathrm{ms}}^{-2}$
It is straight line till x = 200.
For x>200, V = constant, a = 0.
Therefore, most appropriate option will be