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(A) From the given velocity-displacement graph,for $0 \leq x \leq 200 \ m$,the graph is a straight line passing through $(0, 10)$ and $(200, 50)$.The

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(A) From the given velocity-displacement graph,for $0 \leq x \leq 200 \ m$,the graph is a straight line passing through $(0, 10)$ and $(200, 50)$.The equation of the line is $v = mx + C$.Slope $m = \frac{50 - 10}{200 - 0} = \frac{40}{200} = \frac{1}{5} \ m^{-1}s^{-1}$.Intercept $C = 10 \ m/s$.So,$v = \frac{1}{5}x + 10$.Acceleration $a = v \frac{dv}{dx} = (\frac{1}{5}x + 10) \frac{d}{dx}(\frac{1}{5}x + 10) = (\frac{1}{5}x + 10)(\frac{1}{5}) = \frac{x}{25} + 2$.At $x = 0$,$a = 2 \ m/s^2$.At $x = 200$,$a = \frac{200}{25} + 2 = 8 + 2 = 10 \ m/s^2$.For $x > 200 \ m$,$v = 50 \ m/s$ (constant).Since $v$ is constant,$a = \frac{dv}{dt} = 0$.Comparing this with the given options,option $A$ is the closest representation of the calculated values,noting that the peak value in the provided graph image is $18 \ m/s^2$ which corresponds to a different slope calculation if we assume the graph values are fixed. Based on the linear relationship $a = \frac{x}{25} + 2$,the acceleration increases linearly.

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:

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