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:

We need to determine the acceleration-displacement ($$a-x$$) graph from the given velocity-displacement ($$v-x$$) graph of the bicycle's motion.

The acceleration $$a$$ as a function of velocity $$v$$ and displacement $$x$$ is given by the relation: $$a = v \frac{dv}{dx}$$, where $$\frac{dv}{dx}$$ represents the slope of the $$v-x$$ graph.

Let's analyze the motion in two distinct regions based on the graph:

Region 1: From $$x = 0$$ to $$x = 200\text{ m}$$

The $$v-x$$ graph is a straight line starting from $$(0, 10)$$ and ending at $$(200, 50)$$. The equation of this straight line is of the form $$v = mx + c$$.

The slope of this line is: $$\frac{dv}{dx} = \frac{50 - 10}{200 - 0} = \frac{40}{200} = 0.2\text{ s}^{-1}$$.

The intercept on the vertical axis is $$c = 10$$, so the velocity equation is: $$v = 0.2x + 10$$.

Substituting $$v$$ and $$\frac{dv}{dx}$$ into the acceleration formula gives: $$a = (0.2x + 10) \times 0.2 = 0.04x + 2$$.

This is a straight-line equation for acceleration with a positive slope ($$0.04$$) and a vertical intercept of $$2\text{ m s}^{-2}$$.

• At $$x = 0\text{ m}$$, the acceleration is: $$a = 0.04(0) + 2 = 2\text{ m s}^{-2}$$.

• At $$x = 200\text{ m}$$, the acceleration is: $$a = 0.04(200) + 2 = 8 + 2 = 10\text{ m s}^{-2}$$.

Region 2: From $$x = 200\text{ m}$$ to $$x = 400\text{ m}$$

The velocity is constant at $$v = 50\text{ m s}^{-1}$$. Therefore, the slope of the graph is zero: $$\frac{dv}{dx} = 0$$.

Substituting this into the acceleration formula gives: $$a = 50 \times 0 = 0\text{ m s}^{-2}$$.

Combining both regions, the acceleration $$a$$ increases linearly from $$2\text{ m s}^{-2}$$ to $$10\text{ m s}^{-2}$$ in the first 200 meters, and then drops instantly and remains at $$0\text{ m s}^{-2}$$ from 200 meters to 400 meters.

Therefore, the correct description matches the graph where $$a$$ starts at 2, rises to 10 at $$x = 200\text{ m}$$, and becomes 0 thereafter.