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

Q: Why can't I just take the slope of the $v-x$ graph to find acceleration?

The slope of a $v-x$ graph gives you $\frac{dv}{dx}$, not acceleration. Acceleration is the rate of change of velocity with respect to time ($\frac{dv}{dt}$). To connect them, we use the chain rule: $a = \frac{dv}{dt} = \frac{dx}{dt} \cdot \frac{dv}{dx} = v \frac{dv}{dx}$.

Q: If the velocity is increasing linearly, shouldn't the acceleration be constant?

That's only true for a velocity-time ($v-t$) graph! On a $v-x$ graph, if $v = mx + c$, then $a = v \frac{dv}{dx} = (mx + c)m$. Since $v$ is increasing and the slope $m$ is constant, their product (acceleration) must also be increasing.

Q: Why does the acceleration suddenly drop to zero at $x = 200$?

At $x = 200$, the graph becomes horizontal, meaning the velocity is constant at $50 \text{ m/s}$. When velocity is constant, there is no change in velocity, so the acceleration is exactly zero.

Q: How did we get the equation $v = \frac{x}{5} + 10$?

We used the standard equation of a line, $y = mx + c$. The y-intercept $c$ is $10$. The slope $m$ is the 'rise over run', which is $\frac{50 - 10}{200 - 0} = \frac{40}{200} = \frac{1}{5}$.

Q: What would the $a-x$ graph look like if the $v-x$ graph was a parabola?

If the $v-x$ graph was a parabola (e.g., $v \propto x^2$), then its derivative $\frac{dv}{dx}$ would be proportional to $x$. The acceleration $a = v \frac{dv}{dx}$ would then be proportional to $x^2 \cdot x = x^3$, resulting in a cubic curve!

Question

The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by

Sigma Percentile · Accepted Answer

### Decoding the Velocity Graph

Imagine you are tracking the motion of a bicycle. You are given a graph that plots its velocity $v$ against its displacement $x$. Your mission is to translate this information into an acceleration-displacement ($a-x$) graph.

Looking closely at the given $v-x$ graph, we can clearly see that the motion is divided into two distinct phases. The first phase is a steady climb from $x = 0$ to $x = 200	ext{ m}$, where the velocity increases linearly. The second phase, from $x = 200	ext{ m}$ to $x = 400	ext{ m}$, shows the bicycle cruising at a constant velocity. To find the acceleration, we must analyze these two phases separately.

### The Master Equation of Kinematics

When dealing with velocity as a function of time, acceleration is simply the slope of the graph, $a = \frac{dv}{dt}$. However, our graph plots velocity against **displacement**. We cannot just take the slope!

Instead, we must use the chain rule to link acceleration, velocity, and displacement. This gives us our master equation:

$$a = v \frac{dv}{dx}$$

This equation tells us that the acceleration at any point is the product of the velocity itself and the slope of the $v-x$ graph at that point.

### Analyzing the First Phase ($0 \le x \le 200$)

In this region, the $v-x$ graph is a perfect straight line. Let's find its mathematical equation using the standard slope-intercept form, $v = mx + c$.

The line starts at a velocity of $10	ext{ m/s}$, so our y-intercept $c$ is $10$.

To find the slope $m$, we calculate the 'rise over run':

$$m = \frac{50 - 10}{200 - 0} = \frac{40}{200} = \frac{1}{5}$$

So, the equation for velocity is:

$$v = \frac{x}{5} + 10$$

Now, let's plug this into our master equation. We need the derivative of velocity with respect to displacement, $\frac{dv}{dx}$, which is simply the slope we just calculated: $\frac{1}{5}$.

Substituting these into $a = v \frac{dv}{dx}$:

$$a = \left(\frac{x}{5} + 10ight) 	imes \frac{1}{5}$$

$$a = \frac{x}{25} + 2$$

**Look at this beautiful result!** The acceleration $a$ is a linear function of $x$. This means the $a-x$ graph for this phase will also be a straight line.

Let's check the boundary values to see where this line starts and ends:
- At $x = 0$, $a = \frac{0}{25} + 2 = 2	ext{ m/s}^2$.
- At $x = 200$, $a = \frac{200}{25} + 2 = 8 + 2 = 10	ext{ m/s}^2$.

So, the acceleration increases linearly from $2$ to $10$.

### Analyzing the Second Phase ($x > 200$)

Beyond $200	ext{ m}$, the bicycle stops speeding up and maintains a constant velocity of $50	ext{ m/s}$.

Because the velocity is constant, its rate of change with respect to displacement is zero:

$$\frac{dv}{dx} = 0$$

Plugging this into our master equation:

$$a = 50 	imes 0 = 0$$

The acceleration instantly drops to zero and stays there for the remainder of the journey.

### The Final Picture

Combining our findings, the true $a-x$ graph must show a straight line rising from $2$ to $10$ between $x=0$ and $x=200$, followed by an abrupt drop to $0$ for the rest of the motion.

Comparing this with our options, **Option (a)** perfectly captures this dynamic reality!

Animated Solution for Physics - Kinematics: The velocity-displacement graph describing the motion of a bicycle is shown in the following figure. The acceleration-displacement graph of the bicycle's motion is best described by

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Visualized Solution

\text{Analyzing the } v-x \text{ Graph}

\text{The Master Equation}

\text{Part 1: } 0 \le x \le 200

\text{Finding the Equation}

\text{Calculating Acceleration}

\text{Executing the Derivative}

\text{Simplifying Acceleration}

\text{Boundary Values}

\text{Part 2: } x > 200

\text{Acceleration for Part 2}

\text{Conclusion}

The Sigma Insight: Motion Graphs

Decoding the Velocity Graph

The Master Equation of Kinematics

Analyzing the First Phase ( $0 \leq x \leq 200$ )

Analyzing the Second Phase ( $x > 200$ )

The Final Picture

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