Animated Solution for Physics - Kinematics: The velocity- displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by

Decoding the Given Graph

Every great physics problem starts with a careful observation of the given data. In this problem, we are presented with a velocity-displacement graph, commonly referred to as a $v-x$ graph.

Look closely at the shape of the curve. It is a perfect straight line. It begins at a positive value on the vertical axis, which we can call $v_0$, and it terminates at a positive value on the horizontal axis, which we can call $x_0$.

What does this tell us physically? It tells us that as the particle moves forward (as $x$ increases), its velocity $v$ is steadily decreasing. Because the graph is a straight line, this decrease is linear with respect to displacement. However, we must be very careful here: a linear decrease with respect to displacement does not mean a linear decrease with respect to time. This is the core trap of the problem!

Connecting Kinematic Variables

Our ultimate goal is to find the acceleration-displacement graph, or the $a-x$ graph. To do this, we need a mathematical bridge that connects acceleration $a$, velocity $v$, and displacement $x$.

Most students immediately think of the definition of acceleration:

While this is fundamentally correct, it relates acceleration to time $t$, not displacement $x$. We need to eliminate time from our equation. This is where the elegant chain rule of calculus comes to our rescue. We can rewrite the derivative as:

Now, recognize that the rate of change of displacement with respect to time, $\frac{dx}{dt}$, is the very definition of velocity $v$. Substituting this back in, we arrive at our master kinematic equation for this problem:

This equation is incredibly powerful. It tells us that the acceleration at any point is the product of the velocity at that point and the spatial gradient of the velocity (the slope of the $v-x$ graph).

To use our master equation, we need explicit mathematical expressions for $v$ and $\frac{dv}{dx}$. Let's extract these from the given $v-x$ graph.

We know the graph is a straight line. The equation of any straight line can be written in the slope-intercept form:

In our context, the y-axis represents velocity $v$, and the x-axis represents displacement $x$. The y-intercept $c$ is simply the initial velocity $v_0$.

Next, we need the slope $m$. The slope is the "rise over run". The line goes from $(0,v_0)$ to $(x_0,0)$. Therefore, the slope is:

Substituting the slope and intercept back into the equation of the line, we get the algebraic representation of our velocity:

Finding the Derivative

Now that we have $v$ as a function of $x$, we need to find the term $\frac{dv}{dx}$. This requires us to differentiate our velocity equation with respect to $x$.

Since $v_0$ and $x_0$ are constants, the derivative of the constant term $v_0$ is zero. The derivative of the linear term $-\frac{v_0}{x_0}x$ is simply its coefficient.

Notice that the derivative is a constant negative value. This makes perfect sense geometrically, as the slope of a straight line is constant everywhere.

We are now ready to bring everything together. We have our master equation $a=v\frac{dv}{dx}$, and we have the explicit expressions for both $v$ and $\frac{dv}{dx}$. Let's substitute them in:

Now, we must carefully expand this algebraic expression. We will distribute the constant term $-\frac{v_0}{x_0}$ into the parentheses.

First, multiplying it by the $x$ term: Notice how the two negative signs cancel out, resulting in a positive term.

Next, multiplying it by the constant term:

Combining these results, we obtain the final equation for acceleration as a function of displacement:

Matching the Graph

Our final task is to translate this algebraic equation back into a geometric graph. Let's analyze the structure of our acceleration equation:

This equation is clearly in the form of a straight line, $a=mx+c$.

Let's look at the slope $m$. The slope is ${\left(\frac{v_0}{x_0}\right)}^2$. Because this term is a perfect square, it must be strictly positive (assuming $v_0$ and $x_0$ are non-zero). Therefore, the $a-x$ graph must be a straight line that tilts upwards.

Now let's look at the y-intercept $c$. The intercept is $-\frac{v_0^2}{x_0}$. Since $v_0^2$ and $x_0$ are positive, the negative sign in front ensures that the entire intercept is strictly negative. Therefore, the $a-x$ graph must cross the vertical acceleration axis below the origin.

We are looking for a graph that is a straight line, starts from a negative value on the y-axis, and has a positive slope. Looking at our options, only option (c) perfectly matches this exact geometric profile.

Why Does Acceleration Change?

Before we conclude, let's take a moment to appreciate the profound physical story this math is telling us. Why does the acceleration start highly negative and then increase towards zero?

At $x=0$, the particle is moving at its maximum speed, $v_0$. The velocity is decreasing at a constant rate with respect to distance ($-v_0/x_0$). Because the particle is moving very fast, it covers distance very quickly. Therefore, it is losing velocity very rapidly with respect to time. This rapid loss of velocity means a very high deceleration, which corresponds to the large negative intercept on our $a-x$ graph.

As the particle moves forward, it slows down. It still loses the same amount of velocity for every meter it travels, but because it is moving slower, it takes longer to travel that meter. Consequently, the rate of velocity loss with respect to time (the deceleration) becomes less severe.

Finally, at $x=x_0$, the particle comes to a complete stop ($v=0$). At this exact instant, it is no longer covering any distance, so its velocity is no longer changing with respect to time. The acceleration becomes zero. This is beautifully reflected in our $a-x$ graph, which crosses the x-axis exactly at $x_0$. Physics and mathematics are in perfect harmony!