Animated Solution for Physics - Kinematics: The given graph shows the variation of velocity with displacement. Which one of the graph given below correctly represents the variation of acceleration with displacement? (2005)

The problem presents us with a velocity-displacement ($v-x$) graph, which is a straight line with a negative slope. Our goal is to determine the corresponding acceleration-displacement ($a-x$) graph. To bridge the gap between these two graphical representations, we must rely on the fundamental kinematic equations.

We know that acceleration is the rate of change of velocity with respect to time, $a=\frac{dv}{dt}$. However, our given graph does not involve time; it relates velocity directly to displacement. This is where the chain rule comes to our rescue. We can rewrite the acceleration as:

Since $\frac{dx}{dt}$ is simply the velocity $v$, we arrive at the crucial relation:

This equation is the golden key to solving the problem. It tells us that to find the acceleration at any point $x$, we need to multiply the velocity at that point by the slope of the $v-x$ graph at that same point.

Let's translate the visual information of the $v-x$ graph into a mathematical equation. The graph is a straight line, so it must follow the form $y=mx+c$.

Here, the $y$-axis represents velocity $v$, and the $x$-axis represents displacement $x$.

The line intersects the $v$-axis at $v_0$, so the $y$-intercept is $c=v_0$.

The line intersects the $x$-axis at $x_0$, so the slope $m$ is the "rise over run", which is $\frac{0-v_0}{x_0-0}=-\frac{v_0}{x_0}$.

Therefore, the equation of the velocity line is:

Now, we need to find $\frac{dv}{dx}$, which is simply the derivative of our velocity equation with respect to $x$. Since it's a linear equation, the derivative is just the constant slope:

Substituting both $v$ and $\frac{dv}{dx}$ into our master equation $a=v\frac{dv}{dx}$, we get:

Expanding this expression by multiplying the terms, we obtain:

Look closely at this final equation. It is also a linear equation of the form $a=Mx+C$.

The slope of this new line is $M={\left(\frac{v_0}{x_0}\right)}^2$. Because it is a squared term, the slope is strictly positive.

The $y$-intercept of this line is $C=-\frac{v_0^2}{x_0}$, which is strictly negative.

Therefore, the $a-x$ graph must be a straight line with a positive slope starting from a negative intercept on the vertical axis. This perfectly matches option (a).