JEE PYQ: Motion in a Straight Line - Question ID 468afadeb4a1 (JEE Main 2021)

To convert a velocity-displacement ($v$ vs $x$) graph into an acceleration-displacement ($a$ vs $x$) graph, we use the chain rule of calculus: $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \cdot \frac{dv}{dx}.$ Here, $v$ is the velocity at displacement $x$, and $\frac{dv}{dx}$ is the slope of the $v$–$x$ graph at that point.

Key observations from the given $v$–$x$ graph (a straight line with negative slope):

• The slope $\frac{dv}{dx}$ is constant and negative.
• The velocity $v$ decreases linearly from a positive value at $x=0$ to zero at some $x=x_0$.

Hence, $a = v \cdot \frac{dv}{dx}$ will be negative throughout and will vary linearly with $x$ (since $v$ itself is linear in $x$).

Step 1: Express $v$ as a function of $x$.

From the graph, $v$ decreases linearly from $v_0$ at $x=0$ to $0$ at $x=x_0$. Thus $v(x) = v_0 \Bigl(1 - \frac{x}{x_0}\Bigr).$

Step 2: Compute $\frac{dv}{dx}$.

Differentiating, $\frac{dv}{dx} = -\frac{v_0}{x_0}.$ This slope is constant and negative.

Step 3: Compute $a(x)$.

Using $a = v\,\frac{dv}{dx}$, $a(x) = v(x)\,\frac{dv}{dx} = v_0\Bigl(1 - \frac{x}{x_0}\Bigr)\Bigl(-\frac{v_0}{x_0}\Bigr) = -\frac{v_0^2}{x_0}\Bigl(1 - \frac{x}{x_0}\Bigr).$ This is a linear function of $x$ that starts at $a(0)=-\frac{v_0^2}{x_0}$ and increases to $a(x_0)=0$.

Step 4: Sketch $a$ vs $x$.

The acceleration-displacement graph is a straight line with negative intercept at $x=0$ and rising to zero at $x=x_0$. This matches option A.

1. Misapplying $a = \frac{dv}{dt}$ directly. Students often forget to use the chain rule $a = v\,\frac{dv}{dx}$ and instead try to read $\frac{dv}{dt}$ from the $v$–$x$ graph, which is incorrect. 2. Sign errors. The negative slope of $v$ vs $x$ leads to a negative acceleration, but one must track how $a$ varies with $x$. 3. Confusing intercepts. The acceleration does not start at zero; it starts at a negative value and increases linearly to zero.

Exam Tip: Always write down $a = v\,\frac{dv}{dx}$ explicitly when converting $v$–$x$ to $a$–$x$. Then substitute the algebraic form of $v(x)$ and differentiate.