JEE PYQ: Motion in a Straight Line - Question ID 97eae75bef14 (JEE Main 2021)

To convert a velocity-displacement ($v$ vs $x$) graph into an acceleration-displacement ($a$ vs $x$) graph, we rely on the chain rule of calculus:

• Acceleration is the time derivative of velocity: $a = \frac{dv}{dt}.$
• By the chain rule, $a = \frac{dv}{dt} = \frac{dv}{dx}\,\frac{dx}{dt} = v\,\frac{dv}{dx}.$
• Geometrically, $\displaystyle \frac{dv}{dx}$ is the slope of the $v$–$x$ curve at each point.
• Therefore, at any displacement $x$, $a(x) = v(x)\times \bigl[\text{slope of }v\text{ vs }x\text{ at }x\bigr].$

Step 1: Analyze the given $v$–$x$ graph

The figure shows a straight line segment from $(x=0,\,v=v_{0})$ to $(x=x_{1},\,v=0)$. Its equation is $v(x)=v_{0}\Bigl(1-\frac{x}{x_{1}}\Bigr),\quad 0\le x\le x_{1}.$

Step 2: Compute the slope $\displaystyle \frac{dv}{dx}$

Differentiating, $\frac{dv}{dx}=-\frac{v_{0}}{x_{1}}.$ This slope is constant and negative.

Step 3: Express acceleration $a(x)$

Using $a=v\,\frac{dv}{dx}$, $a(x)=v(x)\times\Bigl(-\frac{v_{0}}{x_{1}}\Bigr) =v_{0}\Bigl(1-\frac{x}{x_{1}}\Bigr)\times\Bigl(-\frac{v_{0}}{x_{1}}\Bigr) =-\frac{v_{0}^{2}}{x_{1}}\Bigl(1-\frac{x}{x_{1}}\Bigr).$

Step 4: Interpret the resulting $a$–$x$ graph

• At $x=0$, $a=-\,\dfrac{v_{0}^{2}}{x_{1}}$ (a negative constant).
• At $x=x_{1}$, $a=0$.
• The relation $a(x)$ is linear in $x$, starting at a negative value and rising to zero.

This matches the straight line in option A, which begins below the $x$–axis and ends at the origin.

1. Misapplying $a=dv/dt$ directly. Students often forget to use the chain rule and try to read $dv/dt$ from the $v$–$x$ graph without converting to $dv/dx$. 2. Sign errors. The slope $dv/dx$ is negative, so $a=v\,(dv/dx)$ is negative for positive $v$, but its magnitude decreases linearly with $x$. 3. Confusing $a$–$x$ with $v$–$x$. The acceleration graph is not a copy of the velocity graph; it is a new linear function derived from the product $v\,(dv/dx)$.

Tip: Always write $a=v\,\frac{dv}{dx}$ explicitly and compute the slope of the $v$–$x$ curve at each point.