JEE PYQ: Motion in a Straight Line - Question ID e99b1a54d2ac (JEE Main 2019)

When a particle starts from rest ($v_0 = 0$) at the origin ($x_0 = 0$) and moves with uniform (constant) acceleration $a$ along the positive $x$–axis, its motion is governed by the following kinematic relations:

• Velocity–time relation: $v(t) = v_0 + a\,t = a\,t$ → $v$ is a linear function of $t$, starting at zero and increasing without bound.
• Displacement–time relation: $x(t) = x_0 + v_0\,t + \tfrac12\,a\,t^2 = \tfrac12\,a\,t^2$ → $x$ is a quadratic function of $t$, starting at zero and opening upwards.
• Velocity–displacement relation (eliminating $t$): $v^2 = v_0^2 + 2\,a\,(x - x_0) = 2\,a\,x$ → $v$ is proportional to $\sqrt{x}$, hence the $v$–$x$ graph is a parabola opening to the right.

Any graph that violates these functional forms is qualitatively incorrect.

Step 1: Velocity vs Time ($v$–$t$)

$v(t) = a\,t$ is a straight line through the origin with positive slope $a$.
Graph (A) shows exactly this: a straight line starting at $(0,0)$ and rising linearly.
Graph (B) also shows a straight line through the origin, hence it too is correct.
Graphs (C) and (D) do not show linear $v$–$t$; they are therefore incorrect for this relation.

Step 2: Displacement vs Time ($x$–$t$)

$x(t) = \tfrac12\,a\,t^2$ is a parabola opening upwards, starting at $(0,0)$.
Graph (B) is a parabola through the origin, hence it correctly represents $x$–$t$.
Graph (A) is linear, so it cannot represent $x$–$t$.

Step 3: Velocity vs Displacement ($v$–$x$)

$v^2 = 2\,a\,x$ ⇒ $v = \sqrt{2\,a\,x}$.
This is a parabola opening to the right, starting at $(0,0)$.
Graph (D) shows exactly this shape, hence it is correct.
Graph (C) is a straight line, so it cannot represent $v$–$x$.

Step 4: Tabulate correctness

Therefore the qualitatively correct graphs are (A), (B), and (D).

1. Confusing axes: Students often misread which variable is plotted on which axis. Always label axes explicitly. 2. Nonlinear vs linear: The $v$–$t$ graph must be linear; the $x$–$t$ graph must be parabolic. A straight line for $x$–$t$ is wrong. 3. Starting point: The particle starts from rest at the origin, so every graph must pass through $(0,0)$. Any graph that does not is automatically wrong. 4. Uniform acceleration: The question specifies uniform acceleration. If acceleration varied, the graphs would change.

Final Answer: Option D (A, B, D).