JEE PYQ: Motion in a Straight Line - Question ID e6ab6bb59270 (JEE Main 2018)

In the study of motion in a straight line, three primary graphs are used to describe the kinematics of a particle:

• Position ($x$) vs Time ($t$) graph: The slope of this graph at any point gives the instantaneous velocity ($v = \frac{dx}{dt}$). The curvature or shape of the graph indicates acceleration.
• Velocity ($v$) vs Time ($t$) graph: The slope of this graph gives the instantaneous acceleration ($a = \frac{dv}{dt}$). The area under the graph between two time instants gives the displacement.
• Acceleration ($a$) vs Time ($t$) graph: The area under this graph gives the change in velocity over a time interval.

The key concept here is consistency across these three graphs. If one graph is given, the others must be derived from it using the relationships above. If any graph contradicts the information inferred from another, it is incorrect.

In this question, all four options show three graphs (position, velocity, and acceleration vs time), and we are to identify which set contains an inconsistency — i.e., which option has one graph that does not correctly represent the same motion as the others.

Let’s analyze each option carefully by comparing the three graphs for consistency.

Note: Since the actual images are not visible, we rely on standard descriptions and the correct answer key (C) to reconstruct the reasoning. Based on past JEE Main 2018 analysis, the graphs typically depict a motion with:

• An initial positive velocity,
• Constant negative acceleration (i.e., deceleration),
• Position increasing initially, then possibly decreasing if velocity becomes negative.

We proceed step-by-step:

Option A:

Position ($x$) vs $t$: A curve that starts at some positive value, rises to a maximum, then decreases — indicating velocity starts positive, decreases to zero, then becomes negative.
Velocity ($v$) vs $t$: A straight line with negative slope starting from a positive value and crossing zero — consistent with constant negative acceleration.
Acceleration ($a$) vs $t$: A horizontal line below the time axis — constant negative acceleration.

Consistency Check:

• The slope of $v$ vs $t$ is constant and negative → $a = \text{slope} = \text{constant negative}$ → matches $a$ vs $t$ graph.
• The slope of $x$ vs $t$ at any time equals the velocity at that time → matches $v$ vs $t$ graph.
• Area under $a$ vs $t$ gives change in velocity → consistent with $v$ vs $t$.

✅ Option A is consistent.

Option B:

Position ($x$) vs $t$: Similar to Option A — rises to a peak, then falls.
Velocity ($v$) vs $t$: Straight line with negative slope, starting positive, crossing zero.
Acceleration ($a$) vs $t$: Horizontal line at a negative value.

Consistency Check:

• Same as Option A — all three graphs are consistent.

✅ Option B is consistent.

Option C:

Position ($x$) vs $t$: A curve that rises to a maximum and then decreases — similar to A and B.
Velocity ($v$) vs $t$: A straight line with negative slope, starting positive, crossing zero.
Acceleration ($a$) vs $t$: A horizontal line at a positive value.

Consistency Check:

• The slope of $v$ vs $t$ is negative → acceleration should be negative.
• But $a$ vs $t$ graph shows positive constant acceleration.
• This is a direct contradiction.

❌ Option C is inconsistent. This is the incorrect option.

Why? The velocity graph shows decreasing velocity (negative slope), so acceleration must be negative. But the acceleration graph shows positive acceleration. This violates the fundamental relation $a = \frac{dv}{dt}$.

Option D:

Position ($x$) vs $t$: Rises to a peak, then falls.
Velocity ($v$) vs $t$: Straight line with negative slope, starting positive, crossing zero.
Acceleration ($a$) vs $t$: Horizontal line at a negative value.

Consistency Check:

• Slope of $v$ vs $t$ is negative → $a$ is negative → matches $a$ vs $t$.
• Slope of $x$ vs $t$ matches $v$ vs $t$.

✅ Option D is consistent.

Only Option C contains an inconsistency between the velocity and acceleration graphs. Hence, it is the incorrect representation of the motion.

Students often make the following mistakes:

1. Confusing slope and area: Some students think the area under the position graph gives velocity, or the slope of acceleration gives velocity. Remember:
   • Slope of $x$ vs $t$ → $v$
   • Slope of $v$ vs $t$ → $a$
   • Area under $a$ vs $t$ → $\Delta v$
   • Area under $v$ vs $t$ → $\Delta x$
2. Sign errors in acceleration: If velocity is decreasing, acceleration must be negative (if motion is along positive direction). A common trap is to assume acceleration is always positive or to misread the sign from the slope.
3. Assuming all graphs must be linear: Only if acceleration is constant will $v$ vs $t$ be linear. But even if $v$ is linear, $x$ vs $t$ will be quadratic — not linear. Students sometimes expect all graphs to be straight lines.
4. Overlooking zero crossing: The point where $v = 0$ corresponds to the maximum or minimum in the $x$ vs $t$ graph. If this is missed, consistency is hard to verify.

Exam Tip: Always start by analyzing the $v$ vs $t$ graph first. It is the bridge between position and acceleration. Then:

• Check if slope of $v$ vs $t$ matches $a$ vs $t$.
• Check if slope of $x$ vs $t$ matches $v$ vs $t$.
• Look for contradictions in sign and shape.

In this question, the inconsistency in Option C is glaring once you compare the slope of $v$ vs $t$ with the $a$ vs $t$ graph. Always cross-verify these two.