JEE PYQ: Motion in a Straight Line - Question ID 5257614cd9df (JEE Main 2023)

Core Formula & Concept:

In one-dimensional motion, the velocity \( v \) at any instant is the slope of the position-time (\( x \)-\( t \)) graph at that point. Mathematically, \[ v = \frac{dx}{dt}. \] Key observations:

• If the \( x \)-\( t \) graph is a straight line, its slope is constant, so the \( v \)-\( t \) graph is a horizontal line.
• If the \( x \)-\( t \) graph is a curve whose slope increases (becomes steeper), the velocity increases, giving a rising \( v \)-\( t \) graph.
• If the \( x \)-\( t \) graph is a curve whose slope decreases (becomes less steep), the velocity decreases, giving a falling \( v \)-\( t \) graph.
• A sudden change in the slope of the \( x \)-\( t \) graph corresponds to a jump (discontinuity) in the \( v \)-\( t \) graph.

Step-by-Step Derivation:

Entry A:

The \( x \)-\( t \) graph is a straight line with positive slope, then a straight line with zero slope, then a straight line with negative slope.

• Positive slope \(\Rightarrow\) positive constant velocity \(\Rightarrow\) horizontal line above the time axis.
• Zero slope \(\Rightarrow\) zero velocity \(\Rightarrow\) horizontal line on the time axis.
• Negative slope \(\Rightarrow\) negative constant velocity \(\Rightarrow\) horizontal line below the time axis.

This matches the \( v \)-\( t \) graph in II.

Entry B:

The \( x \)-\( t \) graph is a curve whose slope starts positive and decreases continuously to zero.

• Initially steep positive slope \(\Rightarrow\) high positive velocity.
• Slope decreases smoothly \(\Rightarrow\) velocity decreases smoothly to zero.

This matches the \( v \)-\( t \) graph in IV.

Entry C:

The \( x \)-\( t \) graph is a straight line with positive slope, then a curve whose slope decreases to zero.

• Straight line \(\Rightarrow\) constant positive velocity.
• Curve with decreasing slope \(\Rightarrow\) velocity decreases from that constant value to zero.

This matches the \( v \)-\( t \) graph in III.

Entry D:

The \( x \)-\( t \) graph is a straight line with positive slope, then a straight line with negative slope.

• Positive slope \(\Rightarrow\) positive constant velocity.
• Negative slope \(\Rightarrow\) negative constant velocity.
• The abrupt change in slope produces a jump in the \( v \)-\( t \) graph.

This matches the \( v \)-\( t \) graph in I.

Putting it all together:
A \(\to\) II, B \(\to\) IV, C \(\to\) III, D \(\to\) I.
This corresponds to option B.

Common Traps & Exam Tip:

1. Confusing slope with height: Students sometimes think the value of \( x \) itself gives velocity, but velocity is always the slope of the \( x \)-\( t \) graph. 2. Missing discontinuities: A sharp corner in the \( x \)-\( t \) graph means an instantaneous jump in velocity, which must appear as a vertical step in the \( v \)-\( t \) graph. 3. Sign errors: A negative slope in \( x \)-\( t \) gives a negative velocity, so the \( v \)-\( t \) graph must lie below the time axis.

Exam tip: Always draw a quick sketch of the slope at key points on the \( x \)-\( t \) graph before matching to the \( v \)-\( t \) options.