JEE PYQ: Motion in a Straight Line - Question ID 0ff985c0843f (JEE Main 2026)

Core Formula & Concept:

To find the acceleration \( a \) as a function of distance \( x \) from a given velocity-distance (\( v \)-\( x \)) 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, \( \frac{dv}{dx} \) is the slope of the \( v \)-\( x \) graph at any point, and \( v \) is the velocity at that point. Thus, acceleration is the product of velocity and the slope of the \( v \)-\( x \) curve.

Step-by-Step Derivation:

Step 1: Analyze the given \( v \)-\( x \) graph

The provided \( v \)-\( x \) graph consists of three distinct linear segments:

• From \( x = 0 \) to \( x = x_1 \): A straight line with positive slope \( m_1 \).
• From \( x = x_1 \) to \( x = x_2 \): A straight line with zero slope (horizontal line).
• From \( x = x_2 \) to \( x = x_3 \): A straight line with negative slope \( m_2 \).

Step 2: Compute acceleration in each segment

Using \( a = v \cdot \frac{dv}{dx} \):

• Segment 1 (\( 0 \leq x < x_1 \)):
  \( \frac{dv}{dx} = m_1 \) (constant positive slope).
  Since \( v \) increases linearly with \( x \), \( v = m_1 x \).
  Thus, $a = v \cdot \frac{dv}{dx} = (m_1 x) \cdot m_1 = m_1^2 x.$ This means acceleration increases linearly with \( x \) in this segment.
• Segment 2 (\( x_1 \leq x < x_2 \)):
  \( \frac{dv}{dx} = 0 \) (horizontal line).
  Thus, $a = v \cdot 0 = 0.$ Acceleration is zero throughout this segment.
• Segment 3 (\( x_2 \leq x \leq x_3 \)):
  \( \frac{dv}{dx} = m_2 \) (constant negative slope).
  Since \( v \) decreases linearly with \( x \), \( v = v_2 + m_2 (x - x_2) \), where \( v_2 \) is the velocity at \( x = x_2 \).
  Thus, $a = v \cdot \frac{dv}{dx} = [v_2 + m_2 (x - x_2)] \cdot m_2.$ Since \( m_2 < 0 \), \( a \) is negative and decreases linearly with \( x \).

Step 3: Sketch the \( a \)-\( x \) graph based on the above analysis

• From \( x = 0 \) to \( x = x_1 \): \( a \) increases linearly from 0 to \( a_1 = m_1^2 x_1 \).
• From \( x = x_1 \) to \( x = x_2 \): \( a = 0 \).
• From \( x = x_2 \) to \( x = x_3 \): \( a \) decreases linearly from 0 to a negative value.

This behavior matches the shape of option D, where:

• The acceleration starts at zero, increases linearly,
• Drops abruptly to zero,
• Then decreases linearly into negative values.

Common Traps & Exam Tip:

Students often confuse the \( v \)-\( x \) graph with the \( v \)-\( t \) graph and mistakenly assume that the slope of the \( v \)-\( x \) graph directly gives acceleration. This is incorrect because acceleration depends on both the slope \( \frac{dv}{dx} \) and the velocity \( v \) itself.

Another common error is overlooking the abrupt change in slope at \( x = x_1 \) and \( x = x_2 \), leading to incorrect conclusions about the continuity of acceleration. Remember that acceleration can change discontinuously if the slope of the \( v \)-\( x \) graph changes abruptly.

Exam Tip: Always use the relation \( a = v \cdot \frac{dv}{dx} \) when dealing with \( v \)-\( x \) graphs to avoid confusion.