JEE PYQ: Motion in a Straight Line - Question ID f28a6671e406 (JEE Main 2020)

In one-dimensional motion, the position \( x \) of a particle is given as a function of time \( t \). To find the acceleration, we follow these steps:

• Velocity (\( v \)): The first derivative of position with respect to time, \( v = \frac{dx}{dt} \).
• Acceleration (\( a \)): The first derivative of velocity with respect to time, or the second derivative of position with respect to time, \( a = \frac{dv}{dt} = \frac{d^2x}{dt^2} \).
• Given Relationship: The acceleration depends on \( x \) as \( a \propto x^{-n} \), where \( n \) is an integer. Our goal is to find \( n \).

The key concept here is to express acceleration in terms of \( x \) using the given position-time relationship \( x^2 = at^2 + 2bt + c \), and then compare it with the given form \( a \propto x^{-n} \).

Step 1: Differentiate the given position-time equation implicitly.
Given: $x^2 = at^2 + 2bt + c$ Differentiate both sides with respect to \( t \): $2x \frac{dx}{dt} = 2at + 2b$ Simplify: $x \frac{dx}{dt} = at + b \quad \text{(Equation 1)}$ Here, \( \frac{dx}{dt} = v \) (velocity).

Step 2: Differentiate again to find acceleration.
Differentiate Equation 1 with respect to \( t \): $\frac{d}{dt} \left( x \frac{dx}{dt} \right) = \frac{d}{dt} (at + b)$ Using the product rule on the left side: $\frac{dx}{dt} \cdot \frac{dx}{dt} + x \cdot \frac{d^2x}{dt^2} = a$ Substitute \( \frac{dx}{dt} = v \) and \( \frac{d^2x}{dt^2} = a \) (acceleration): $v^2 + x a = a$ Rearrange to solve for acceleration \( a \): $a = \frac{a - v^2}{x} \quad \text{(Equation 2)}$ This expression is not yet in terms of \( x \) alone. We need to eliminate \( v \).

Step 3: Express \( v \) in terms of \( x \).
From Equation 1: $v = \frac{at + b}{x}$ Square both sides: $v^2 = \frac{(at + b)^2}{x^2}$ Substitute \( v^2 \) into Equation 2: $a = \frac{a - \frac{(at + b)^2}{x^2}}{x} = \frac{a x^2 - (at + b)^2}{x^3}$ Simplify the numerator using the given \( x^2 = at^2 + 2bt + c \): $a x^2 - (at + b)^2 = a(at^2 + 2bt + c) - (a^2 t^2 + 2abt + b^2)$ Expand: $= a^2 t^2 + 2abt + a c - a^2 t^2 - 2abt - b^2$ Cancel terms: $= a c - b^2$ Thus, acceleration becomes: $a = \frac{a c - b^2}{x^3}$ This shows that acceleration is inversely proportional to \( x^3 \): $a \propto x^{-3}$

Step 4: Compare with the given form \( a \propto x^{-n} \).
From the derived relationship \( a \propto x^{-3} \), we see that \( n = 3 \).

1. Implicit Differentiation Mistake: Students often forget to apply the product rule when differentiating \( x \frac{dx}{dt} \). This leads to incorrect expressions for acceleration.
2. Substitution Error: Failing to substitute \( x^2 = at^2 + 2bt + c \) correctly when simplifying the numerator can result in an incorrect expression for acceleration.
3. Algebraic Simplification: Not expanding and canceling terms properly in the numerator can make the problem seem more complicated than it is.
4. Exam Tip: Always express acceleration in terms of \( x \) alone before comparing with the given form. This ensures clarity and avoids confusion with intermediate variables like \( v \) or \( t \).

Final Answer: The value of \( n \) is \( \boxed{3} \).